4 Discrete Random Variables 离散随机变量

本页译自 OpenStax《Introductory Statistics》第 4 章 Discrete Random Variables。公式经本地 MathJax 渲染，自定义宏已注入。

Introduction 引言

A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, he or she could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%?

一名学生参加一份十道题的判断题测验。由于日程繁忙，他/她无暇复习，只好随机猜答每道题。该学生以至少 70% 的得分通过测验的概率是多少？

Small companies might be interested in the number of long-distance phone calls their employees make during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees make more than 20 long-distance phone calls during the peak time?

小公司可能关心员工在一天的高峰时段拨打的长途电话数量。假设平均数为 20 通。员工在高峰时段拨打超过 20 通长途电话的概率是多少？

These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. A random variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment.

这两个例子说明了涉及离散随机变量的两类不同的概率问题。回顾一下，离散数据是你能够计数的数据。随机变量用文字描述统计实验的结果。随机变量的值在每次重复实验时都可能不同。

Random Variable Notation 随机变量记号

Upper case letters such as *X* or *Y* denote a random variable. Lower case letters like *x* or *y* denote the value of a random variable. If ***X* is a random variable, then *X* is written in words, and *x* is given as a number.**

大写字母如 *X* 或 *Y* 表示随机变量。小写字母如 *x* 或 *y* 表示随机变量的值。如果 ***X* 是随机变量，那么 *X* 用文字写出，而 *x* 以数字给出。**

For example, let *X* = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is *TTT*; *THH*; *HTH*; *HHT*; *HTT*; *THT*; *TTH*; *HHH*. Then, *x* = 0, 1, 2, 3. *X* is in words and *x* is a number. Notice that for this example, the *x* values are countable outcomes. Because you can count the possible values that *X* can take on and the outcomes are random (the *x* values 0, 1, 2, 3), *X* is a discrete random variable.

例如，令 *X* = 你抛掷三枚均匀硬币得到的正面数。抛掷三枚均匀硬币的样本空间为 *TTT*; *THH*; *HTH*; *HHT*; *HTT*; *THT*; *TTH*; *HHH*。于是 *x* = 0, 1, 2, 3。*X* 用文字表示，*x* 是数字。注意在此例中，*x* 的值是可计数的结果。因为你能数出 *X* 可能取的值，且结果是随机的（*x* 的取值 0, 1, 2, 3），所以 *X* 是一个离散随机变量。

Toss a coin ten times and record the number of heads. After all members of the class have completed the experiment (tossed a coin ten times and counted the number of heads), fill in Table 4.1. Let *X* = the number of heads in ten tosses of the coin.

抛掷硬币十次，记录正面的个数。在全班同学都完成实验（抛掷硬币十次并数出正面个数）后，填写表 4.1。令 *X* = 十次抛掷中正面的个数。

| *x* | **Frequency of *x* | Relative Frequency of *x*** |

| *x* | **频数 *x* | 相对频率 *x*** |

|---------|----------------------|-------------------------------|

| | | |

Table 4.1

表 4.1

1. Which value(s) of *x* occurred most frequently?

1. *x* 的哪些取值出现得最频繁？

2. If you tossed the coin 1,000 times, what values could *x* take on? Which value(s) of *x* do you think would occur most frequently?

2. 如果你抛掷硬币 1,000 次，*x* 可能取哪些值？你认为 *x* 的哪些取值会出现得最频繁？

3. What does the relative frequency column sum to?

3. 相对频率这一列的总和是多少？

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量的概率分布函数（PDF）

A discrete probability distribution function has two characteristics:

离散型概率分布函数有两个特征：

1. Each probability is between zero and one, inclusive.

1. 每个概率都在零与一之间（含端点）。

2. The sum of the probabilities is one.

2. 所有概率之和为 1。

A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let *X* = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, *x* = 0, 1, 2, 3, 4, 5.

一位儿童心理学家关注新生儿啼哭在午夜后唤醒母亲的次数。通过对 50 位母亲的一个随机样本，得到以下信息。令 *X* = 每周新生儿啼哭在午夜后唤醒母亲的次数。在此例中，*x* = 0, 1, 2, 3, 4, 5。

*P*(*x*) = probability that *X* takes on a value *x*.

*P*(*x*) = *X* 取值为 *x* 的概率。

| *x* | *P*(*x*) |

|-----|--------------------------------|

| 0 | *P*(*x* = 0) = $\frac{2}{50}$ |

| 1 | *P*(*x* = 1) = $\frac{11}{50}$ |

| 2 | *P*(*x* = 2) = $\frac{23}{50}$ |

| 3 | *P*(*x* = 3) = $\frac{9}{50}$ |

| 4 | *P*(*x* = 4) = $\frac{4}{50}$ |

| 5 | *P*(*x* = 5) = $\frac{1}{50}$ |

Table 4.2

表 4.2

*X* takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because:

*X* 取值 0, 1, 2, 3, 4, 5。这是一个离散 PDF，因为：

1. Each *P*(*x*) is between zero and one, inclusive.

1. 每个 *P*(*x*) 都在零与一之间（含端点）。

2. The sum of the probabilities is one, that is,

2. 所有概率之和为 1，即，

$$\frac{2}{50} + \frac{11}{50} + \frac{23}{50} + \frac{9}{50} + \frac{4}{50} + \frac{1}{50} = 1$$

A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. Let *X* = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, *x* = 0, 1, 2, 3, 4, 5. *P*(*x*) = the probability that *X* takes on value *x*. Why is this a discrete probability distribution function (two reasons)?

一位医院研究人员关注平均每位术后患者在 12 小时轮班期间按铃呼叫护士的次数。通过对 50 名患者的一个随机样本，得到以下信息。令 *X* = 一名患者在 12 小时轮班期间按铃呼叫护士的次数。在本练习中，*x* = 0, 1, 2, 3, 4, 5。*P*(*x*) = *X* 取值为 *x* 的概率。为什么这是一个离散概率分布函数（两个理由）？

| *x* | *P*(*x*) |

|-----|--------------------------------|

| 0 | *P*(*x* = 0) = $\frac{4}{50}$ |

| 1 | *P*(*x* = 1) = $\frac{8}{50}$ |

| 2 | *P*(*x* = 2) = $\frac{16}{50}$ |

| 3 | *P*(*x* = 3) = $\frac{14}{50}$ |

| 4 | *P*(*x* = 4) = $\frac{6}{50}$ |

| 5 | *P*(*x* = 5) = $\frac{2}{50}$ |

Table 4.3

表 4.3

Suppose Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected.

假设 Nancy 每周上课三天。她一周上课三天的情况占 80%，上课两天占 15%，上课一天占 4%，不上课占 1%。假设随机选取某一周。

Problem 问题

a\. Let *X* = the number of days Nancy \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.

a\. 令 *X* = Nancy _______________ 的天数。

b\. *X* takes on what values?

b\. *X* 取哪些值？

c\. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table) like the one in Example 4.1. The table should have two columns labeled *x* and *P*(*x*). What does the *P*(*x*) column sum to?

c\. 假设随机选取某一周。构建一个概率分布表（称为 PDF 表），类似于示例 4.1 中的表。该表应有两列，分别标为 *x* 和 *P*(*x*)。*P*(*x*) 这一列的总和是多少？

Solution 解答

a\. Let *X* = the number of days Nancy attends class per week.

a\. 令 *X* = Nancy 每周上课的天数。

b\. 0, 1, 2, and 3

b\. 0, 1, 2, 和 3

c\.

| *x* | *P*(*x*) |

|-----|----------|

| 0 | 0.01 |

| 1 | 0.04 |

| 2 | 0.15 |

| 3 | 0.80 |

Table 4.4

表 4.4

Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is *X* and what values does it take on?

Jeremiah 每周有两次篮球训练。百分之九十的时间他两次都参加；百分之八的时间他参加一次；百分之二的时间他一次也不参加。*X* 是什么？它取哪些值？

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.

期望值常被称为"长期"平均值或均值。这意味着，在反复进行一个试验的长期过程中，你会期望得到这个平均值。

You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. As you learned in Chapter 3 Probability Topics, probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers.

你抛一枚硬币并记录结果。结果是正面的概率是多少？如果你抛硬币两次，概率是否会告诉你这两次抛掷会得到一次正面和一次反面？你可能抛一枚均匀的硬币十次并记录到九次正面。正如你在第 3 章"概率主题"中学到的，概率并不描述试验的短期结果。它提供的是关于长期内可期望发生什么的信息。为了证明这一点，卡尔·皮尔逊曾抛一枚均匀的硬币 24,000 次！他记录了每次抛掷的结果，得到正面 12,012 次。在他的实验中，皮尔逊阐释了大数定律。

The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter *μ*. In other words, after conducting many trials of an experiment, you would expect this average value.

大数定律指出，随着概率试验中试验次数的增加，事件的理论概率与相对频率之间的差异趋近于零（理论概率与相对频率越来越接近）。在评估统计试验的长期结果时，我们常常想知道"平均"结果。这个"长期平均"被称为试验的均值或期望值，用希腊字母 *μ* 表示。换言之，在进行一个试验的多次重复之后，你会期望得到这个平均值。

The mean, *μ*, of a discrete probability function is the expected value.

离散型概率函数的均值 *μ* 就是期望值。

$$\mu = {\sum\left( x \bullet P(x) \right)}$$

The standard deviation, Σ, of the PDF is the square root of the variance.

概率分布函数的标准差 Σ 是方差的平方根。

$$\sigma = \sqrt{\sum{\left\lbrack \left( {x~–~\mu} \right)^{2}~ \bullet ~Ρ(x) \right\rbrack~}}$$$

When all outcomes in the probability distribution are equally likely, these formulas coincide with the mean and standard deviation of the set of possible outcomes.

当概率分布中的所有结果等可能时，这些公式与可能结果集合的均值和标准差一致。

To find the expected value or long term average, *μ*, simply multiply each value of the random variable by its probability and add the products.

要求期望值或长期平均 *μ*，只需将随机变量的每个取值乘以其概率，再将这些乘积相加。

A men's soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, *μ*, of the number of days per week the men's soccer team plays soccer.

一支男子足球队每周踢球零天、一天或两天。他们踢零天的概率是 0.2，踢一天的概率是 0.5，踢两天的概率是 0.3。求这支男子足球队每周踢球天数的长期平均或期望值 *μ*。

To do the problem, first let the random variable *X* = the number of days the men's soccer team plays soccer per week. *X* takes on the values 0, 1, 2. Construct a PDF table adding a column *x*\**P*(*x*). In this column, you will multiply each *x* value by its probability.

要解决这个问题，首先令随机变量 *X* = 男子足球队每周踢球的天数。*X* 取值 0、1、2。构造一个概率分布函数表，并增加一列 *x*\**P*(*x*)。在这一列中，你将把每个 *x* 取值乘以其概率。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|----------------|

| 0 | 0.2 | (0)(0.2) = 0 |

| 1 | 0.5 | (1)(0.5) = 0.5 |

| 2 | 0.3 | (2)(0.3) = 0.6 |

Table 4.5 Expected Value Table This table is called an expected value table. The table helps you calculate the expected value or long-term average.

表 4.5 期望值表这个表称为期望值表。该表帮助你计算期望值或长期平均。

Add the last column *x*\**P*(*x*) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1.

把最后一列 *x*\**P*(*x*) 相加，得到长期平均或期望值：(0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1。

The expected value is 1.1. The men's soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week. We say *μ* = 1.1.

期望值为 1.1。平均而言，这支男子足球队预期每周踢球 1.1 天。如果从周复一周的长期来看，1.1 就是该队踢球的长期平均或期望值。我们说 *μ* = 1.1。

Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times per week a newborn baby's crying wakes its mother after midnight. Calculate the standard deviation of the variable as well.

求新生儿在午夜后哭闹唤醒母亲的次数的期望值。期望值就是每周新生儿哭闹唤醒母亲的平均次数。同时计算该变量的标准差。

| *x* | *P*(*x*) | *x*\**P*(*x*) | (*x* – *μ*)² ⋅ *P*(*x*) |

|-----|-----------------------------------------------|-----------------------------------------------------|---------------------------------------|

| 0 | *P*(*x* = 0) = $\frac{2}{50}$ | (0)$\left( \frac{2}{50} \right)$ = 0 | (0 – 2.1)² ⋅ 0.04 = 0.1764 |

| 1 | *P*(*x* = 1) = $\left( \frac{11}{50} \right)$ | (1)$\left( \frac{11}{50} \right)$ = $\frac{11}{50}$ | (1 – 2.1)² ⋅ 0.22 = 0.2662 |

| 2 | *P*(*x* = 2) = $\frac{23}{50}$ | (2)$\left( \frac{23}{50} \right)$ = $\frac{46}{50}$ | (2 – 2.1)² ⋅ 0.46 = 0.0046 |

| 3 | *P*(*x* = 3) = $\frac{9}{50}$ | (3)$\left( \frac{9}{50} \right)$ = $\frac{27}{50}$ | (3 – 2.1)² ⋅ 0.18 = 0.1458 |

| 4 | *P*(*x* = 4) = $\frac{4}{50}$ | (4)$\left( \frac{4}{50} \right)$ = $\frac{16}{50}$ | (4 – 2.1)² ⋅ 0.08 = 0.2888 |

| 5 | *P*(*x* = 5) = $\frac{1}{50}$ | (5)$\left( \frac{1}{50} \right)$ = $\frac{5}{50}$ | (5 – 2.1)² ⋅ 0.02 = 0.1682 |

Table 4.6 You expect a newborn to wake its mother after midnight 2.1 times per week, on the average.

表 4.6 平均而言，你预期新生儿每周在午夜后唤醒母亲 2.1 次。

Add the values in the third column of the table to find the expected value of *X*:

把表中第三列的各值相加，求 *X* 的期望值：

*μ* = Expected Value = $\frac{105}{50}$ = 2.1

*μ* = 期望值 = $\frac{105}{50}$ = 2.1

Use *μ* to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value *x*, multiply the square of its deviation by its probability. (Each deviation has the format *x* – *μ*).

用 *μ* 补全该表。该表的第四列将给出计算标准差所需的值。对于每个 *x* 取值，将其偏差的平方乘以其概率。（每个偏差都具有 *x* – *μ* 的形式。）

Add the values in the fourth column of the table:

把表中第四列的各值相加：

0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05

The standard deviation of *X* is the square root of this sum: *σ* = $\sqrt{1.05}$ ≈ 1.0247

*X* 的标准差是这个和的平方根：*σ* = $\sqrt{1.05}$ ≈ 1.0247

一位医院研究者关注平均每位术后病人在 12 小时轮班期间按呼叫铃招呼护士的次数。对 50 名病人进行随机抽样，得到以下信息。期望值是多少？

| *x* | *P*(*x*) |

|-----|--------------------------------|

| 0 | *P*(*x* = 0) = $\frac{4}{50}$ |

| 1 | *P*(*x* = 1) = $\frac{8}{50}$ |

| 2 | *P*(*x* = 2) = $\frac{16}{50}$ |

| 3 | *P*(*x* = 3) = $\frac{14}{50}$ |

| 4 | *P*(*x* = 4) = $\frac{6}{50}$ |

| 5 | *P*(*x* = 5) = $\frac{2}{50}$ |

Table 4.7

表 4.7

Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay \$2 to play and could profit \$100,000 if you match all five numbers in order (you get your \$2 back plus \$100,000). Over the long term, what is your expected profit of playing the game?

假设你在玩一个机会游戏，从 0、1、2、3、4、5、6、7、8、9 中选出五个数字。一台计算机从 0 到 9 中有放回地随机选出五个数字。你付 \$2 来玩，如果你按顺序全部猜中这五个数字，可获利 \$100,000（你拿回 \$2 另加 \$100,000）。从长远来看，你玩这个游戏的期望利润是多少？

To do this problem, set up an expected value table for the amount of money you can profit.

要解决这个问题，为你能赚取的金额建立一个期望值表。

Let *X* = the amount of money you profit. The values of *x* are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of *x* are 100,000 dollars and −2 dollars.

令 *X* = 你赚取的金额。*x* 的取值不是 0、1、2、3、4、5、6、7、8、9。因为你关心的是利润（或亏损），*x* 的取值是 100,000 美元和 −2 美元。

To win, you must get all five numbers correct, in order. The probability of choosing one correct number is $\frac{1}{10}$ because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is

要获胜，你必须按顺序全部猜中五个数字。选对一个数字的概率是 $\frac{1}{10}$，因为有十个数字。你可以多次选同一个数字。按顺序且全部猜中五个数字的概率是

$$\left( \frac{1}{10} \right)\left( \frac{1}{10} \right)\left( \frac{1}{10} \right)\left( \frac{1}{10} \right)\left( \frac{1}{10} \right) = (1)(10^{- 5}) = 0.00001.$$

Therefore, the probability of winning is 0.00001 and the probability of losing is

因此，获胜的概率是 0.00001，失败的概率是

$$1 - 0.00001 = 0.99999.$$

The expected value table is as follows:

期望值表如下：

| | *x* | *P*(*x*) | *x*\**P*(*x*) |

|--------|---------|----------|--------------------------|

| Loss | –2 | 0.99999 | (–2)(0.99999) = –1.99998 |

| 输 | –2 | 0.99999 | (–2)(0.99999) = –1.99998 |

| Profit | 100,000 | 0.00001 | (100000)(0.00001) = 1 |

| 赢 | 100,000 | 0.00001 | (100000)(0.00001) = 1 |

Table 4.8 Αdd the last column. –1.99998 + 1 = –0.99998

表 4.8 把最后一列相加。–1.99998 + 1 = –0.99998

Since –0.99998 is about –1, you would, on average, expect to lose approximately \$1 for each game you play. However, each time you play, you either lose \$2 or profit \$100,000. The \$1 is the average or expected LOSS per game after playing this game over and over.

由于 –0.99998 约等于 –1，平均而言，你每玩一局预期大约亏损 \$1。然而，每次你玩的时候，要么亏损 \$2，要么获利 \$100,000。这个 \$1 是反复玩这个游戏之后每局的平均或期望亏损。

You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay \$1 to play. If you guess the right suit every time, you get your money back and \$256. What is your expected profit of playing the game over the long term?

你在玩一个机会游戏，从一副 52 张的标准扑克牌中抽出四张牌。在每张牌被抽出之前，你猜测它的花色。每次抽牌后牌都放回牌堆。你付 \$1 来玩。如果你每次都猜对花色，你拿回本金并得 \$256。从长远来看，你玩这个游戏的期望利润是多少？

Suppose you play a game with a biased coin. You play each game by tossing the coin once. *P*(heads) = $\frac{2}{3}$ and *P*(tails) = $\frac{1}{3}$. If you toss a head, you pay \$6. If you toss a tail, you win \$10. If you play this game many times, will you come out ahead?

假设你用一个有偏的硬币玩游戏。每局游戏抛掷硬币一次。*P*(正面) = $\frac{2}{3}$，*P*(反面) = $\frac{1}{3}$。如果你抛到正面，你付 \$6。如果你抛到反面，你赢 \$10。如果你多次玩这个游戏，你会盈利吗？

Problem 问题

a\. Define a random variable *X*.

a. 定义一个随机变量 *X*。

b\. Complete the following expected value table.

b. 补全下列期望值表。

| | *x* | | |

|------|-----|---------------|-----------------|

| WIN | 10 | $\frac{1}{3}$ | |

| 赢 | 10 | $\frac{1}{3}$ | |

| LOSE | | | $\frac{–12}{3}$ |

| 输 | | | $\frac{–12}{3}$ |

Table 4.9

表 4.9

c\. What is the expected value, *μ*? Do you come out ahead?

c. 期望值 *μ* 是多少？你会盈利吗？

Solution 解答

a\. *X* = amount of profit

a. *X* = 利润金额

b\.

| | *x* | *P*(*x*) | *xP*(*x*) |

|------|-----|---------------|-----------------|

| WIN | 10 | $\frac{1}{3}$ | $\frac{10}{3}$ |

| 赢 | 10 | $\frac{1}{3}$ | $\frac{10}{3}$ |

| LOSE | –6 | $\frac{2}{3}$ | $\frac{–12}{3}$ |

| 输 | –6 | $\frac{2}{3}$ | $\frac{–12}{3}$ |

Table 4.10

表 4.10

c\. Add the last column of the table. The expected value *μ* = $\frac{\text{–}2}{3}$. You lose, on average, about 67 cents each time you play the game so you do not come out ahead.

c. 把表的最后一列相加。期望值 *μ* = $\frac{\text{–}2}{3}$。平均而言，你每玩一局大约亏损 67 美分，因此你不会盈利。

Suppose you play a game with a spinner. You play each game by spinning the spinner once. *P*(red) = $\frac{2}{5}$, *P*(blue) = $\frac{2}{5}$, and *P*(green) = $\frac{1}{5}$. If you land on red, you pay \$10. If you land on blue, you don't pay or win anything. If you land on green, you win \$10. Complete the following expected value table.

假设你玩一个转盘游戏。每局游戏转动转盘一次。*P*(红) = $\frac{2}{5}$，*P*(蓝) = $\frac{2}{5}$，*P*(绿) = $\frac{1}{5}$。如果停在红色，你付 \$10。如果停在蓝色，你既不付也不赢。如果停在绿色，你赢 \$10。补全下列期望值表。

| | *x* | *P*(*x*) | |

|-------|-----|---------------|------------------------|

| Red | | | $\text{–}\frac{20}{5}$ |

| 红 | | | $\text{–}\frac{20}{5}$ |

| Blue | | $\frac{2}{5}$ | |

| 蓝 | | $\frac{2}{5}$ | |

| Green | 10 | | |

| 绿 | 10 | | |

Table 4.11

表 4.11

Like data, probability distributions have standard deviations. To calculate the standard deviation (*σ*) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled (*x* – *μ*)²*P*(*x*) and take the square root.

与数据一样，概率分布也有标准差。要计算概率分布的标准差 (*σ*)，先求每个取值与期望值的偏差，将其平方，乘以其概率，把这些乘积相加，再开平方。要了解如何进行这一计算，请看男子足球队每周踢球天数那张表。要求标准差，把标有 (*x* – *μ*)²*P*(*x*) 的那一列各项相加，再开平方。

| *x* | *P*(*x*) | *x*\**P*(*x*) | (*x* – *μ*)²*P*(*x*) |

|-----|----------|----------------|------------------------------------|

| 0 | 0.2 | (0)(0.2) = 0 | (0 – 1.1)²(0.2) = 0.242 |

| 1 | 0.5 | (1)(0.5) = 0.5 | (1 – 1.1)²(0.5) = 0.005 |

| 2 | 0.3 | (2)(0.3) = 0.6 | (2 – 1.1)²(0.3) = 0.243 |

Table 4.12

表 4.12

Add the last column in the table. 0.242 + 0.005 + 0.243 = 0.490. The standard deviation is the square root of 0.49, or *σ* = $\sqrt{0.49}$ = 0.7

把表中最后一列相加。0.242 + 0.005 + 0.243 = 0.490。标准差是 0.49 的平方根，即 *σ* = $\sqrt{0.49}$ = 0.7

Generally for probability distributions, we use a calculator or a computer to calculate *μ* and *σ* to reduce roundoff error. For some probability distributions, there are short-cut formulas for calculating *μ* and *σ*.

一般来说，对于概率分布，我们使用计算器或计算机来计算 *μ* 和 *σ*，以减少舍入误差。对某些概率分布，存在计算 *μ* 和 *σ* 的快捷公式。

Problem 问题

Toss a fair, six-sided die twice. Let *X* = the number of faces that show an even number. Construct a table like Table 4.11 and calculate the mean *μ* and standard deviation *σ* of *X*.

抛一枚均匀的六面骰子两次。令 *X* = 出现偶数的面的个数。构造一张类似表 4.11 的表，并计算 *X* 的均值 *μ* 和标准差 *σ*。

Solution 解答

Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes:

抛一枚均匀的六面骰子两次，与抛两枚均匀的六面骰子具有相同的样本空间。该样本空间有 36 个结果：

| | | | | | |

|--------|--------|--------|--------|--------|--------|

| (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |

| (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |

| (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |

| (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |

| (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |

| (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |

Table 4.13

表 4.13

Use the sample space to complete the following table:

利用样本空间补全下列表格：

| *x* | *P*(*x*) | *xP*(*x*) | (*x* – *μ*)² $\cdot$ *P*(*x*) |

|-----|-----------------|-----------------|-------------------------------------------------------|

| 0 | $\frac{9}{36}$ | 0 | (0 – 1)² ⋅ $\frac{9}{36}$ = $\frac{9}{36}$ |

| 1 | $\frac{18}{36}$ | $\frac{18}{36}$ | (1 – 1)² ⋅ $\frac{18}{36}$ = 0 |

| 2 | $\frac{9}{36}$ | $\frac{18}{36}$ | (1 – 1)² ⋅ $\frac{9}{36}$ = $\frac{9}{36}$ |

Table 4.14 Calculating *μ* and *σ*.

表 4.14 计算 *μ* 和 *σ*。

Add the values in the third column to find the expected value: *μ* = $\frac{36}{36}$ = 1. Use this value to complete the fourth column.

把第三列各值相加求期望值：*μ* = $\frac{36}{36}$ = 1。用这个值补全第四列。

Add the values in the fourth column and take the square root of the sum: σ = $\sqrt{\frac{18}{36}}$ ≈ 0.7071.

把第四列各值相加并对该和开平方：σ = $\sqrt{\frac{18}{36}}$ ≈ 0.7071。

Problem 问题

On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win \$50. If you lose the bet, you pay \$20. Let *X* = the amount of profit from a bet.

2013年5月11日晚9:30，伊朗在未来48小时内发生中等强度地震活动（一次中等强度地震）的概率约为21.42%。假设你打赌在此期间伊朗会发生一次中等强度地震。若赌赢，你赢得\$50；若赌输，你付出\$20。令 *X* = 一次打赌的利润。

*P*(win) = *P*(one moderate earthquake will occur) = 21.42%

*P*(赢) = *P*(发生一次中等强度地震) = 21.42%

*P*(loss) = *P*(one moderate earthquake will *not* occur) = 100% – 21.42%

*P*(输) = *P*(不发生中等强度地震) = 100% – 21.42%

If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of *X*? Construct a table similar to Table 4.12 and Table 4.13 to help you answer these questions.

若你多次打赌，是否会盈利？请用数字以完整的句子解释你的答案。*X* 的标准差是多少？构造一个类似于表 4.12 和表 4.13 的表格来帮助你回答这些问题。

Solution 解答

| | *x* | *P(x)* | *x(Px)* | (*x* – *μ*)²*P*(*x*) |

|------|-----|--------|---------|---------------------------------------------------|

| win | 50 | 0.2142 | 10.71 | \[50 – (–5.006)\]²(0.2142) = 648.0964 |

| 赢 | 50 | 0.2142 | 10.71 | \[50 – (–5.006)\]²(0.2142) = 648.0964 |

| loss | –20 | 0.7858 | –15.716 | \[–20 – (–5.006)\]²(0.7858) = 176.6636 |

| 输 | –20 | 0.7858 | –15.716 | \[–20 – (–5.006)\]²(0.7858) = 176.6636 |

Table 4.15

表 4.15

Mean = Expected Value = 10.71 + (–15.716) = –5.006.

均值 = 期望值 = 10.71 + (–15.716) = –5.006。

If you make this bet many times under the same conditions, your long term outcome will be an average *loss* of \$5.01 per bet.

若你在相同条件下多次打赌，你的长期结果将是平均每次打赌*亏损* \$5.01。

$\text{Standard~Deviation~=~}\sqrt{648.0964 + 176.6636} \approx 28.7186$

On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about 1.08%. As in Example 4.8, you bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win \$100. If you lose the bet, you pay \$10. Let *X* = the amount of profit from a bet. Find the mean and standard deviation of *X*.

2013年5月11日晚9:30，日本在未来48小时内发生中等强度地震活动（一次中等强度地震）的概率约为1.08%。如示例 4.8，你打赌在此期间日本会发生一次中等强度地震。若赌赢，你赢得\$100；若赌输，你付出\$10。令 *X* = 一次打赌的利润。求 *X* 的均值与标准差。

Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson. Your instructor will let you know if he or she wishes to cover these distributions.

一些较常见的离散概率函数有二项分布、几何分布、超几何分布和泊松分布。大多数初级课程不涵盖几何分布、超几何分布和泊松分布。若你的教师希望涵盖这些分布，他会告知你。

A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution in order to perform the necessary calculations. These distributions are tools to make solving probability problems easier. Each distribution has its own special characteristics. Learning the characteristics enables you to distinguish among the different distributions.

概率分布函数是一种模式。你试图将一个概率问题套入某种模式或分布，以便进行必要的计算。这些分布是使概率问题求解更简便的工具。每一种分布都有其自身的特殊特征。掌握这些特征能让你区分不同的分布。

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4.3 Binomial Distribution 4.3 二项分布

There are three characteristics of a binomial experiment.

二项试验有三个特征。

1. There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter *n* denotes the number of trials.

1. 试验次数固定。可将试验视为一个实验的重复。字母 *n* 表示试验次数。

2. There are only two possible outcomes, called "success" and "failure," for each trial. The letter *p* denotes the probability of a success on one trial, and *q* denotes the probability of a failure on one trial. *p* + *q* = 1.

2. 每次试验只有两种可能结果，称为“成功”与“失败”。字母 *p* 表示一次试验成功的概率，*q* 表示一次试验失败的概率。*p* + *q* = 1。

3. The *n* trials are independent and are repeated using identical conditions. Because the *n* trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, *p*, of a success and probability, *q*, of a failure remain the same. For example, randomly guessing at a true-false statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics true-false question with probability *p* = 0.6. Then, *q* = 0.4. This means that for every true-false statistics question Joe answers, his probability of success (*p* = 0.6) and his probability of failure (*q* = 0.4) remain the same.

3. 这 *n* 次试验相互独立，且在相同条件下重复。由于 *n* 次试验相互独立，一次试验的结果无助于预测另一次试验的结果。换言之，对每一次单独试验而言，成功的概率 *p* 与失败的概率 *q* 保持不变。例如，对一道判断对错的统计题随机猜测，只有两种结果。若成功定义为猜对，则失败就是猜错。假设乔在任意一道统计判断题上猜对的概率恒为 *p* = 0.6，则 *q* = 0.4。这意味着乔所回答的每一道统计判断题，其成功概率（*p* = 0.6）与失败概率（*q* = 0.4）都保持不变。

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable *X* = the number of successes obtained in the *n* independent trials.

二项试验的结果符合二项概率分布。随机变量 *X* = 在 *n* 次独立试验中所获得的成功次数。

The mean, *μ*, and variance, *σ*², for the binomial probability distribution are *μ* = *np* and *σ*² = *npq*. The standard deviation, *σ*, is then *σ* = $\sqrt{npq}$.

二项概率分布的均值 *μ* 与方差 *σ*² 分别为 *μ* = *np* 与 *σ*² = *npq*。标准差 *σ* 则为 *σ* = $\sqrt{npq}$。

Any experiment that has characteristics two and three and where *n* = 1 is called a Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.

任何满足特征二与特征三且 *n* = 1 的试验称为伯努利试验（得名于雅各布·伯努利，他在17世纪末对其作了深入研究）。当在一个或多个伯努利试验中统计成功次数时，就构成了二项试验。

At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could be defined as an individual who withdrew. The random variable *X* = the number of students who withdraw from the randomly selected elementary physics class.

在 ABC 学院，某学期初等物理课程的退课率为30%。这意味着，对任意一个学期，有70%的学生全程留在课堂上。可将“成功”定义为退课的学生。随机变量 *X* = 从随机选取的初等物理班级中退课的学生人数。

The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case?

州卫生委员会关注学校午餐中水果的供应情况。该州48%的学校每天在午餐中提供水果，这意味着52%的学校不提供。在此情形下，“成功”是什么？

Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define *X* as the number of wins, then *X* takes on the values 0, 1, 2, 3, ..., 20. The probability of a success is *p* = 0.55. The probability of a failure is *q* = 0.45. The number of trials is *n* = 20. The probability question can be stated mathematically as *P*(*x* = 15).

假设你玩一个游戏，结果非赢即输。你任意一局获胜的概率为55%，失败的概率为45%。每局游戏相互独立。若你玩20局，写出描述你在20局中赢15局的概率的函数。此处若将 *X* 定义为获胜次数，则 *X* 取值 0, 1, 2, 3, ..., 20。成功概率为 *p* = 0.55，失败概率为 *q* = 0.45，试验次数为 *n* = 20。该概率问题可表述为 *P*(*x* = 15)。

A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.

一名驯养员正在教海豚做技巧动作。海豚成功完成该技巧的概率为35%，未能成功完成的概率为65%。在20次尝试中，你想求海豚成功12次的概率。请用数学语言表述该概率问题。

Problem 问题

A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than ten heads? Let *X* = the number of heads in 15 flips of the fair coin. *X* takes on the values 0, 1, 2, 3, ..., 15. Since the coin is fair, *p* = 0.5 and *q* = 0.5. The number of trials is *n* = 15. State the probability question mathematically.

一枚均匀的硬币抛掷15次，每次抛掷相互独立。出现超过10次正面的概率是多少？令 *X* = 均匀硬币15次抛掷中正面的次数。*X* 取值 0, 1, 2, 3, ..., 15。由于硬币均匀，*p* = 0.5，*q* = 0.5，试验次数 *n* = 15。请用数学语言表述该概率问题。

A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.

一枚均匀的六面骰子掷10次，每次掷骰相互独立。你想求掷出1点超过3次的概率。请用数学语言表述该概率问题。

Approximately 70% of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.

约70%的统计学学生能及时完成作业以供收取和批改。每位学生独立完成作业。在一个50人的统计班上，至少有40人按时完成作业的概率是多少？学生为随机选取。

Problem 问题

a\. This is a binomial problem because there is only a success or a \_\_\_\_\_\_\_\_\_\_, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.

a\. 这是一个二项分布问题，因为结果只有成功或\_\_\_\_\_\_\_\_\_\_，试验次数固定，且每次试验成功的概率为0.70。

b\. If we are interested in the number of students who do their homework on time, then how do we define *X*?

b\. 若我们关注按时完成作业的学生人数，那么如何定义 *X*？

c\. What values does *x* take on?

c\. *x* 取哪些值？

d\. What is a "failure," in words?

d\. 用文字表述，“失败”是什么？

e\. If *p* + *q* = 1, then what is *q*?

e\. 若 *p* + *q* = 1，则 *q* 是多少？

f\. The words "at least" translate as what kind of inequality for the probability question *P*(*x* \_\_\_\_ 40).

f\. “至少”一词对应于概率问题 *P*(*x* \_\_\_\_ 40) 中的何种不等式？

Solution 解答

a\. failure

a\. 失败

b\. *X* = the number of statistics students who do their homework on time

b\. *X* = 按时完成统计学作业的学生人数

c\. 0, 1, 2, …, 50

d\. Failure is defined as a student who does not complete his or her homework on time.

d\. 失败定义为未能按时完成作业的学生。

The probability of a success is *p* = 0.70. The number of trials is *n* = 50.

成功概率为 *p* = 0.70。试验次数为 *n* = 50。

e\. *q* = 0.30

f\. greater than or equal to (≥)

f\. 大于或等于（≥）

The probability question is *P*(*x* ≥ 40).

该概率问题为 *P*(*x* ≥ 40)。

Sixty-five percent of people pass the state driver’s exam on the first try. A group of 50 individuals who have taken the driver’s exam is randomly selected. Give two reasons why this is a binomial problem.

65%的人首次尝试即通过州驾驶考试。随机选取一组50名已参加过驾驶考试的人。给出两个理由说明为何这是一个二项分布问题。

Notation for the Binomial: B = Binomial Probability Distribution Function 二项分布的记号：B = 二项概率分布函数

*X* ~ *B*(*n*, *p*)

Read this as "*X* is a random variable with a binomial distribution." The parameters are *n* and *p*; *n* = number of trials, *p* = probability of a success on each trial.

读作“*X* 是一个服从二项分布的随机变量”。参数为 *n* 和 *p*；*n* = 试验次数，*p* = 每次试验成功的概率。

It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?

据称，约41%的成年劳动者具有高中文凭但未继续接受任何高等教育。若随机选取20名成年劳动者，求其中至多12人具有高中文凭但未继续接受高等教育的概率。你预期有多少名成年劳动者具有高中文凭但未继续接受高等教育？

Let *X* = the number of workers who have a high school diploma but do not pursue any further education.

令 *X* = 具有高中文凭但未继续接受高等教育的劳动者人数。

*X* takes on the values 0, 1, 2, ..., 20 where *n* = 20, *p* = 0.41, and *q* = 1 – 0.41 = 0.59. *X* ~ *B*(20, 0.41)

*X* 取值 0, 1, 2, ..., 20，其中 *n* = 20，*p* = 0.41，*q* = 1 – 0.41 = 0.59。*X* ~ *B*(20, 0.41)

Find *P*(*x* ≤ 12). *P*(*x* ≤ 12) = 0.9738. (calculator or computer)

求 *P*(*x* ≤ 12)。*P*(*x* ≤ 12) = 0.9738。（用计算器或计算机）

Go into 2^nd DISTR. The syntax for the instructions are as follows:

进入 2^nd DISTR。指令的语法如下：

**To calculate (*x* = value): binompdf(*n*, *p*, number)** if "number" is left out, the result is the binomial probability table.

**要计算 (*x* = 某值)：binompdf(*n*, *p*, number)** 若省略“number”，结果为二项概率表。

**To calculate *P*(*x* ≤ value): binomcdf(*n*, *p*, number)** if "number" is left out, the result is the cumulative binomial probability table.

**要计算 *P*(*x* ≤ 某值)：binomcdf(*n*, *p*, number)** 若省略“number”，结果为累积二项概率表。

**For this problem: After you are in 2^nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is *P*(*x* ≤ 12) = 0.9738.**

**针对本题：进入 2^nd DISTR 后，向下箭头选择 binomcdf，按 ENTER。输入 20,0.41,12)。结果为 *P*(*x* ≤ 12) = 0.9738。**

If you want to find *P*(*x* = 12), use the pdf (binompdf). If you want to find *P*(*x* \> 12), use 1 - binomcdf(20,0.41,12).

若要求 *P*(*x* = 12)，使用 pdf（binompdf）。若要求 *P*(*x* \> 12)，使用 1 - binomcdf(20,0.41,12)。

The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.

至多12名劳动者具有高中文凭但未继续接受高等教育的概率为0.9738。

The graph of *X* ~ *B*(20, 0.41) is as follows:

*X* ~ *B*(20, 0.41) 的图像如下：

The *y*-axis contains the probability of *x*, where *X* = the number of workers who have only a high school diploma.

纵轴为 *x* 的概率，其中 *X* = 仅具有高中文凭的劳动者人数。

The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, *μ* = *np* = (20)(0.41) = 8.2.

你预期具有高中文凭但未继续接受高等教育的成年劳动者人数为均值 *μ* = *np* = (20)(0.41) = 8.2。

The formula for the variance is σ² = *npq*. The standard deviation is *σ* = $\sqrt{npq}$.

方差公式为 σ² = *npq*。标准差为 *σ* = $\sqrt{npq}$。

*σ* = $\sqrt{(20)(0.41)(0.59)}$ = 2.20.

*σ* = $\sqrt{(20)(0.41)(0.59)}$ = 2.20。

About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.

约32%的学生参与校外的社区志愿者项目。若随机选取30名学生，求其中至多14人参与校外社区志愿者项目的概率。使用 TI-83+ 或 TI-84 计算器求答案。

Problem 问题

In the 2013 *Jerry’s Artarama* art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let *X* = the number of pages that feature signature artists.

在2013年的 *Jerry’s Artarama* 美术用品目录中，共有560页。其中有8页刊登了署名艺术家的作品。假设我们随机抽取100页。令 *X* = 刊登署名艺术家作品的页数。

1. What values does *x* take on?

1. *x* 取哪些值？

2. What is the probability distribution? Find the following probabilities:

2. 概率分布是什么？求下列概率：

1. the probability that two pages feature signature artists

1. 恰有两页刊登署名艺术家作品的概率

2. the probability that at most six pages feature signature artists

2. 至多六页刊登署名艺术家作品的概率

3. the probability that more than three pages feature signature artists.

3. 超过三页刊登署名艺术家作品的概率。

3. Using the formulas, calculate the (i) mean and (ii) standard deviation.

3. 利用公式，计算（i）均值与（ii）标准差。

Solution 解答

1. *x* = 0, 1, 2, 3, 4, 5, 6, 7, 8

2. *X* ~ *B*$\left( {100,\frac{8}{560}} \right)$

1. *P*(*x* = 2) = binompdf$\left( {100,\frac{8}{560},2} \right)$ = 0.2466

2. *P*(*x* ≤ 6) = binomcdf$\left( {100,\frac{8}{560},6} \right)$ = 0.9994

3. *P*(*x* \> 3) = 1 – *P*(*x* ≤ 3) = 1 – binomcdf$\left( {100,\frac{8}{560},3} \right)$ = 1 – 0.9443 = 0.0557

3. 1. Mean = *np* = (100)$\left( \frac{8}{560} \right)$ = $\frac{800}{560}$ ≈ 1.4286

3. 1. 均值 = *np* = (100)$\left( \frac{8}{560} \right)$ = $\frac{800}{560}$ ≈ 1.4286

2. Standard Deviation = $\sqrt{npq}$ = $\sqrt{(100)\left( \frac{8}{560} \right)\left( \frac{552}{560} \right)}$ ≈ 1.1867

2. 标准差 = $\sqrt{npq}$ = $\sqrt{(100)\left( \frac{8}{560} \right)\left( \frac{552}{560} \right)}$ ≈ 1.1867

According to a Gallup poll, 60% of American adults prefer saving over spending. Let *X* = the number of American adults out of a random sample of 50 who prefer saving to spending.

据盖洛普民意调查，60%的美国成年人偏好储蓄而非消费。令 *X* = 在随机抽样的50名美国成年人中偏好储蓄而非消费的人数。

1. What is the probability distribution for *X*?

1. *X* 的概率分布是什么？

2. Use your calculator to find the following probabilities:

2. 使用计算器求下列概率：

1. the probability that 25 adults in the sample prefer saving over spending

1. 样本中25名成年人偏好储蓄而非消费的概率

2. the probability that at most 20 adults prefer saving

2. 至多20名成年人偏好储蓄的概率

3. the probability that more than 30 adults prefer saving

3. 超过30名成年人偏好储蓄的概率

3. Using the formulas, calculate the (i) mean and (ii) standard deviation of *X*.

3. 利用公式，计算 *X* 的（i）均值与（ii）标准差。

The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let *X* = the number of people who will develop pancreatic cancer.

罹患胰腺癌的终生风险约为七十八分之一（1.28%）。假设我们随机抽取200人。令 *X* = 将罹患胰腺癌的人数。

Problem 问题

1. What is the probability distribution for *X*?

1. *X* 的概率分布是什么？

2. Using the formulas, calculate the (i) mean and (ii) standard deviation of *X*.

2. 利用公式，计算 *X* 的（i）均值与（ii）标准差。

3. Use your calculator to find the probability that at most eight people develop pancreatic cancer

3. 使用计算器求出至多八人罹患胰腺癌的概率

4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.

4. 五人还是六人更可能罹患胰腺癌？用数值说明你的答案。

During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let *X* = the number of shots that scored points.

在 2013 年 NBA 常规赛中，洛杉矶快船队的 DeAndre Jordan 拥有联盟最高的投篮命中率。DeAndre 有 61.3% 的投篮命中得分。假设你在 2013 赛季中随机抽取 DeAndre 命中的 80 次投篮。设 *X* = 得分的投篮次数。

1. What is the probability distribution for *X*?

1. *X* 的概率分布是什么？

2. Using the formulas, calculate the (i) mean and (ii) standard deviation of *X*.

2. 利用公式，计算 *X* 的（i）均值与（ii）标准差。

3. Use your calculator to find the probability that DeAndre scored with 60 of these shots.

3. 使用计算器求出 DeAndre 在这 80 次投篮中 60 次命中的概率。

4. Find the probability that DeAndre scored with more than 50 of these shots.

4. 求 DeAndre 在这 80 次投篮中命中超过 50 次的概率。

The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is $\frac{6}{16}$. The probability of a student on the second draw is $\frac{5}{15}$, when the first draw selects a student. The probability is $\frac{6}{15}$, when the first draw selects a staff member. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.

下面的例子说明了一个非二项分布的问题。它违反了独立性条件。ABC 学院有一个学生咨询委员会，由十名教职员工和六名学生组成。委员会希望选出一名主席和一名记录员。主席和记录员都是学生的概率是多少？所有委员会成员的姓名被放入一个盒子中，无放回地抽取两个姓名。第一个抽出的姓名决定主席，第二个抽出的姓名决定记录员。共有两次试验。然而，这两次试验并不独立，因为第一次试验的结果影响第二次试验的结果。第一次抽到学生的概率是 $\frac{6}{16}$。当第一次抽到学生时，第二次抽到学生的概率是 $\frac{5}{15}$。当第一次抽到教职员工时，该概率为 $\frac{6}{15}$。每次试验中抽到学生姓名的概率都会改变，因此违反了独立性条件。

A lacrosse team is selecting a captain. The names of all the seniors are put into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why.

一支长曲棍球队正在选拔队长。所有高年级学生的姓名被放入帽子中，最先抽出的三个姓名将成为队长。姓名一旦被抽出就不再放回（一个人不能同时担任两名队长）。你想知道这些队长是否都打同一个位置。说明这是否为二项分布，并说明原因。

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4.4 Geometric Distribution 4.4 几何分布

There are three main characteristics of a geometric experiment.

几何试验有三个主要特征。

1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.

1. 存在一个或多个伯努利试验，除最后一次为成功外，其余均为失败。换句话说，你不断重复所做之事，直到第一次成功，然后停止。例如，你向靶心掷飞镖，直到命中靶心。第一次命中靶心即为一次"成功"，于是你停止掷飞镖。你可能需要六次尝试才能命中靶心。你可以将这些试验看作：失败、失败、失败、失败、失败、成功、停止。

2. In theory, the number of trials could go on forever. There must be at least one trial.

2. 理论上，试验次数可以无限进行下去。但至少必须有一个试验。

3. The probability, *p*, of a success and the probability, *q*, of a failure is the same for each trial. *p* + *q* = 1 and *q* = 1 − *p*. For example, the probability of rolling a three when you throw one fair die is $\frac{1}{6}$. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is *q* = $\frac{\text{5}}{\text{6}}$, the probability of a failure. The probability of getting a three on the fifth roll is $\left( \frac{5}{6} \right)\left( \frac{5}{6} \right)\left( \frac{5}{6} \right)\left( \frac{5}{6} \right)\left( \frac{1}{6} \right)$ = 0.0804

3. 每次试验成功的概率 *p* 与失败的概率 *q* 都相同。*p* + *q* = 1，且 *q* = 1 − *p*。例如，掷一枚均匀骰子得到三点概率是 $\frac{1}{6}$。无论你掷多少次骰子，这都成立。假设你想知道第五次掷出第一个三点的概率。在第一到第四次掷骰中，你没有掷出带三点的面。每次掷骰的概率为 *q* = $\frac{\text{5}}{\text{6}}$，即失败的概率。第五次掷出三点的概率为 $\left( \frac{5}{6} \right)\left( \frac{5}{6} \right)\left( \frac{5}{6} \right)\left( \frac{5}{6} \right)\left( \frac{1}{6} \right)$ = 0.0804

*X* = the number of independent trials until the first success.

*X* = 直到第一次成功为止的独立试验次数。

You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is *p* = 0.57. What is the probability that it takes five games until you lose? Let *X* = the number of games you play until you lose (includes the losing game). Then *X* takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is *P*(*x* = 5).

你玩一个胜负游戏（只有赢或输两种可能），一直玩到输为止。你输的概率是 *p* = 0.57。玩到第五次才输的概率是多少？设 *X* = 你玩到输为止进行的游戏次数（包含输的那一局）。则 *X* 取值 1, 2, 3, ...（可以无限延续）。其概率问题是 *P*(*x* = 5)。

You throw darts at a board until you hit the center area. Your probability of hitting the center area is *p* = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does *X* take on?

你向靶板掷飞镖，直到命中中心区域。你命中中心区域的概率是 *p* = 0.17。你想求出掷八次才命中中心的概率。*X* 取哪些值？

A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions?

一位安全工程师认为，她所在工厂中所有工业事故的 35% 是由员工未遵守操作规程引起的。她决定查看事故报告（随机抽取，读后放回原堆），直到找到一份显示事故由员工未遵守操作规程引起的报告为止。平均而言，这位安全工程师需要查看多少份报告才能找到一份显示事故由员工未遵守操作规程引起的报告？这位安全工程师在找到一份显示事故由员工未遵守操作规程引起的报告之前，至少必须检查三份报告的概率是多少？

Let *X* = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. *X* takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find *P*(*x* ≥ 3). ("At least" translates to a "greater than or equal to" symbol).

设 *X* = 安全工程师必须检查、直到找到一份显示事故由员工未遵守操作规程引起的报告为止的事故数。*X* 取值 1, 2, 3, ...。第一个问题要求你求期望值或均值。第二个问题要求你求 *P*(*x* ≥ 3)（"至少"对应于"大于或等于"符号）。

An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically?

一位教师认为有 15% 的学生期末考试成绩低于 C。她决定查看期末试卷（随机抽取，读后放回原堆），直到找到一份成绩低于 C 的试卷。我们想知道这位教师必须检查至少十份试卷才能找到一份成绩低于 C 的试卷的概率。用数学表达式写出这个概率问题。

Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?

假设你在你的大学里寻找一名住在距你五英里以内的学生。你知道 25,000 名学生中有 55% 确实住在距你五英里以内。你从该校随机联系学生，直到有人说他或她住在距你五英里以内。你需要联系四个人的概率是多少？

This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).

这是一个几何分布问题，因为在你得到所期望的那一次成功之前，可能会有若干次失败。而且，每次你询问学生是否住在距你五英里以内时，成功的概率保持不变。试验次数（你询问学生的次数）没有确定的数值。

Problem 问题

a\. Let *X* = the number of \_\_\_\_\_\_\_\_\_\_\_\_ you must ask \_\_\_\_\_\_\_\_\_\_\_\_ one says yes.

a. 设 *X* = 你必须询问的 \_\_\_\_\_\_\_\_\_\_\_\_ 的数量，直到 \_\_\_\_\_\_\_\_\_\_\_\_ 有人说"是"。

b\. What values does *X* take on?

b. *X* 取哪些值？

c\. What are *p* and *q*?

c. *p* 和 *q* 各是多少？

d\. The probability question is *P*(\_\_\_\_\_\_\_).

d. 概率问题是 *P*(\_\_\_\_\_\_\_)。

Solution 解答

a\. Let *X* = the number of students you must ask until one says yes.

a. 设 *X* = 你必须询问、直到有人说"是"为止的学生人数。

b\. 1, 2, 3, …, (total number of students)

b. 1, 2, 3, …，（学生总人数）

c\. *p* = 0.55; *q* = 0.45

c. *p* = 0.55；*q* = 0.45

d\. *P*(*x* = 4)

d. *P*(*x* = 4)

You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are *p* and *q*?

你需要找一家有某种特殊打印机墨水的商店。你知道在售卖打印机墨水的商店中，有 10% 售卖这种特殊墨水。你随机给每家商店打电话，直到有一家有所需的墨水。*p* 和 *q* 各是多少？

Notation for the Geometric: G = Geometric Probability Distribution Function 几何分布记号：G = 几何概率分布函数

*X* ~ *G*(*p*)

Read this as "*X* is a random variable with a geometric distribution." The parameter is *p*; *p* = the probability of a success for each trial.

读作"*X* 是一个服从几何分布的随机变量"。参数为 *p*；*p* = 每次试验成功的概率。

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?

假设某计算机元件的缺陷概率为 0.02。随机抽取元件。求第一个缺陷由被检测的第 7 个元件造成的概率。你预计要检测多少个元件才能发现一个有缺陷的？

Let *X* = the number of computer components tested until the first defect is found.

设 *X* = 检测到第一个缺陷为止所检测的计算机元件数。

*X* takes on the values 1, 2, 3, ... where *p* = 0.02. *X* ~ G(0.02)

*X* 取值 1, 2, 3, ...，其中 *p* = 0.02。*X* ~ G(0.02)

Find *P*(*x* = 7). *P*(*x* = 7) = 0.0177.

求 *P*(*x* = 7)。*P*(*x* = 7) = 0.0177。

To find the probability that *x* = 7,

要求 *x* = 7 的概率，

Enter 2^nd, DISTR
Scroll down and select geometpdf(
Press ENTER
Enter 0.02, 7); press ENTER to see the result: *P*(*x* = 7) = 0.0177

按 2^nd，DISTR
向下滚动并选择 geometpdf(
按 ENTER
输入 0.02, 7)；按 ENTER 查看结果：*P*(*x* = 7) = 0.0177

To find the probability that *x* ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function.

要求 *x* ≤ 7 的概率，遵循相同的步骤，但选择 E:geometcdf( 作为分布函数。

The probability that the seventh component is the first defect is 0.0177.

第 7 个元件是第一个缺陷的概率为 0.0177。

The graph of *X* ~ G(0.02) is:

*X* ~ G(0.02) 的图像如下：

The *y*-axis contains the probability of *x*, where *X* = the number of computer components tested.

*y* 轴表示 *x* 的概率，其中 *X* = 所检测的计算机元件数。

The number of components that you would expect to test until you find the first defective one is the mean, $\mu\text{~=~50}$.

你预计检测到第一个缺陷元件所需检测的元件数即为均值，$\mu\text{~=~50}$。

The formula for the mean is *μ* = $\frac{1}{p}$ = $\frac{1}{0.02}$ = 50

均值的公式为 *μ* = $\frac{1}{p}$ = $\frac{1}{0.02}$ = 50

The formula for the variance is *σ*² = $\left( \frac{1}{p} \right)\left( {\frac{1}{p} - 1} \right)$ = $\left( \frac{1}{0.02} \right)\left( {\frac{1}{0.02} - 1} \right)$ = 2,450

方差的公式为 *σ*² = $\left( \frac{1}{p} \right)\left( {\frac{1}{p} - 1} \right)$ = $\left( \frac{1}{0.02} \right)\left( {\frac{1}{0.02} - 1} \right)$ = 2,450

The standard deviation is *σ* = $\sqrt{\left( \frac{1}{p} \right)\left( {\frac{1}{p} - 1} \right)}$ = $\sqrt{\left( \frac{1}{0.\text{02}} \right)\left( {\frac{1}{0.\text{02}} - 1} \right)}$ = 49.5

标准差为 *σ* = $\sqrt{\left( \frac{1}{p} \right)\left( {\frac{1}{p} - 1} \right)}$ = $\sqrt{\left( \frac{1}{0.\text{02}} \right)\left( {\frac{1}{0.\text{02}} - 1} \right)}$ = 49.5

The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.

一根钢筋有缺陷的概率为 0.01。随机抽取钢筋。求第一个缺陷出现在第 9 根钢筋上的概率。使用 TI-83+ 或 TI-84 计算器求答案。

Problem 问题

The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let *X* = the number of people you ask until one says he or she has pancreatic cancer. Then *X* is a discrete random variable with a geometric distribution: *X* ~ *G*$\left( \frac{1}{78} \right)$ or *X* ~ *G*(0.0128).

罹患胰腺癌的终身风险约为 1/78（1.28%）。设 *X* = 你询问的人数，直到有人说他或她罹患胰腺癌。则 *X* 是一个服从几何分布的离散随机变量：*X* ~ *G*$\left( \frac{1}{78} \right)$ 或 *X* ~ *G*(0.0128)。

1. What is the probability of that you ask ten people before one says he or she has pancreatic cancer?

1. 在有人说他或她罹患胰腺癌之前，你询问了十个人的概率是多少？

2. What is the probability that you must ask 20 people?

2. 你必须询问 20 个人的概率是多少？

3. Find the (i) mean and (ii) standard deviation of *X*.

3. 求 *X* 的（i）均值与（ii）标准差。

Solution 解答

1. *P*(*x* = 10) = geometpdf(0.0128, 10) = 0.0114

2. *P*(*x* = 20) = geometpdf(0.0128, 20) = 0.01

3. 1. Mean = *μ* = $\frac{1}{p}$ = $\frac{1}{0.0128}$ = 78

3. 1. 均值 = *μ* = $\frac{1}{p}$ = $\frac{1}{0.0128}$ = 78

2. Standard Deviation = *σ* = $\sqrt{\frac{1 - p}{p^{2}}}$ = $\sqrt{\frac{1 - 0.0128}{0.0128^{2}}}$ ≈ 77.6234

2. 标准差 = *σ* = $\sqrt{\frac{1 - p}{p^{2}}}$ = $\sqrt{\frac{1 - 0.0128}{0.0128^{2}}}$ ≈ 77.6234

The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let *X* = the number of Afghani women you ask until one says that she is literate.

一个国家的识字率衡量的是 15 岁及以上人口中能够读写的比例。阿富汗女性的识字率为 12%。设 *X* = 你询问的阿富汗女性人数，直到有人说她识字。

1. What is the probability distribution of *X*?

1. *X* 的概率分布是什么？

2. What is the probability that you ask five women before one says she is literate?

2. 在有人说她识字之前，你询问了五位女性的概率是多少？

3. What is the probability that you must ask ten women?

3. 你必须询问十位女性的概率是多少？

4. Find the (i) mean and (ii) standard deviation of *X*.

4. 求 *X* 的（i）均值与（ii）标准差。

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4.5 Hypergeometric Distribution 4.5 超几何分布

There are five characteristics of a hypergeometric experiment.

超几何试验有五个特征。

1. You take samples from two groups.

1. 你从两个组中抽取样本。

2. You are concerned with a group of interest, called the first group.

2. 你关注一个感兴趣的总体，称为第一组。

3. You sample without replacement from the combined groups. For example, you want to choose a softball team from a combined group of 11 men and 13 women. The team consists of ten players.

3. 你从合并后的组中无放回抽样。例如，你想从 11 名男性和 13 名女性合并而成的总体中选出一支垒球队。该队由十名队员组成。

4. Each pick is not independent, since sampling is without replacement. In the softball example, the probability of picking a woman first is $\frac{13}{24}$. The probability of picking a man second is $\frac{11}{23}$ if a woman was picked first. It is $\frac{10}{23}$ if a man was picked first. The probability of the second pick depends on what happened in the first pick.

4. 由于是无放回抽样，每次抽取都不独立。在垒球例子中，第一次抽到女性的概率是 $\frac{13}{24}$。如果第一次抽到的是女性，则第二次抽到男性的概率是 $\frac{11}{23}$；如果第一次抽到的是男性，则为 $\frac{10}{23}$。第二次抽取的概率取决于第一次抽取的结果。

5. You are not dealing with Bernoulli Trials.

5. 你处理的不是伯努利试验。

The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable *X* = the number of items from the group of interest.

超几何试验的结果符合超几何概率分布。随机变量 *X* = 来自感兴趣总体的物品数量。

Problem 问题

A candy dish contains 100 jelly beans and 80 gumdrops. Fifty candies are picked at random. What is the probability that 35 of the 50 are gumdrops? The two groups are jelly beans and gumdrops. Since the probability question asks for the probability of picking gumdrops, the group of interest (first group) is gumdrops. The size of the group of interest (first group) is 80. The size of the second group is 100. The size of the sample is 50 (jelly beans or gumdrops). Let *X* = the number of gumdrops in the sample of 50. *X* takes on the values *x* = 0, 1, 2, ..., 50. What is the probability statement written mathematically?

一个糖罐中有 100 颗软糖和 80 颗橡皮糖。随机抽取 50 颗糖果。其中 35 颗是橡皮糖的概率是多少？这两个组分别是软糖和橡皮糖。由于概率问题要求的是抽到橡皮糖的概率，感兴趣的总体（第一组）是橡皮糖。感兴趣总体（第一组）的大小为 80。第二组的大小为 100。样本量为 50（软糖或橡皮糖）。设 *X* = 这 50 颗样本中橡皮糖的数量。*X* 取值 *x* = 0, 1, 2, ..., 50。用数学表达式写出该概率陈述。

Solution 解答

*P*(*x* = 35)

A bag contains letter tiles. Forty-four of the tiles are vowels, and 56 are consonants. Seven tiles are picked at random. You want to know the probability that four of the seven tiles are vowels. What is the group of interest, the size of the group of interest, and the size of the sample?

一个袋子中装有字母块。其中 44 个为元音字母，56 个为辅音字母。随机抽取 7 个字母块。你想知道这 7 个字母块中有 4 个是元音字母的概率。感兴趣的总体是什么？感兴趣总体的大小是多少？样本量是多少？

Problem 问题

Suppose a shipment of 100 DVD players is known to have ten defective players. An inspector randomly chooses 12 for inspection. He is interested in determining the probability that, among the 12 players, at most two are defective. The two groups are the 90 non-defective DVD players and the 10 defective DVD players. The group of interest (first group) is the defective group because the probability question asks for the probability of at most two defective DVD players. The size of the sample is 12 DVD players. (They may be non-defective or defective.) Let *X* = the number of defective DVD players in the sample of 12. *X* takes on the values 0, 1, 2, ..., 10. *X* may not take on the values 11 or 12. The sample size is 12, but there are only 10 defective DVD players. Write the probability statement mathematically.

假设一批 100 台 DVD 播放器中已知有 10 台次品。一名检验员随机抽取 12 台进行检查。他想要确定这 12 台中至多有 2 台为次品的概率。两组分别是 90 台非次品 DVD 播放器和 10 台次品 DVD 播放器。关注组（第一组）是次品组，因为该概率问题要求的是至多有 2 台次品 DVD 播放器的概率。样本量为 12 台 DVD 播放器。（它们可能是非次品，也可能是次品。）令 *X* = 抽取的 12 台样本中的次品 DVD 播放器数量。*X* 取值 0, 1, 2, ..., 10。*X* 不可能取值 11 或 12。样本量为 12，但总共只有 10 台次品 DVD 播放器。用算式写出该概率。

Solution 解答

*P*(*x* ≤ 2)

A gross of eggs contains 144 eggs. A particular gross is known to have 12 cracked eggs. An inspector randomly chooses 15 for inspection. She wants to know the probability that, among the 15, at most three are cracked. What is *X*, and what values does it take on?

一罗（gross）鸡蛋有 144 枚。已知某一罗中有 12 枚裂壳蛋。一名检验员随机抽取 15 枚检查。她想知道这 15 枚中至多有 3 枚裂壳的概率。*X* 是什么？它取哪些值？

You are president of an on-campus special events organization. You need a committee of seven students to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. If the members of the committee are randomly selected, what is the probability that your committee has more than four men?

你是一个校内特别活动组织的主席。你需要一个由七名学生组成的委员会来策划学院院长的特别生日聚会。你的组织由 18 名女性和 15 名男性组成。你关注的是委员会中男性的数量。如果委员会成员是随机选出的，那么你的委员会中男性多于四人的概率是多少？

This is a hypergeometric problem because you are choosing your committee from two groups (men and women).

这是一个超几何分布问题，因为你是从两个组（男性和女性）中挑选委员会成员。

Problem 问题

a\. Are you choosing with or without replacement?

a. 你是在有放回还是无放回地选取？

b\. What is the group of interest?

b. 关注组是什么？

c\. How many are in the group of interest?

c. 关注组有多少人？

d\. How many are in the other group?

d. 另一组有多少人？

e\. Let *X* = \_\_\_\_\_\_\_\_\_ on the committee. What values does *X* take on?

e. 令 *X* = 委员会中的 \_\_\_\_\_\_\_\_\_。*X* 取哪些值？

f\. The probability question is *P*(\_\_\_\_\_\_\_).

f. 该概率问题是 *P*(\_\_\_\_\_\_\_)。

Solution 解答

a\. without

a. 无放回

b\. the men

b. 男性

c\. 15 men

c. 15 名男性

d\. 18 women

d. 18 名女性

e\. Let *X* = the number of men on the committee. *x* = 0, 1, 2, …, 7.

e. 令 *X* = 委员会中的 男性人数。*x* = 0, 1, 2, …, 7.

f\. *P*(*x* \> 4)

f. *P*(*x* \> 4)

A palette has 200 milk cartons. Of the 200 cartons, it is known that ten of them have leaked and cannot be sold. A stock clerk randomly chooses 18 for inspection. He wants to know the probability that among the 18, no more than two are leaking. Give five reasons why this is a hypergeometric problem.

一托盘有 200 个牛奶纸盒。在这 200 个纸盒中，已知有 10 个已经渗漏、无法出售。一名理货员随机抽取 18 个检查。他想知道这 18 个中不超过 2 个渗漏的概率。给出五条理由说明为什么这是一个超几何分布问题。

Notation for the Hypergeometric: H = Hypergeometric Probability Distribution Function 超几何分布的记号：H = 超几何概率分布函数

*X* ~ *H*(*r*, *b*, *n*)

Read this as "*X* is a random variable with a hypergeometric distribution." The parameters are *r*, *b*, and *n*; *r* = the size of the group of interest (first group), *b* = the size of the second group, *n* = the size of the chosen sample.

读作"*X* 是一个服从超几何分布的随机变量"。参数为 *r*、*b* 和 *n*；*r* = 关注组（第一组）的大小，*b* = 第二组的大小，*n* = 抽取的样本量。

A school site committee is to be chosen randomly from six men and five women. If the committee consists of four members chosen randomly, what is the probability that two of them are men? How many men do you expect to be on the committee?

要从 6 名男性和 5 名女性中随机选出学校场地委员会。如果委员会由随机选出的 4 名成员组成，那么其中恰有 2 名男性的概率是多少？你预计委员会中有多少名男性？

Let *X* = the number of men on the committee of four. The men are the group of interest (first group).

令 *X* = 四人选出的委员会中的男性人数。男性是关注组（第一组）。

*X* takes on the values 0, 1, 2, 3, 4, where *r = 6*, *b = 5*, and *n = 4*. *X ~ H*(6, 5, 4)

*X* 取值 0, 1, 2, 3, 4，其中 *r = 6*，*b = 5*，*n = 4*。*X ~ H*(6, 5, 4)

Find *P*(*x* = 2). *P*(*x* = 2) = 0.4545 (calculator or computer)

求 *P*(*x* = 2)。*P*(*x* = 2) = 0.4545（用计算器或计算机）。

Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft Excel, that do.

目前，TI-83+ 和 TI-84 没有超几何概率函数。但有许多计算机软件包有，包括 Microsoft Excel。

The probability that there are two men on the committee is about 0.45.

委员会中恰有两名男性的概率约为 0.45。

The graph of *X* ~ *H*(6, 5, 4) is:

*X* ~ *H*(6, 5, 4) 的图像为：

The *y*-axis contains the probability of *X*, where *X* = the number of men on the committee.

纵轴表示 *X* 的概率，其中 *X* = 委员会中的男性人数。

You would expect *m* = 2.18 (about two) men on the committee.

你预计委员会中男性人数 *m* = 2.18（约两人）。

The formula for the mean is $\mu = \frac{nr}{r + b} = \frac{(4)(6)}{6 + 5} = 2.18$

均值的公式为 $\mu = \frac{nr}{r + b} = \frac{(4)(6)}{6 + 5} = 2.18$

An intramural basketball team is to be chosen randomly from 15 boys and 12 girls. The team has ten slots. You want to know the probability that eight of the players will be boys. What is the group of interest and the sample?

要从 15 名男生和 12 名女生中随机选出一支校内篮球队。球队有 10 个位置。你想知道其中恰有 8 名男生的概率。关注组和样本分别是什么？

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4.6 Poisson Distribution 4.6 泊松分布

There are two main characteristics of a Poisson experiment.

泊松试验有两个主要特征。

1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.

1. 泊松概率分布给出了在固定区间的时间或空间内发生若干事件的概率，前提是这些事件以已知的平均速率发生，且与距上一事件的时间无关。例如，一本书的编辑可能关注某本书中拼写错误的单词数。平均而言，可能每 100 页有 5 个拼写错误的单词。这里的区间是这 100 页。

2. The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000). You will verify the relationship in the homework exercises. *n* is the number of trials, and *p* is the probability of a "success."

2. 当成功概率"很小"（如 0.01）且试验次数"很大"（如 1,000）时，可用泊松分布近似二项分布。你将在作业习题中验证这一关系。*n* 是试验次数，*p* 是"成功"的概率。

The random variable *X* = the number of occurrences in the interval of interest.

随机变量 *X* = 关注区间内发生的次数。

The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?

一家面包店在半小时内摆上货架的面包平均数量是 12 条。我们关注的是五分钟内摆上货架的面包数量。关注的时间区间是五分钟。随机摆上货架的面包中，五分钟内恰有 3 条的概率是多少？

Let *X* = the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is $\left( \frac{5}{30} \right)$(12) = 2 loaves of bread.

令 *X* = 五分钟内摆上货架的面包条数。如果在 30 分钟（半小时）内摆上货架的面包平均数量是 12 条，那么五分钟内摆上货架的面包平均数量为 $\left( \frac{5}{30} \right)$(12) = 2 条面包。

The probability question asks you to find *P*(*x* = 3).

该概率问题要求你求 *P*(*x* = 3)。

The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes?

每小时钓到的鱼的平均数量是 8 条。我们关注的是 15 分钟内钓到的鱼的数量。关注的时间区间是 15 分钟。15 分钟内钓到的鱼的平均数量是多少？

Problem 问题

A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let *X* = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.

一家银行平均预期每天收到 6 张空头支票。该银行在任意一天收到少于 5 张空头支票的概率是多少？关注的是该银行一天内收到的支票数量，因此关注的时间区间是 1 天。令 *X* = 该银行一天内收到的空头支票数量。如果该银行预期每天收到 6 张空头支票，那么平均每天 6 张。用算式写出该概率问题。

Solution 解答

*P*(*x* \< 5)

An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically.

一家电子产品商店平均预期每天有 10 件退货。经理想知道该店在任意一天退货少于 8 件的概率。用算式写出该概率问题。

You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast.

你注意到一位新闻记者平均每次播报会说"呃"两次。该新闻记者每次播报说"呃"超过两次的概率是多少？

This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast.

这是一个泊松分布问题，因为你关注的是该新闻记者在一次播报中说"呃"的次数。

Problem 问题

a\. What is the interval of interest?

a. 关注的区间是什么？

b\. What is the average number of times the news reporter says "uh" during one broadcast?

b. 该新闻记者在一次播报中说"呃"的平均次数是多少？

c\. Let *X* = \_\_\_\_\_\_\_\_\_\_\_\_. What values does *X* take on?

c. 令 *X* = \_\_\_\_\_\_\_\_\_\_\_\_。*X* 取哪些值？

d\. The probability question is *P*(\_\_\_\_\_\_).

d. 该概率问题是 *P*(\_\_\_\_\_\_)。

Solution 解答

a\. one broadcast

a. 一次播报

b\. 2

b. 2

c\. Let *X* = the number of times the news reporter says "uh" during one broadcast.

c. 令 *X* = 该新闻记者在一次播报中说"呃"的次数。

*x* = 0, 1, 2, 3, ...

d\. *P*(*x* \> 2)

d. *P*(*x* \> 2)

An emergency room at a particular hospital gets an average of five patients per hour. A doctor wants to know the probability that the ER gets more than five patients per hour. Give the reason why this would be a Poisson distribution.

某家医院急诊室平均每小时接诊 5 名患者。一位医生想知道该急诊室每小时接诊超过 5 名患者的概率。给出理由说明为什么这是一个泊松分布。

Notation for the Poisson: P = Poisson Probability Distribution Function 泊松分布的记号：P = 泊松概率分布函数

*X* ~ *P*(*μ*)

Read this as "*X* is a random variable with a Poisson distribution." The parameter is *μ* (or *λ*); *μ* (or *λ*) = the mean for the interval of interest. The standard deviation of the Poisson distribution with mean *µ* is *Σ*=√*μ*

读作"*X* 是一个服从泊松分布的随机变量"。参数为 *μ*（或 *λ*）；*μ*（或 *λ*）= 关注区间的均值。均值为 *µ* 的泊松分布的标准差为 *Σ*=√*μ*

Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?

莉亚的答录机在上午 8 点到 10 点之间大约接到 6 个电话。莉亚在接下来的 15 分钟内接到超过 1 个电话的概率是多少？

Let *X* = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $\frac{1}{4}$ hour.)

令 *X* = 莉亚在 15 分钟内接到的电话数量。（关注区间为 15 分钟，即 $\frac{1}{4}$ 小时。）

*x* = 0, 1, 2, 3, ...

If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives

如果莉亚平均两小时接到 6 个电话，而两小时中有八个 15 分钟区间，那么莉亚接到

$\left( \frac{1}{8} \right)$(6) = 0.75 calls in 15 minutes, on average. So, *μ* = 0.75 for this problem.

$\left( \frac{1}{8} \right)$(6) = 0.75 个电话，平均而言，即 15 分钟内 0.75 个。因此，本问题中 *μ* = 0.75。

*X* ~ *P*(0.75)

Find *P*(*x* \> 1). *P*(*x* \> 1) = 0.1734 (calculator or computer)

求 *P*(*x* \> 1)。*P*(*x* \> 1) = 0.1734（用计算器或计算机）。

Press 1 – and then press 2^nd DISTR.
Arrow down to poissoncdf. Press ENTER.
Enter (.75,1).
The result is *P*(*x* \> 1) = 0.1734.

按 1 –，然后按 2^nd DISTR。
向下箭头选择 poissoncdf。按 ENTER。
输入 (.75,1)。
结果为 *P*(*x* \> 1) = 0.1734。

The TI calculators use *λ* (lambda) for the mean.

TI 计算器用 *λ*（lambda）表示均值。

The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:

莉亚在接下来 15 分钟内接到超过 1 个电话的概率约为 0.1734：

*P*(*x* \> 1) = 1 − poissoncdf(0.75, 1).

*P*(*x* \> 1) = 1 − poissoncdf(0.75, 1)。

The graph of *X* ~ *P*(0.75) is:

*X* ~ *P*(0.75) 的图像为：

The *y*-axis contains the probability of *x* where *X* = the number of calls in 15 minutes.

纵轴表示 *x* 的概率，其中 *X* = 15 分钟内的电话数量。

A customer service center receives about ten emails every half-hour. What is the probability that the customer service center receives more than four emails in the next six minutes? Use the TI-83+ or TI-84 calculator to find the answer.

一家客服中心每半小时大约收到 10 封电子邮件。该客服中心在接下来 6 分钟内收到超过 4 封电子邮件的概率是多少？用 TI-83+ 或 TI-84 计算器求出答案。

According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let *X* = the number of emails an email user receives per day. The discrete random variable *X* takes on the values *x* = 0, 1, 2 …. The random variable *X* has a Poisson distribution: *X* ~ *P*(147). The mean is 147 emails.

据电子邮件管理公司 Baydin 统计，一名电子邮件用户平均每天收到 147 封邮件。令 *X* = 一名电子邮件用户每天收到的邮件数。离散随机变量 *X* 取值 *x* = 0, 1, 2 …。该随机变量 *X* 服从泊松分布：*X* ~ *P*(147)。均值为 147 封邮件。

Problem 问题

1. What is the probability that an email user receives exactly 160 emails per day?

1. 一名电子邮件用户每天恰好收到 160 封邮件的概率是多少？

2. What is the probability that an email user receives at most 160 emails per day?

2. 一名电子邮件用户每天至多收到 160 封邮件的概率是多少？

3. What is the standard deviation?

3. 标准差是多少？

Solution 解答

1. *P*(*x* = 160) = poissonpdf(147, 160) ≈ 0.0180

2. *P*(*x* ≤ 160) = poissoncdf(147, 160) ≈ 0.8666

3. Standard Deviation = $\sigma = \sqrt{\mu} = \sqrt{147} \approx 12.1244$

3. 标准差 = $\sigma = \sqrt{\mu} = \sqrt{147} \approx 12.1244$

According to a recent poll by the Pew Internet Project, girls between the ages of 14 and 17 send an average of 187 text messages each day. Let *X* = the number of texts that a girl aged 14 to 17 sends per day. The discrete random variable *X* takes on the values *x* = 0, 1, 2 …. The random variable *X* has a Poisson distribution: *X* ~ *P*(187). The mean is 187 text messages.

据皮尤互联网项目（Pew Internet Project）近期的一项调查，14 至 17 岁的女孩平均每天发送 187 条短信。令 *X* = 一名 14 至 17 岁女孩每天发送的短信数。离散随机变量 *X* 取值 *x* = 0, 1, 2 …。该随机变量 *X* 服从泊松分布：*X* ~ *P*(187)。均值为 187 条短信。

1. What is the probability that a teen girl sends exactly 175 texts per day?

1. 一名少女每天恰好发送 175 条短信的概率是多少？

2. What is the probability that a teen girl sends at most 150 texts per day?

2. 一名少女每天至多发送 150 条短信的概率是多少？

3. What is the standard deviation?

3. 标准差是多少？

Text message users receive or send an average of 41.5 text messages per day.

短信用户每天平均收发 41.5 条短信。

Problem 问题

1. How many text messages does a text message user receive or send per hour?

1. 一名短信用户每小时收发多少条短信？

2. What is the probability that a text message user receives or sends two messages per hour?

2. 一名短信用户每小时恰好收发 2 条短信的概率是多少？

3. What is the probability that a text message user receives or sends more than two messages per hour?

3. 一名短信用户每小时收发超过 2 条短信的概率是多少？

Solution 解答

1. Let *X* = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is $\frac{41.5}{24}$ ≈ 1.7292.

1. 令 *X* = 一名用户一小时内收发（发送或接收）的短信数。每小时收到的短信平均数量为 $\frac{41.5}{24}$ ≈ 1.7292。

2. *X* ~ *P*(1.7292), so *P*(*x* = 2) = poissonpdf(1.7292, 2) ≈ 0.2653

2. *X* ~ *P*(1.7292)，因此 *P*(*x* = 2) = poissonpdf(1.7292, 2) ≈ 0.2653

3. *P*(*x* \> 2) = 1 – *P*(*x* ≤ 2) = 1 – poissoncdf(1.7292, 2) ≈ 1 – 0.7495 = 0.2505

Atlanta's Hartsfield-Jackson International Airport is the busiest airport in the world. On average there are 2,500 arrivals and departures each day.

亚特兰大的哈兹菲尔德-杰克逊国际机场是世界上最繁忙的机场。平均每天有 2,500 架次起降。

1. How many airplanes arrive and depart the airport per hour?

1. 该机场每小时有多少架次飞机起降？

2. What is the probability that there are exactly 100 arrivals and departures in one hour?

2. 在一小时内恰好有 100 架次起降的概率是多少？

3. What is the probability that there are at most 100 arrivals and departures in one hour?

3. 在一小时内至多 100 架次起降的概率是多少？

Problem 问题

On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

2013 年 5 月 13 日下午 4:30 起，阿拉斯加未来 48 小时发生低强度地震活动的概率被报道约为 1.02%。利用这一信息对接下来 200 天进行计算，求在接下来 200 天中有 10 天出现低强度地震活动的概率。同时用二项分布和泊松分布计算概率。两者是否接近？

Solution 解答

Let *X* = the number of days with low seismic activity.

令 *X* = 出现低强度地震活动的天数。

Using the binomial distribution:

使用二项分布：

*P*(*x* = 10) = binompdf(200, .0102, 10) ≈ 0.000039

*P*(*x* = 10) = binompdf(200, .0102, 10) ≈ 0.000039

Using the Poisson distribution:

使用泊松分布：

Calculate *μ* = *np* = 200(0.0102) ≈ 2.04
*P*(*x* = 10) = poissonpdf(2.04, 10) ≈ 0.000045

计算 *μ* = *np* = 200(0.0102) ≈ 2.04
*P*(*x* = 10) = poissonpdf(2.04, 10) ≈ 0.000045

We expect the approximation to be good because *n* is large (greater than 20) and *p* is small (less than 0.05). The results are close—both probabilities reported are almost 0.

由于 *n* 较大（大于 20）且 *p* 较小（小于 0.05），我们预计近似效果良好。两个结果很接近——给出的概率都几乎为 0。

On May 13, 2013, starting at 4:30 PM, the probability of moderate seismic activity for the next 48 hours in the Kuril Islands off the coast of Japan was reported at about 1.43%. Use this information for the next 100 days to find the probability that there will be low seismic activity in five of the next 100 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?

2013年5月13日下午4:30起，日本沿海千岛群岛未来48小时内发生中等强度地震活动的概率据报道约为1.43%。利用这一信息，对接下来100天进行推算，求在接下来100天中有5天发生低强度地震活动的概率。分别用二项分布和泊松分布计算概率。二者是否接近？

4.7 Discrete Distribution (Playing Card Experiment) 4.7 离散分布（扑克牌实验）

Discrete Distribution (Playing Card Experiment) 离散分布（扑克牌实验）

Class Time:

上课时间：

Names:

姓名：

Student Learning Outcomes

学生学习目标

The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution.

学生将比较经验数据与理论分布，以判断一个日常实验是否符合离散分布。

Supplies

用品

One full deck of playing cards
One programming calculator

一副完整的扑克牌
一台可编程计算器

ProcedureThe experimental procedure for empirical data is to pick one card from a deck of shuffled cards.

经验数据的实验步骤是从一副洗好的牌中抽出一张牌。

1. The theoretical probability of picking a diamond from a deck is \_\_\_\_\_\_\_\_\_.

1. 从一副牌中抽到方片的理论概率为 \_\_\_\_\_\_\_\_\_。

2. Shuffle a deck of cards.

2. 洗一副牌。

3. Pick one card from it.

3. 从中抽出一张牌。

4. Record whether it was a diamond or not a diamond.

4. 记录抽到的是方片还是非方片。

5. Put the card back and reshuffle.

5. 将牌放回并重新洗牌。

6. Do this a total of ten times.

6. 共进行十次。

7. Record the number of diamonds picked.

7. 记录抽到方片的次数。

8. Let *X* = number of diamonds. Theoretically, *X* ~ *B*(\_\_\_\_\_,\_\_\_\_\_)

8. 令 *X* = 方片数。理论上，*X* ~ *B*(\_\_\_\_\_,\_\_\_\_\_)

Organize the Data

整理数据

1. Record the number of diamonds picked for your class with playing cards in Table 4.16. Then calculate the relative frequency.

1. 将你们班用扑克牌抽到方片的次数记录在表 4.16 中，然后计算相对频率。

| *x* | Frequency | Relative Frequency |

| *x* | 频数 | 相对频率 |

|-----|-----------|----------------------|

| 0 | | \_\_\_\_\_\_\_\_\_\_ |

| 1 | | \_\_\_\_\_\_\_\_\_\_ |

| 2 | | \_\_\_\_\_\_\_\_\_\_ |

| 3 | | \_\_\_\_\_\_\_\_\_\_ |

| 4 | | \_\_\_\_\_\_\_\_\_\_ |

| 5 | | \_\_\_\_\_\_\_\_\_\_ |

| 6 | | \_\_\_\_\_\_\_\_\_\_ |

| 7 | | \_\_\_\_\_\_\_\_\_\_ |

| 8 | | \_\_\_\_\_\_\_\_\_\_ |

| 9 | | \_\_\_\_\_\_\_\_\_\_ |

| 10 | | \_\_\_\_\_\_\_\_\_\_ |

Table 4.16

表 4.16

2. Calculate the following:

2. 计算以下内容：

1. $\overline{x}$ = \_\_\_\_\_\_\_\_

2. *s* = \_\_\_\_\_\_\_\_

3. Construct a histogram of the empirical data.

3. 绘制经验数据的直方图。

Theoretical Distribution

理论分布

1. Build the theoretical PDF chart based on the distribution in the Procedure section.

1. 根据"步骤"一节中的分布构建理论概率分布函数图表。

| *x* | *P*(*x*) |

|-----|----------|

| 0 | |

| 1 | |

| 2 | |

| 3 | |

| 4 | |

| 5 | |

| 6 | |

| 7 | |

| 8 | |

| 9 | |

| 10 | |

2. Calculate the following:

2. 计算以下内容：

1. *μ* = \_\_\_\_\_\_\_\_\_\_\_\_

2. *σ* = \_\_\_\_\_\_\_\_\_\_\_\_

3. Construct a histogram of the theoretical distribution.

3. 绘制理论分布的直方图。

Using the Data

使用数据

*RF* = relative frequency

*RF* = 相对频率

Use the table from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.

使用"理论分布"一节中的表格计算下列答案。将答案四舍五入保留四位小数。

*P*(*x* = 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*P*(1 \< *x* \< 4) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*P*(*x* ≥ 8) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

*P*(*x* = 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*P*(1 \< *x* \< 4) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*P*(*x* ≥ 8) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.

使用"整理数据"一节中的数据计算下列答案。将答案四舍五入保留四位小数。

*RF*(*x* = 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*RF*(1 \< *x* \< 4) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*RF*(*x* ≥ 8) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

*RF*(*x* = 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*RF*(1 \< *x* \< 4) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
*RF*(*x* ≥ 8) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Discussion QuestionsFor questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations.

讨论问题：针对问题 1 和 2，思考两张图的形状、概率、相对频率、均值与标准差。

1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical, empirical, and simulation distributions. Use complete sentences.

1. 已知数据会变化，请用完整的句子描述理论分布、经验分布与模拟分布在图形和分布上的三点相似之处。

2. Describe the three most significant differences between the graphs or distributions of the theoretical, empirical, and simulation distributions.

2. 描述理论分布、经验分布与模拟分布在图形或分布上最显著的三个不同之处。

3. Using your answers from questions 1 and 2, does it appear that the two sets of data fit the theoretical distribution? In complete sentences, explain why or why not.

3. 结合你在问题 1 和 2 中的回答，这两组数据是否看起来符合理论分布？请用完整的句子说明原因。

4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.16 or [link] to change, and how would it change? Why? Why wouldn’t the other table(s) change?

4. 假设该实验重复了 500 次。你认为表 4.16 还是 [link] 会发生变化，会如何变化？为什么？为什么另一张（或几张）表不会变化？

4.8 Discrete Distribution (Lucky Dice Experiment) 4.8 离散分布（幸运骰子实验）

Discrete Distribution (Lucky Dice Experiment) 离散分布（幸运骰子实验）

Class Time:

上课时间：

Names:

姓名：

Student Learning Outcomes

学生学习目标

The student will compare empirical data and a theoretical distribution to determine if a Tet gambling game fits a discrete distribution.
The student will demonstrate an understanding of long-term probabilities.

学生将比较经验数据与理论分布，以判断一种 Tet 赌博游戏是否符合离散分布。
学生将展示对长期概率的理解。

Supplies

用品

one “Lucky Dice” game or three regular dice

一套“幸运骰子”游戏或三颗普通骰子

Procedure

步骤

Round answers to relative frequency and probability problems to four decimal places.

将相对频率与概率问题的答案四舍五入保留四位小数。

1. The experimental procedure is to bet on one object. Then, roll three Lucky Dice and count the number of matches. The number of matches will decide your profit.

1. 实验步骤是押一个物体。然后掷三颗幸运骰子，统计匹配的数量。匹配的数量将决定你的收益。

2. What is the theoretical probability of one die matching the object?

2. 一颗骰子与所押物体匹配的理论概率是多少？

3. Choose one object to place a bet on. Roll the three Lucky Dice. Count the number of matches.

3. 选择一个物体下注。掷出三颗幸运骰子。统计匹配的数量。

4. Let *X* = number of matches. Theoretically, *X* ~ *B*(\_\_\_\_\_\_,\_\_\_\_\_\_)

4. 令 *X* = 匹配数。理论上，*X* ~ *B*(\_\_\_\_\_\_,\_\_\_\_\_\_)

5. Let *Y* = profit per game.

5. 令 *Y* = 每局收益。

Organize the DataIn Table 4.17, fill in the *y* value that corresponds to each *x* value. Next, record the number of matches picked for your class. Then, calculate the relative frequency.

整理数据：在表 4.17 中，填入与每个 *x* 值对应的 *y* 值。然后，记录你们班抽到的匹配数量。接着计算相对频率。

1. Complete the table.

1. 完成表格。

| x | y | Frequency | Relative Frequency |

| x | y | 频数 | 相对频率 |

|-----|-----|-----------|--------------------|

| 0 | | | |

| 1 | | | |

| 2 | | | |

| 3 | | | |

Table 4.17

表 4.17

2. Calculate the following:

2. 计算以下内容：

1. $\overline{x}$ = \_\_\_\_\_\_\_

2. *s_x* = \_\_\_\_\_\_\_\_

3. $\overline{y}$ = \_\_\_\_\_\_\_

4. *s_y* = \_\_\_\_\_\_\_

3. Explain what $\overline{x}$ represents.

3. 解释 $\overline{x}$ 的含义。

4. Explain what $\overline{y}$ represents.

4. 解释 $\overline{y}$ 的含义。

5. Based upon the experiment:

5. 根据实验结果：

1. What was the average profit per game?

1. 每局的平均收益是多少？

2. Did this represent an average win or loss per game?

2. 这代表每局平均盈利还是亏损？

3. How do you know? Answer in complete sentences.

3. 你如何判断？请用完整的句子作答。

6. Construct a histogram of the empirical data.

6. 绘制经验数据的直方图。

Theoretical DistributionBuild the theoretical PDF chart for *x* and *y* based on the distribution from the Procedure section.

理论分布：根据"步骤"一节中的分布，为 *x* 和 *y* 构建理论概率分布函数图表。

1. | *x* | *y* | *P*(*x*) = *P*(*y*) |

|-----|-----|---------------------|

| 0 | | |

| 1 | | |

| 2 | | |

| 3 | | |

Table 4.18

表 4.18

2. Calculate the following:

2. 计算以下内容：

1. *μ_x* = \_\_\_\_\_\_\_

2. *σ_x* = \_\_\_\_\_\_\_

3. *μ_x* = \_\_\_\_\_\_\_

3. Explain what *μ_x* represents.

3. 解释 *μ_x* 的含义。

4. Explain what *μ_y* represents.

4. 解释 *μ_y* 的含义。

5. Based upon theory:

5. 根据理论：

1. What was the expected profit per game?

1. 每局的期望收益是多少？

2. Did the expected profit represent an average win or loss per game?

2. 期望收益代表每局平均盈利还是亏损？

3. How do you know? Answer in complete sentences.

3. 你如何判断？请用完整的句子作答。

6. Construct a histogram of the theoretical distribution.

6. 绘制理论分布的直方图。

Use the Data

使用数据

*RF* = relative frequency

*RF* = 相对频率

Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.

使用"理论分布"一节中的数据计算下列答案。将答案四舍五入保留四位小数。

1. *P*(*x* = 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

2. *P*(0 \< *x* \< 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

3. *P*(*x* ≥ 2) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.

使用"整理数据"一节中的数据计算下列答案。将答案四舍五入保留四位小数。

1. *RF*(x = 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

2. *RF*(0 \< *x* \< 3) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

3. *RF*(*x* ≥ 2) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Discussion QuestionFor questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations.

讨论问题：针对问题 1 和 2，考虑图形、概率、相对频率、均值与标准差。

1. Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences.

1. 已知数据会变化，请用完整的句子描述理论分布与经验分布在图形和分布上的三点相似之处。

2. Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions.

2. 描述理论分布与经验分布在图形或分布上最显著的三个不同之处。

3. Thinking about your answers to questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not.

3. 结合你对问题 1 和 2 的回答，这些数据是否看起来符合理论分布？请用完整的句子说明原因。

4. Suppose that the experiment had been repeated 500 times. Would you expect Table 4.17 or Table 4.18 to change, and how would it change? Why? Why wouldn’t the other table change?

4. 假设该实验重复了 500 次。你认为表 4.17 还是表 4.18 会发生变化，会如何变化？为什么？为什么另一张表不会变化？

Key Terms 关键术语

Bernoulli Trials

伯努利试验

an experiment with the following characteristics:

具有以下特征的试验：

1. There are only two possible outcomes called "success" and "failure" for each trial.

1. 每次试验只有两种可能结果，称为"成功"和"失败"。

2. The probability *p* of a success is the same for any trial (so the probability *q* = 1 − *p* of a failure is the same for any trial).

2. 成功的概率 *p* 对任一次试验都相同（因此失败的概率 *q* = 1 − *p* 对任一次试验也都相同）。

Binomial Experiment

二项试验

a statistical experiment that satisfies the following three conditions:

满足以下三个条件的统计试验：

1. There are a fixed number of trials, *n*.

1. 试验次数固定，为 *n*。

2. There are only two possible outcomes, called "success" and, "failure," for each trial. The letter *p* denotes the probability of a success on one trial, and *q* denotes the probability of a failure on one trial.

2. 每次试验只有两种可能结果，称为"成功"和"失败"。字母 *p* 表示一次试验成功的概率，*q* 表示一次试验失败的概率。

3. The *n* trials are independent and are repeated using identical conditions.

3. 这 *n* 次试验相互独立，并在相同条件下重复进行。

Binomial Probability Distribution

二项概率分布

a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, *n*, of independent trials. "Independent" means that the result of any trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV *X* is defined as the number of successes in *n* trials. The notation is: *X* ~ *B*(*n*, *p*). The mean is *μ* = *np* and the standard deviation is *σ* = $\sqrt{npq}$. The probability of exactly *x* successes in *n* trials is

由伯努利试验产生的离散随机变量（RV）；共有固定 *n* 次相互独立的试验。"独立"是指任一次试验（例如第一次试验）的结果不影响后续试验的结果，且所有试验在相同条件下进行。在此情形下，二项随机变量 *X* 定义为 *n* 次试验中的成功次数。记号为：*X* ~ *B*(*n*, *p*)。均值为 *μ* = *np*，标准差为 *σ* = $\sqrt{npq}$。*n* 次试验中恰好 *x* 次成功的概率为

*P*(*X* = *x*) = $\left( \begin{array}{l} n \\ x \end{array} \right)$*p*^x*q*^{n − x}.

Expected Value

期望值

expected arithmetic average when an experiment is repeated many times; also called the mean. Notations: *μ*. For a discrete random variable (RV) with probability distribution function *P*(*x*),the definition can also be written in the form *μ* = $\sum{}$*xP*(*x*).

当试验重复多次时的期望算术平均；也称均值。记号：*μ*。对于具有概率分布函数 *P*(*x*) 的离散随机变量（RV），该定义也可写作 *μ* = $\sum{}$*xP*(*x*)。

Geometric Distribution

几何分布

a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated until the first success. The geometric variable *X* is defined as the number of trials until the first success. Notation: *X* ~ *G*(*p*). The mean is *μ* = $\frac{1}{p}$ and the standard deviation is *σ* = $\sqrt{\frac{1}{p}\left( {\frac{1}{p} - 1} \right)}$. The probability of exactly *x* failures before the first success is given by the formula: *P*(*X* = *x*) = *p*(1 – *p*)^{*x* – 1}.

由伯努利试验产生的离散随机变量（RV）；试验重复进行直到首次成功。几何变量 *X* 定义为直到首次成功所需的试验次数。记号：*X* ~ *G*(*p*)。均值为 *μ* = $\frac{1}{p}$，标准差为 *σ* = $\sqrt{\frac{1}{p}\left( {\frac{1}{p} - 1} \right)}$。首次成功之前恰好 *x* 次失败的概率由公式给出：*P*(*X* = *x*) = *p*(1 – *p*)^{*x* – 1}。

Geometric Experiment

几何试验

a statistical experiment with the following properties:

具有以下性质的统计试验：

1. There are one or more Bernoulli trials with all failures except the last one, which is a success.

1. 有一个或多个伯努利试验，除最后一次为成功外，其余均为失败。

2. In theory, the number of trials could go on forever. There must be at least one trial.

2. 从理论上讲，试验次数可以无限延续。但至少必须有一个试验。

3. The probability, *p*, of a success and the probability, *q*, of a failure do not change from trial to trial.

3. 成功的概率 *p* 与失败的概率 *q* 在各次试验之间保持不变。

Hypergeometric Experiment

超几何试验

a statistical experiment with the following properties:

具有以下性质的统计试验：

1. You take samples from two groups.

1. 从两个组中抽取样本。

2. You are concerned with a group of interest, called the first group.

2. 关注的是一个感兴趣的组，称为第一组。

3. You sample without replacement from the combined groups.

3. 从合并后的两组中无放回抽样。

4. Each pick is not independent, since sampling is without replacement.

4. 由于是无放回抽样，每次抽取并不独立。

5. You are not dealing with Bernoulli Trials.

5. 处理的不是伯努利试验。

Hypergeometric Probability

超几何概率

a discrete random variable (RV) that is characterized by:

具有以下特征的离散随机变量（RV）：

1. A fixed number of trials.

1. 试验次数固定。

2. The probability of success is not the same from trial to trial.

2. 成功的概率在各次试验之间并不相同。

We sample from two groups of items when we are interested in only one group. *X* is defined as the number of successes out of the total number of items chosen. Notation: *X* ~ *H*(*r*, *b*, *n*), where *r* = the number of items in the group of interest, *b* = the number of items in the group not of interest, and *n* = the number of items chosen.

当我们只对其中一组感兴趣时，从两组物品中抽样。*X* 定义为在所选项目总数中的成功次数。记号：*X* ~ *H*(*r*, *b*, *n*)，其中 *r* = 感兴趣组中的物品数，*b* = 不感兴趣组中的物品数，*n* = 被选中的物品数。

Mean

均值

a number that measures the central tendency; a common name for mean is 'average.' The term 'mean' is a shortened form of 'arithmetic mean.' By definition, the mean for a sample (detonated by $\overline{x}$) is $\overline{x} = \frac{{Sum}~{of}~{all}~{values}~{in}~{the}~{sample}}{{Number}~{of}~{values}~{in}~{the}~{sample}}$ and the mean for a population (denoted by *μ*) is *μ* = $\frac{{Sum}~{of}~{all}~{values}~{in}~{the}~{population}}{{Number}~{of}~{values}~{in}~{the}~{population}}$.

衡量集中趋势的一个数值；均值通常称为"平均"。"mean"是"arithmetic mean（算术平均）"的缩写。按定义，样本均值（由 $\overline{x}$ 表示）为 $\overline{x} = \frac{{Sum}~{of}~{all}~{values}~{in}~{the}~{sample}}{{Number}~{of}~{values}~{in}~{the}~{sample}}$，总体均值（由 *μ* 表示）为 *μ* = $\frac{{Sum}~{of}~{all}~{values}~{in}~{the}~{population}}{{Number}~{of}~{values}~{in}~{the}~{population}}$。

Mean of a Probability Distribution

概率分布的均值

the long-term average of many trials of a statistical experiment

多次重复统计试验的长期平均

Poisson Probability Distribution

泊松概率分布

a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable:

统计某一特定区间内某事件发生的次数的离散随机变量（RV）；该变量的特征如下：

The probability that the event occurs in a given interval is the same for all intervals.
The events occur with a known mean and independently of the time since the last event.

事件在给定区间内发生的概率对所有区间都相同。
事件以已知均值发生，且与距上一事件的时间相互独立。

The distribution is defined by the mean *μ* of the event in the interval. Notation: *X* ~ *P*(*μ*). The mean is *μ* = *np*. The standard deviation is $\sigma\text{~=~}\sqrt{\mu}$. The probability of having exactly *x* successes in *r* trials is *P*(*X* = x ) = $(e^{- \mu})\frac{\mu^{x}}{x!}$. The Poisson distribution is often used to approximate the binomial distribution, when *n* is "large" and *p* is "small" (a general rule is that *n* should be greater than or equal to 20 and *p* should be less than or equal to 0.05).

该分布由区间内事件的均值 *μ* 定义。记号：*X* ~ *P*(*μ*)。均值为 *μ* = *np*。标准差为 $\sigma\text{~=~}\sqrt{\mu}$。*r* 次试验中恰好 *x* 次成功的概率为 *P*(*X* = x ) = $(e^{- \mu})\frac{\mu^{x}}{x!}$。当 *n* "较大"且 *p* "较小"时，泊松分布常用来近似二项分布（一般规则是 *n* 应大于或等于 20，且 *p* 应小于或等于 0.05）。

Probability Distribution Function (PDF)

概率分布函数（PDF）

a mathematical description of a discrete random variable (*RV*), given either in the form of an equation (formula) or in the form of a table listing all the possible outcomes of an experiment and the probability associated with each outcome.

对离散随机变量（*RV*）的数学描述，可给出为方程（公式）形式，或列出试验所有可能结果及每个结果对应概率的表格形式。

Random Variable (RV)

随机变量（RV）

a characteristic of interest in a population being studied; common notation for variables are upper case Latin letters *X*, *Y*, *Z*,...; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters *x, y,* and *z*. For example, if *X* is the number of children in a family, then *x* represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in the two following ways.

所研究总体中感兴趣的特征；变量的常用记号为大写拉丁字母 *X*、*Y*、*Z*，…；来自定义域（变量所有可能取值的集合）的特定值的常用记号为小写拉丁字母 *x*、*y*、*z*。例如，若 *X* 为家庭中的孩子数，则 *x* 表示某个特定的整数 0、1、2、3，…。统计学中的变量与中级代数中的变量在以下两方面不同。

The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if *X* = hair color then the domain is {black, blond, gray, green, orange}.
We can tell what specific value *x* the random variable *X* takes only after performing the experiment.

随机变量（RV）的定义域不一定是一个数值集合；定义域也可以用文字表示；例如，若 *X* = 发色，则定义域为 {黑、金、灰、绿、橙}。
只有在进行了试验之后，我们才能知道随机变量 *X* 取哪个特定值 *x*。

Standard Deviation of a Probability Distribution

概率分布的标准差

a number that measures how far the outcomes of a statistical experiment are from the mean of the distribution $\sigma = \sqrt{\sum{\left\lbrack \left( {x~–~\mu} \right)^{2}~ \bullet ~Ρ(x) \right\rbrack~}}$

衡量统计试验的结果偏离分布均值的程度的数值 $\sigma = \sqrt{\sum{\left\lbrack \left( {x~–~\mu} \right)^{2}~ \bullet ~Ρ(x) \right\rbrack~}}$

The Law of Large Numbers

大数定律

As the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero.

随着概率试验中试验次数的增加，事件的理论概率与相对频率概率之间的差异趋近于零。

Chapter Review 本章回顾

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量的概率分布函数（PDF）

The characteristics of a probability distribution function (PDF) for a discrete random variable are as follows:

离散随机变量的概率分布函数（PDF）的特征如下：

1. Each probability is between zero and one, inclusive (*inclusive* means to include zero and one).

1. 每个概率都在 0 与 1 之间（含端点）（*inclusive* 意为包含 0 和 1）。

2. The sum of the probabilities is one.

2. 各概率之和等于 1。

4.2 Mean or Expected Value and Standard Deviation 4.2 均值（期望值）与标准差

The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes.

离散随机变量的期望值（即均值）预测了重复多次的统计试验的长期结果。概率分布的标准差用于衡量可能结果的变异性。

4.3 Binomial Distribution 4.3 二项分布

A statistical experiment can be classified as a binomial experiment if the following conditions are met:

若满足以下条件，一个统计试验可归类为二项试验：

1. There are a fixed number of trials, *n*.

1. 试验次数固定，为 *n*。

2. There are only two possible outcomes, called "success" and, "failure" for each trial. The letter *p* denotes the probability of a success on one trial and *q* denotes the probability of a failure on one trial.

2. 每次试验只有两种可能结果，称为"成功"和"失败"。字母 *p* 表示一次试验成功的概率，*q* 表示一次试验失败的概率。

3. The *n* trials are independent and are repeated using identical conditions.

3. 这 *n* 次试验相互独立，并以相同条件重复进行。

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable *X* = the number of successes obtained in the *n* independent trials. The mean of *X* can be calculated using the formula *μ* = *np*, and the standard deviation is given by the formula σ = $\ \sqrt{npq}$.

二项试验的结果符合二项概率分布。随机变量 *X* = 在 *n* 次独立试验中获得的成功次数。*X* 的均值可用公式 *μ* = *np* 计算，标准差由公式 σ = $\ \sqrt{npq}$ 给出。

4.4 Geometric Distribution 4.4 几何分布

There are three characteristics of a geometric experiment:

几何试验有三个特征：

1. There are one or more Bernoulli trials with all failures except the last one, which is a success.

1. 有一个或多个伯努利试验，除最后一次为成功外，其余均为失败。

2. In theory, the number of trials could go on forever. There must be at least one trial.

2. 从理论上讲，试验次数可以无限延续。但至少必须有一个试验。

3. The probability, *p*, of a success and the probability, *q*, of a failure are the same for each trial.

3. 成功的概率 *p* 与失败的概率 *q* 对各次试验都相同。

In a geometric experiment, define the discrete random variable *X* as the number of independent trials until the first success. We say that X has a geometric distribution and write *X* ~ *G*(*p*) where *p* is the probability of success in a single trial.

在几何试验中，将离散随机变量 *X* 定义为直到首次成功所需的独立试验次数。我们说 X 服从几何分布，并记作 *X* ~ *G*(*p*)，其中 *p* 为单次试验成功的概率。

The mean of the geometric distribution *X* ~ *G*(*p*) is *μ* = $\frac{1}{p}$ and the standard deviation is $\sigma\sqrt{\frac{\left( \text{1} - p \right)}{p^{2}}}$ = $\sqrt{\frac{1}{p}\left( {\frac{1}{p} - 1} \right)}$.

几何分布 *X* ~ *G*(*p*) 的均值为 *μ* = $\frac{1}{p}$，标准差为 $\sigma\sqrt{\frac{\left( \text{1} - p \right)}{p^{2}}}$ = $\sqrt{\frac{1}{p}\left( {\frac{1}{p} - 1} \right)}$。

4.5 Hypergeometric Distribution 4.5 超几何分布

A hypergeometric experiment is a statistical experiment with the following properties:

超几何试验是一种具有以下性质的统计试验：

1. You take samples from two groups.

1. 你从两个组中抽取样本。

2. You are concerned with a group of interest, called the first group.

2. 你关注的是一个感兴趣组，称为第一组。

3. You sample without replacement from the combined groups.

3. 你从合并的组中无放回地抽样。

4. Each pick is not independent, since sampling is without replacement.

4. 由于抽样是无放回的，每次抽取不是独立的。

5. You are not dealing with Bernoulli Trials.

5. 你处理的不是伯努利试验。

The outcomes of a hypergeometric experiment fit a hypergeometric probability distribution. The random variable *X* = the number of items from the group of interest. The distribution of *X* is denoted *X* ~ *H*(*r*, *b*, *n*), where *r* = the size of the group of interest (first group), *b* = the size of the second group, and *n* = the size of the chosen sample. It follows that

超几何试验的结果服从超几何概率分布。随机变量 *X* = 来自感兴趣组的物品数量。*X* 的分布记为 *X* ~ *H*(*r*, *b*, *n*)，其中 *r* = 感兴趣组（第一组）的规模，*b* = 第二组的规模，*n* = 所选样本的规模。于是

*n* ≤ *r* + *b*. The mean of *X* is *μ* = $\frac{nr}{r\text{~+~}b}$ and the standard deviation is *σ* = $\sqrt{\frac{rbn(r\text{~+~}b\text{~−~}n)}{{(r\text{~+~}b)}^{2}\,(r\text{~+~}b - \text{1})}}$.

*n* ≤ *r* + *b*。*X* 的均值为 *μ* = $\frac{nr}{r\text{~+~}b}$，标准差为 *σ* = $\sqrt{\frac{rbn(r\text{~+~}b\text{~−~}n)}{{(r\text{~+~}b)}^{2}\,(r\text{~+~}b - \text{1})}}$。

4.6 Poisson Distribution 4.6 泊松分布

A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is "small" (less than or equal to 0.05) and the number of trials is "large" (greater than or equal to 20).

离散随机变量的泊松概率分布给出了在固定的时间或空间区间内发生若干事件的概率，前提是这些事件以已知的平均速率发生，且与上次事件以来的时间无关。当成功概率"很小"（小于或等于 0.05）且试验次数"很大"（大于或等于 20）时，可用泊松分布近似二项分布。

Formula Review 公式回顾

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

Mean or Expected Value: $\mu = \underset{x \in X}{\sum^{}}xP(x)$

均值或期望值：$\mu = \underset{x \in X}{\sum^{}}xP(x)$

Standard Deviation: $\sigma = \sqrt{\underset{x \in X}{\sum^{}}{(x - \mu)}^{2}P(x)}$

标准差：$\sigma = \sqrt{\underset{x \in X}{\sum^{}}{(x - \mu)}^{2}P(x)}$

4.3 Binomial Distribution 4.3 二项分布

*X* ~ *B*(*n*, *p*) means that the discrete random variable *X* has a binomial probability distribution with *n* trials and probability of success *p*.

*X* ~ *B*(*n*, *p*) 表示离散随机变量 *X* 服从二项分布，其中 *n* 为试验次数，*p* 为成功概率。

*X* = the number of successes in *n* independent trials

*X* = *n* 次独立试验中的成功次数

*n* = the number of independent trials

*n* = 独立试验的次数

*X* takes on the values *x* = 0, 1, 2, 3, ..., *n*

*X* 取值 *x* = 0, 1, 2, 3, ..., *n*

*p* = the probability of a success for any trial

*p* = 任意一次试验成功的概率

*q* = the probability of a failure for any trial

*q* = 任意一次试验失败的概率

*p* + *q* = 1

*q* = 1 – *p*

The mean of *X* is *μ* = *np*. The standard deviation of *X* is *σ* = $\sqrt{npq}$.

*X* 的均值为 *μ* = *np*。*X* 的标准差为 *σ* = $\sqrt{npq}$。

4.4 Geometric Distribution 4.4 几何分布

*X* ~ G(*p*) means that the discrete random variable *X* has a geometric probability distribution with probability of success in a single trial *p*.

*X* ~ G(*p*) 表示离散随机变量 *X* 服从几何分布，其中单次试验成功概率为 *p*。

*X* = the number of independent trials until the first success

*X* = 直到首次成功为止的独立试验次数

*X* takes on the values *x* = 1, 2, 3, ...

*X* 取值 *x* = 1, 2, 3, ...

*p* = the probability of a success for any trial

*p* = 任意一次试验成功的概率

*q* = the probability of a failure for any trial *p* + *q* = 1

*q* = 任意一次试验失败的概率 *p* + *q* = 1

*q* = 1 – *p*

The mean is *μ* = $\frac{1}{p}$.

均值为 *μ* = $\frac{1}{p}$。

The standard deviation is *σ* = $\sqrt{\frac{1\ –\ p}{p^{2}}}$ = $\sqrt{\frac{1}{p}\left( {\frac{1}{p} - 1} \right)}$ .

标准差为 *σ* = $\sqrt{\frac{1\ –\ p}{p^{2}}}$ = $\sqrt{\frac{1}{p}\left( {\frac{1}{p} - 1} \right)}$ 。

4.5 Hypergeometric Distribution 4.5 超几何分布

*X* ~ *H*(*r*, *b*, *n*) means that the discrete random variable *X* has a hypergeometric probability distribution with *r* = the size of the group of interest (first group), *b* = the size of the second group, and *n* = the size of the chosen sample.

*X* ~ *H*(*r*, *b*, *n*) 表示离散随机变量 *X* 服从超几何分布，其中 *r* = 感兴趣组（第一组）的规模，*b* = 第二组的规模，*n* = 所选样本的规模。

*X* = the number of items from the group of interest that are in the chosen sample, and *X* may take on the values *x* = 0, 1, ..., up to the size of the group of interest. (The minimum value for *X* may be larger than zero in some instances.)

*X* = 所选样本中来自由感兴趣组的物品数量，*X* 可取值 *x* = 0, 1, ..., 直至感兴趣组的规模。（在某些情况下，*X* 的最小值可能大于零。）

*n* ≤ *r* + *b*

The mean of *X* is given by the formula *μ* = $\frac{nr}{r\text{~+~}b}$ and the standard deviation is = $\sqrt{\frac{rbn(r\text{~+~}b - n)}{{(r\text{~+~}b)}^{2}(r\text{~+~}b - \text{1})}}$.

*X* 的均值由公式 *μ* = $\frac{nr}{r\text{~+~}b}$ 给出，标准差为 = $\sqrt{\frac{rbn(r\text{~+~}b - n)}{{(r\text{~+~}b)}^{2}(r\text{~+~}b - \text{1})}}$。

4.6 Poisson Distribution 4.6 泊松分布

*X* ~ *P*(*μ*) means that *X* has a Poisson probability distribution where *X* = the number of occurrences in the interval of interest.

*X* ~ *P*(*μ*) 表示 *X* 服从泊松分布，其中 *X* = 感兴趣区间内的发生次数。

*X* takes on the values *x* = 0, 1, 2, 3, ...

*X* 取值 *x* = 0, 1, 2, 3, ...

The mean *μ* is typically given.

均值 *μ* 通常已知。

The variance is *σ*² = *μ*, and the standard deviation is

方差为 *σ*² = *μ*，标准差为

$\sigma\text{~=~}\sqrt{\mu}$.

$\sigma\text{~=~}\sqrt{\mu}$。

When *P*(*μ*) is used to approximate a binomial distribution, *μ* = *np* where *n* represents the number of independent trials and *p* represents the probability of success in a single trial.

当用 *P*(*μ*) 近似二项分布时，*μ* = *np*，其中 *n* 表示独立试验次数，*p* 表示单次试验成功的概率。

Practice 练习

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量概率分布函数（PDF）

*Use the following information to answer the next five exercises:* A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution.

用以下信息回答接下来的五个习题：一家公司想评估其人员流失率，也就是说，新员工会在公司留任多久。多年来，他们已经建立了如下概率分布。

Let *X* = the number of years a new hire will stay with the company.

令 *X* = 新员工将在公司留任的年数。

Let *P*(*x*) = the probability that a new hire will stay with the company *x* years.

令 *P*(*x*) = 新员工在公司留任 *x* 年的概率。

Complete Table 4.19 using the data provided.

利用所给数据完成表 4.19。

| *x* | *P*(*x*) |

|-----|----------|

| 0 | 0.12 |

| 1 | 0.18 |

| 2 | 0.30 |

| 3 | 0.15 |

| 4 | |

| 5 | 0.10 |

| 6 | 0.05 |

Table 4.19 2.

表 4.19 2.

*P*(*x* = 4) = \_\_\_\_\_\_\_

*P*(*x* ≥ 5) = \_\_\_\_\_\_\_

4\.

On average, how long would you expect a new hire to stay with the company?

平均而言，你预期一名新员工会在公司留任多久？

What does the column "*P*(*x*)" sum to?

列"*P*(*x*)"的和是多少？

*Use the following information to answer the next six exercises:* A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution.

用以下信息回答接下来的六个习题：一位面包师在决定要在店里烤多少盘松饼来卖。他想烤得刚好够卖，一盘不多。通过长期观察，面包师已经建立了一个概率分布。

| *x* | *P*(*x*) |

|-----|----------|

| 1 | 0.15 |

| 2 | 0.35 |

| 3 | 0.40 |

| 4 | 0.10 |

Table 4.20 6.

表 4.20 6.

Define the random variable *X*.

定义随机变量 *X*。

What is the probability the baker will sell more than one batch? *P*(*x* \> 1) = \_\_\_\_\_\_\_

面包师卖出多于一批的概率是多少？*P*(*x* \> 1) = \_\_\_\_\_\_\_

8\.

What is the probability the baker will sell exactly one batch? *P*(*x* = 1) = \_\_\_\_\_\_\_

面包师恰好卖出一批的概率是多少？*P*(*x* = 1) = \_\_\_\_\_\_\_

On average, how many batches should the baker make?

平均而言，面包师应该烤多少批？

*Use the following information to answer the next four exercises:* Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random.

用以下信息回答接下来的四个习题：艾伦每周练习音乐三天。她三天全都练习的情况占 85% 的时间，练习两天的情况占 8%，练习一天的情况占 4%，不练习的情况占 3%。随机选取一周。

10\.

Define the random variable *X*.

定义随机变量 *X*。

11.

Construct a probability distribution table for the data.

为这些数据构造一个概率分布表。

12\.

We know that for a probability distribution function to be discrete, it must have two characteristics. One is that the sum of the probabilities is one. What is the other characteristic?

我们知道，要使一个概率分布函数是离散的，它必须有两个特征。其一是概率之和为 1。另一个特征是什么？

*Use the following information to answer the next five exercises:* Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five events 35% of the time, four events 25% of the time, three events 20% of the time, two events 10% of the time, one event 5% of the time, and no events 5% of the time.

用以下信息回答接下来的五个习题：哈维尔每个月都参加社区活动。他每个月参加的活动不超过五场。他恰好参加五场的情况占 35% 的时间，四场占 25%，三场占 20%，两场占 10%，一场占 5%，不参加占 5%。

13.

Define the random variable *X*.

定义随机变量 *X*。

14\.

What values does *x* take on?

*x* 取哪些值？

15.

Construct a PDF table.

构造一个概率分布函数（PDF）表。

16\.

Find the probability that Javier volunteers for less than three events each month. *P*(*x* \< 3) = \_\_\_\_\_\_\_

求哈维尔每月参加少于三场活动的概率。*P*(*x* \< 3) = \_\_\_\_\_\_\_

17.

Find the probability that Javier volunteers for at least one event each month. *P*(*x* \> 0) = \_\_\_\_\_\_\_

求哈维尔每月至少参加一场活动的概率。*P*(*x* \> 0) = \_\_\_\_\_\_\_

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

18\.

Complete the expected value table.

完成期望值表。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|---------------|

| 0 | 0.2 | |

| 1 | 0.2 | |

| 2 | 0.4 | |

| 3 | 0.2 | |

Table 4.21 19.

表 4.21 19.

Find the expected value from the expected value table.

由期望值表求期望值。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|---------------|

| 2 | 0.1 | 2(0.1) = 0.2 |

| 4 | 0.3 | 4(0.3) = 1.2 |

| 6 | 0.4 | 6(0.4) = 2.4 |

| 8 | 0.2 | 8(0.2) = 1.6 |

Table 4.22 20.

表 4.22 20.

Find the standard deviation.

求标准差。

| *x* | *P*(*x*) | *x*\**P*(*x*) | (*x* – *μ*)²*P*(*x*) |

|-----|----------|---------------|----------------------------------|

| 2 | 0.1 | 2(0.1) = 0.2 | (2–5.4)²(0.1) = 1.156 |

| 4 | 0.3 | 4(0.3) = 1.2 | (4–5.4)²(0.3) = 0.588 |

| 6 | 0.4 | 6(0.4) = 2.4 | (6–5.4)²(0.4) = 0.144 |

| 8 | 0.2 | 8(0.2) = 1.6 | (8–5.4)²(0.2) = 1.352 |

Table 4.23 21.

表 4.23 21.

Identify the mistake in the probability distribution table.

找出概率分布表中的错误。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|---------------|

| 1 | 0.15 | 0.15 |

| 2 | 0.25 | 0.50 |

| 3 | 0.30 | 0.90 |

| 4 | 0.20 | 0.80 |

| 5 | 0.15 | 0.75 |

Table 4.24 22.

表 4.24 22.

Identify the mistake in the probability distribution table.

找出概率分布表中的错误。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|---------------|

| 1 | 0.15 | 0.15 |

| 2 | 0.25 | 0.40 |

| 3 | 0.25 | 0.65 |

| 4 | 0.20 | 0.85 |

| 5 | 0.15 | 1 |

Table 4.25

表 4.25

*Use the following information to answer the next five exercises:* A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He has the following probability distribution.

用以下信息回答接下来的五个习题：一位物理学教授想知道，有多大比例的物理专业学生将在接下来的几年里从事研究生阶段的研究。他有以下概率分布。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|---------------|

| 1 | 0.35 | |

| 2 | 0.20 | |

| 3 | 0.15 | |

| 4 | | |

| 5 | 0.10 | |

| 6 | 0.05 | |

Table 4.26 23.

表 4.26 23.

Define the random variable *X*.

定义随机变量 *X*。

24\.

Define *P*(*x*), or the probability of *x*.

定义 *P*(*x*)，即 *x* 的概率。

25.

Find the probability that a physics major will do post-graduate research for four years. *P*(*x* = 4) = \_\_\_\_\_\_\_

求一名物理专业学生将从事四年研究生研究的概率。*P*(*x* = 4) = \_\_\_\_\_\_\_

26\.

FInd the probability that a physics major will do post-graduate research for at most three years. *P*(*x* ≤ 3) = \_\_\_\_\_\_\_

求一名物理专业学生从事不超过三年研究生研究的概率。*P*(*x* ≤ 3) = \_\_\_\_\_\_\_

27.

On average, how many years would you expect a physics major to spend doing post-graduate research?

平均而言，你预期一名物理专业学生从事研究生研究多少年？

*Use the following information to answer the next seven exercises:* A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution.

用以下信息回答接下来的七个习题：一位芭蕾舞老师想知道每年班里有多大比例的学生会继续升入下一年，以便她规划开设哪些课程。多年来，她已经建立了如下概率分布。

Let *X* = the number of years a student will study ballet with the teacher.
Let *P*(*x*) = the probability that a student will study ballet *x* years.

令 *X* = 一名学生将与该老师学习芭蕾的年数。
令 *P*(*x*) = 一名学生学习芭蕾 *x* 年的概率。

28\.

Complete Table 4.27 using the data provided.

利用所给数据完成表 4.27。

| *x* | *P*(*x*) | *x*\**P*(*x*) |

|-----|----------|---------------|

| 1 | 0.10 | |

| 2 | 0.05 | |

| 3 | 0.10 | |

| 4 | | |

| 5 | 0.30 | |

| 6 | 0.20 | |

| 7 | 0.10 | |

Table 4.27 29.

表 4.27 29.

In words, define the random variable *X*.

用语言描述，定义随机变量 *X*。

30\.

*P*(*x* = 4) = \_\_\_\_\_\_\_

31.

*P*(*x* \< 4) = \_\_\_\_\_\_\_

32\.

On average, how many years would you expect a child to study ballet with this teacher?

平均而言，你预期一个学生与这位老师学习芭蕾多少年？

33.

What does the column "*P*(*x*)" sum to and why?

列"*P*(*x*)"的和是多少？为什么？

34\.

What does the column "*x*\**P*(*x*)" sum to and why?

列"*x*\**P*(*x*)"的和是多少？为什么？

35.

You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win \$30. If it is not a face card, you pay \$2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?

你在玩一个游戏：从一副标准牌中抽取一张牌并放回。如果是人头牌，你赢得 \$30；如果不是人头牌，你付 \$2。一副 52 张牌中有 12 张人头牌。玩这个游戏的期望值是多少？

36\.

你在玩一个游戏：从一副标准牌中抽取一张牌并放回。如果是人头牌，你赢得 \$30；如果不是人头牌，你付 \$2。一副 52 张牌中有 12 张人头牌。你应该玩这个游戏吗？

4.3 Binomial Distribution 4.3 二项分布

*Use the following information to answer the next eight exercises:* The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status.

用以下信息回答接下来的八个习题：加州大学洛杉矶分校的高等教育研究所收集了来自美国 270 所四年制学院和大学的 203,967 名入学新生（首次、全日制）的数据。其中 71.3% 的学生回答"是"，即他们认为同性伴侣应当拥有合法的婚姻地位。假设你在调查中随机抽取八名首次入学的全日制新生。你关注的是其中认为同性伴侣应当拥有合法婚姻地位的人数。

37.

In words, define the random variable *X*.

用语言描述，定义随机变量 *X*。

38\.

*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

39.

What values does the random variable *X* take on?

随机变量 *X* 取哪些值？

40\.

Construct the probability distribution function (PDF).

构造概率分布函数（PDF）。

| *x* | *P*(*x*) |

|-----|----------|

| | |

Table 4.28 41.

表 4.28 41.

On average (*μ*), how many would you expect to answer yes?

平均而言（*μ*），你预期有多少人回答"是"？

42\.

What is the standard deviation (*σ*)?

标准差（*σ*）是多少？

43.

What is the probability that at most five of the freshmen reply "yes"?

至多五名新生回答"是"的概率是多少？

44\.

What is the probability that at least two of the freshmen reply "yes"?

至少两名新生回答"是"的概率是多少？

4.4 Geometric Distribution 4.4 几何分布

*Use the following information to answer the next six exercises:* The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies "yes." You are interested in the number of freshmen you must ask.

用以下信息回答接下来的六个习题：加州大学洛杉矶分校的高等教育研究所收集了来自美国 270 所四年制学院和大学的 203,967 名入学新生（首次、全日制）的数据。其中 71.3% 的学生回答"是"，即他们认为同性伴侣应当拥有合法的婚姻地位。假设你在研究中随机选取新生，直到找到一名回答"是"的人。你关注的是必须询问的新生人数。

45.

In words, define the random variable *X*.

用语言描述，定义随机变量 *X*。

46\.

*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

47.

What values does the random variable *X* take on?

随机变量 *X* 取哪些值？

48\.

Construct the probability distribution function (PDF). Stop at *x* = 6.

构造概率分布函数（PDF）。到 *x* = 6 为止。

| *x* | *P*(*x*) |

|-----|----------|

| 1 | |

| 2 | |

| 3 | |

| 4 | |

| 5 | |

| 6 | |

Table 4.29 49.

表 4.29 49.

On average (*μ*), how many freshmen would you expect to have to ask until you found one who replies "yes?"

平均而言（*μ*），你预期需要询问多少名新生才能找到一名回答"是"的人？

50\.

What is the probability that you will need to ask fewer than three freshmen?

你需要询问少于三名新生的概率是多少？

4.5 Hypergeometric Distribution 4.5 超几何分布

*Use the following information to answer the next five exercises:* Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample.

用以下信息回答接下来的五个习题：假设一群统计学学生分为两组：商科专业和非商科专业。该群中有 16 名商科专业学生和 7 名非商科专业学生。随机抽取 9 名学生作为样本。我们关注的是样本中商科专业学生的人数。

51.

In words, define the random variable *X*.

用语言描述，定义随机变量 *X*。

52\.

*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

53.

What values does *X* take on?

*X* 取哪些值？

54\.

Find the standard deviation.

求标准差。

55.

On average (*μ*), how many would you expect to be business majors?

平均而言（*μ*），你预期其中有多少名商科专业学生？

4.6 Poisson Distribution 4.6 泊松分布

*Use the following information to answer the next six exercises:* On average, a clothing store gets 120 customers per day.

使用以下信息回答接下来的六道习题：平均而言，一家服装店每天接待 120 位顾客。

56\.

56.

Assume the event occurs independently in any given day. Define the random variable *X*.

假设该事件在任意给定的一天中独立发生。定义随机变量 *X*。

57.

What values does *X* take on?

随机变量 *X* 取哪些值？

58\.

58.

What is the probability of getting 150 customers in one day?

一天内接待 150 位顾客的概率是多少？

59.

What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.

在前四个小时接待 35 位顾客的概率是多少？假设该店每天营业 12 小时。

60\.

60.

What is the probability that the store will have more than 12 customers in the first hour?

该店在第一个小时内顾客数超过 12 人的概率是多少？

61.

What is the probability that the store will have fewer than 12 customers in the first two hours?

该店在前两个小时内顾客数少于 12 人的概率是多少？

62\.

62.

Which type of distribution can the Poisson model be used to approximate? When would you do this?

泊松模型可用于近似哪种类型的分布？你会在什么情况下这样做？

*Use the following information to answer the next six exercises:* On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age.

使用以下信息回答接下来的六道习题：平均而言，美国每天有八名青少年死于机动车伤害。因此，全国各地各州正在就是否提高驾驶年龄展开辩论。

63.

Assume the event occurs independently in any given day. In words, define the random variable *X*.

假设该事件在任意给定的一天中独立发生。用文字定义随机变量 *X*。

64\.

64.

*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

65.

What values does *X* take on?

随机变量 *X* 取哪些值？

66\.

66.

For the given values of the random variable *X*, fill in the corresponding probabilities.

对随机变量 *X* 的给定取值，填入相应的概率。

67.

Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically.

在美国，任意给定的一天中都没有青少年死于机动车伤害，这种情况可能发生吗？用数值说明你的答案。

68\.

68.

Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.

在美国，任意给定的一天中死于机动车伤害的青少年超过 20 人，这种情况可能发生吗？用数值说明你的答案。

Homework 作业

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量的概率分布函数（PDF）

69\.

69.

Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given in Table 4.30.

假设获得理学学士（B.S.）学位所需年数的概率分布函数（PDF）由表 4.30 给出。

| *x* | *P*(*x*) |

|-----|----------|

| 3 | 0.05 |

| 4 | 0.40 |

| 5 | 0.30 |

| 6 | 0.15 |

| 7 | 0.10 |

Table 4.30

表 4.30

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. What does it mean that the values zero, one, and two are not included for *x* in the PDF?

2. 在概率分布函数中，*x* 的取值不包含零、一和二，这意味着什么？

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

70\.

70.

A theater group holds a fund-raiser. It sells 100 raffle tickets for \$5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of \$150.

一个剧团举办筹款活动。它以每张 \$5 的价格出售 100 张抽奖券。假设你购买了四张。奖品是两张百老汇演出的门票，总价值 \$150。

1. What are you interested in here?

1. 你在此关心的是什么？

2. In words, define the random variable *X*.

2. 用文字定义随机变量 *X*。

3. List the values that *X* may take on.

3. 列出 *X* 可能取的值。

4. Construct a PDF.

4. 构造一个概率分布函数（PDF）。

5. If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?

5. 如果这项筹款活动频繁重复进行，而你每次都购买四张抽奖券，那么你每次抽奖的期望平均收益是多少？

71.

A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails.

一个游戏包括从一副普通的 52 张扑克牌中抽取一张牌，并抛掷一枚硬币。这枚硬币是均匀的，正面和反面朝上的可能性相等。

If the card is a face card, and the coin lands on Heads, you win \$6
If the card is a face card, and the coin lands on Tails, you win \$2
If the card is not a face card, you lose \$2, no matter what the coin shows.

如果抽到人头牌，且硬币正面朝上，你赢得 \$6。
如果抽到人头牌，且硬币反面朝上，你赢得 \$2。
如果抽到的不是人头牌，无论硬币哪面朝上，你都输掉 \$2。

1. Find the expected value for this game (expected net gain or loss).

1. 求此游戏的期望值（期望净收益或净损失）。

2. Explain what your calculations indicate about your long-term average profits and losses on this game.

2. 解释你的计算结果说明了此游戏长期平均盈利与亏损的什么情况。

3. Should you play this game to win money?

3. 为了赢钱，你应该玩这个游戏吗？

72\.

72.

You buy a lottery ticket to a lottery that costs \$10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one \$500 prize, two \$100 prizes, and four \$25 prizes. Find your expected gain or loss.

你购买一张彩票，该彩票每张售价 \$10。这次彩票总共只有 100 张可供出售。其中有一个 \$500 奖金、两个 \$100 奖金和四个 \$25 奖金。求你的期望收益或损失。

73.

Complete the PDF and answer the questions.

完成该概率分布函数（PDF）并回答问题。

| *x* | *P*(*x*) | *xP*(*x*) |

|-----|----------|-----------|

| 0 | 0.3 | |

| 1 | 0.2 | |

| 2 | | |

| 3 | 0.4 | |

Table 4.31

表 4.31

1. Find the probability that *x* = 2.

1. 求 *x* = 2 的概率。

2. Find the expected value.

2. 求期望值。

74\.

74.

Suppose that you are offered the following "deal." You roll a die. If you roll a six, you win \$10. If you roll a four or five, you win \$5. If you roll a one, two, or three, you pay \$6.

假设你被提供以下"交易"。你掷一枚骰子。如果掷出六，你赢得 \$10。如果掷出四或五，你赢得 \$5。如果掷出一、二或三，你支付 \$6。

1. What are you ultimately interested in here (the value of the roll or the money you win)?

1. 你在这里最终关心的是什么（掷出的点数还是你赢得的钱）？

2. In words, define the Random Variable *X*.

2. 用文字定义随机变量 *X*。

3. List the values that *X* may take on.

3. 列出 *X* 可能取的值。

4. Construct a PDF.

4. 构造一个概率分布函数（PDF）。

5. Over the long run of playing this game, what are your expected average winnings per game?

5. 从长远来看，反复玩这个游戏，你每局的期望平均收益是多少？

6. Based on numerical values, should you take the deal? Explain your decision in complete sentences.

6. 根据数值，你应该接受这个交易吗？用完整的句子解释你的决定。

75.

A venture capitalist, willing to invest \$1,000,000, has three investments to choose from. The first investment, a software company, has a 10% chance of returning \$5,000,000 profit, a 30% chance of returning \$1,000,000 profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning \$3,000,000 profit, a 40% chance of returning \$1,000,000 profit, and a 40% chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning \$6,000,000 profit, a 70% of no profit or loss, and a 20% chance of losing the million dollars.

一位愿意投资 \$1,000,000 的风险投资家有三种投资可供选择。第一项投资是一家软件公司，有 10% 的概率获得 \$5,000,000 利润，30% 的概率获得 \$1,000,000 利润，60% 的概率损失这百万美元。第二家公司是一家硬件公司，有 20% 的概率获得 \$3,000,000 利润，40% 的概率获得 \$1,000,000 利润，40% 的概率损失这百万美元。第三家公司是一家生物技术公司，有 10% 的概率获得 \$6,000,000 利润，70% 的概率不盈不亏，20% 的概率损失这百万美元。

1. Construct a PDF for each investment.

1. 为每项投资构造一个概率分布函数（PDF）。

2. Find the expected value for each investment.

2. 求每项投资的期望值。

3. Which is the safest investment? Why do you think so?

3. 哪项投资最安全？你为什么这样认为？

4. Which is the riskiest investment? Why do you think so?

4. 哪项投资风险最大？你为什么这样认为？

5. Which investment has the highest expected return, on average?

5. 平均而言，哪项投资的期望回报最高？

76\.

76.

Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let *X* = the number of children married people have.

假设对美国 20,000 名已婚成年人就其子女数量进行了随机调查。调查结果被汇总并用作理论概率。令 *X* = 已婚人士所拥有的子女数量。

| *x* | *P*(*x*) | *xP*(*x*) |

|-------------|----------|-----------|

| 0 | 0.10 | |

| 1 | 0.20 | |

| 2 | 0.30 | |

| 3 | | |

| 4 | 0.10 | |

| 5 | 0.05 | |

| 6 (or more) | 0.05 | |

Table 4.32

表 4.32

1. Find the probability that a married adult has three children.

1. 求一位已婚成年人有三个孩子的概率。

2. In words, what does the expected value in this example represent?

2. 用文字说明，本例中的期望值代表什么？

3. Find the expected value.

3. 求期望值。

4. Is it more likely that a married adult will have two to three children or four to six children? How do you know?

4. 已婚成年人更可能有两到三个孩子，还是四到六个孩子？你是如何知道的？

77.

Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as in Table 4.33.

假设获得理学学士（B.S.）学位所需年数的概率分布函数（PDF）由表 4.33 给出。

| *x* | *P*(*x*) |

|-----|----------|

| 3 | 0.05 |

| 4 | 0.40 |

| 5 | 0.30 |

| 6 | 0.15 |

| 7 | 0.10 |

Table 4.33

表 4.33

On average, how many years do you expect it to take for an individual to earn a B.S.?

平均而言，你预期一个人获得理学学士学位需要多少年？

78\.

78.

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.

光顾音像租赁店的人常常一次租借多张 DVD。Video To Go 店每位顾客租借 DVD 数量的概率分布由下表给出。该店每位顾客最多可租五张，因此没有人会租超过五张 DVD。

| *x* | *P*(*x*) |

|-----|----------|

| 0 | 0.03 |

| 1 | 0.50 |

| 2 | 0.24 |

| 3 | |

| 4 | 0.07 |

| 5 | 0.04 |

Table 4.34

表 4.34

1. Describe the random variable *X* in words.

1. 用文字描述随机变量 *X*。

2. Find the probability that a customer rents three DVDs.

2. 求一位顾客租借三张 DVD 的概率。

3. Find the probability that a customer rents at least four DVDs.

3. 求一位顾客至少租借四张 DVD 的概率。

4. Find the probability that a customer rents at most two DVDs.

4. 求一位顾客至多租借两张 DVD 的概率。

Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer.

另一家店 Entertainment Headquarters 出租 DVD 和视频游戏。该店每位顾客租借 DVD 数量的概率分布如下。他们同样对每位顾客设有五张 DVD 的上限。

| *x* | *P*(*x*) |

|-----|----------|

| 0 | 0.35 |

| 1 | 0.25 |

| 2 | 0.20 |

| 3 | 0.10 |

| 4 | 0.05 |

| 5 | 0.05 |

Table 4.35

表 4.35

5. At which store is the expected number of DVDs rented per customer higher?

5. 哪家店每位顾客租借 DVD 的期望数量更高？

6. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.

6. 如果 Video To Go 估计下周将有 300 位顾客，他们下周预期租出多少张 DVD？用完整的句子回答。

7. If Video To Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.

7. 如果 Video To Go 预期下周有 300 位顾客，而 Entertainment HQ 预计有 420 位顾客，那么哪家店下周 DVD 租借的期望数量更高？请解释。

8. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?

8. 这两家音像店中，哪家每位顾客租借 DVD 数量的变动更大？你是如何知道的？

79.

A "friend" offers you the following "deal." For a \$10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift.

一位"朋友"向你提供以下"交易"。支付 \$10 费用后，你可以从一个装有 100 个看似完全相同的信封的盒子中抽取一个信封。不过，每个信封里都有一张免费礼品的优惠券。

Ten of the coupons are for a free gift worth \$6.
Eighty of the coupons are for a free gift worth \$8.
Six of the coupons are for a free gift worth \$12.
Four of the coupons are for a free gift worth \$40.

其中十张优惠券可兑换价值 \$6 的免费礼品。
八十张优惠券可兑换价值 \$8 的免费礼品。
六张优惠券可兑换价值 \$12 的免费礼品。
四张优惠券可兑换价值 \$40 的免费礼品。

Based upon the financial gain or loss over the long run, should you play the game?

根据长期的财务收益或损失，你应该玩这个游戏吗？

1. Yes, I expect to come out ahead in money.

1. 是的，我预期会在金钱上获利。

2. No, I expect to come out behind in money.

2. 不，我预期会在金钱上亏损。

3. It doesn’t matter. I expect to break even.

3. 无所谓。我预期会不盈不亏。

80\.

80.

Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.

佛罗里达州立大学在 2013 年夏季学期安排了 14 个统计班。其中一个班可容纳 30 名学生，八个班可容纳 60 名学生，一个班可容纳 70 名学生，四个班可容纳 100 名学生。

1. What is the average class size assuming each class is filled to capacity?

1. 假设每个班都满员，平均班级规模是多少？

2. Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable *X* equal the size of the student’s class. Define the PDF for *X*.

2. 共有 980 个名额。假设每个班都满员，随机抽取一名统计学生。令随机变量 *X* 等于该学生所在班级的规模。定义 *X* 的概率分布函数（PDF）。

3. Find the mean of *X*.

3. 求 *X* 的均值。

4. Find the standard deviation of *X*.

4. 求 *X* 的标准差。

81.

In a lottery, there are 250 prizes of \$5, 50 prizes of \$25, and ten prizes of \$100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even?

在一次彩票中，有 250 个 \$5 奖金、50 个 \$25 奖金和十个 \$100 奖金。假设将发行并售出 10,000 张彩票，那么为达到收支平衡，应收取多少公平价格？

4.3 Binomial Distribution 4.3 二项分布

82\.

82.

According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery.

根据最近的一篇文章，在健康的婴儿室中，每 1,000 名婴儿大约有 2 名出生即患有明显听力损失（耳聋）。在重症监护婴儿室中，这一数字上升到平均每 1,000 名婴儿中有 30 名。

Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf.

假设随机调查了来自健康婴儿室的 1,000 名婴儿。求恰好有两名婴儿出生即耳聋的概率。

Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu.

利用以下信息回答接下来的四道习题。最近，一位护士评论说，当患者致电医疗咨询热线自称患流感时，他或她真正患流感（而不只是重感冒）的几率只有约 4%。在接下来致电自称患流感的 25 名患者中，我们关注实际患流感的人数。

83\.

83.

Define the random variable and list its possible values.

定义随机变量并列出其可能取值。

84\.

84.

State the distribution of *X*.

写出 *X* 的分布。

85\.

85.

Find the probability that at least four of the 25 patients actually have the flu.

求这 25 名患者中至少有 4 人实际患流感的概率。

86\.

86.

On average, for every 25 patients calling in, how many do you expect to have the flu?

平均而言，每 25 名致电的患者中，你预计有多少名实际患流感？

87\.

87.

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given Table 4.36. There is five-video limit per customer at this store, so nobody ever rents more than five DVDs.

光顾影像出租店的人常常一次租用多张 DVD。Video To Go 店每位顾客租借 DVD 数量的概率分布由表 4.36 给出。该店每位顾客最多可租 5 张，因此没有人会租超过 5 张 DVD。

| *x* | *P*(*x*) |

|-----|----------|

| 0 | 0.03 |

| 1 | 0.50 |

| 2 | 0.24 |

| 3 | |

| 4 | 0.07 |

| 5 | 0.04 |

Table 4.36

表 4.36

1. Describe the random variable *X* in words.

1. 用文字描述随机变量 *X*。

2. Find the probability that a customer rents three DVDs.

2. 求某顾客租用 3 张 DVD 的概率。

3. Find the probability that a customer rents at least four DVDs.

3. 求某顾客租用至少 4 张 DVD 的概率。

4. Find the probability that a customer rents at most two DVDs.

4. 求某顾客租用至多 2 张 DVD 的概率。

88\.

88.

A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities.

校报记者决定随机调查 12 名学生，了解他们今年是否会参加 Tet（越南新年）庆祝活动。根据往年经验，她知道有 18% 的学生会参加 Tet 庆祝活动。我们关注将参加庆祝活动的学生人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)*

3. 给出 *X* 的分布。*X ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)*

4. How many of the 12 students do we expect to attend the festivities?

4. 预计这 12 名学生中有多少人会参加庆祝活动？

5. Find the probability that at most four students will attend.

5. 求至多 4 名学生参加的概率。

6. Find the probability that more than two students will attend.

6. 求超过 2 名学生参加的概率。

*Use the following information to answer the next three exercises:* The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games.

*利用以下信息回答接下来的三道习题：* 根据截至某日的 13 年战绩（共 1,034 场比赛中获胜 382 场），圣何塞鲨鱼队在任何一场给定比赛中获胜的概率为 0.3694。即将到来的月度赛程包含 12 场比赛。

89\.

89.

The expected number of wins for that upcoming month is:

该月预期的获胜场数为：

1. 1.67

2. 12

3. $\frac{382}{1043}$

4. 4.43

Let *X* = the number of games won in that upcoming month.

令 *X* = 该月获胜的比赛场数。

90\.

90.

What is the probability that the San Jose Sharks win six games in that upcoming month?

圣何塞鲨鱼队在该月恰好赢下 6 场比赛的概率是多少？

1. 0.1476

2. 0.2336

3. 0.7664

4. 0.8903

91\.

91.

What is the probability that the San Jose Sharks win at least five games in that upcoming month

圣何塞鲨鱼队在该月至少赢下 5 场比赛的概率是多少

1. 0.3694

2. 0.5266

3. 0.4734

4. 0.2305

92\.

92.

A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct.

一名学生参加一份十道题目的判断题测验，但他没有复习，每题都随机猜测答案。求该学生以至少 70% 的正确率通过测验的概率。

93\.

93.

A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly.

一名学生参加一份 32 道题目的选择题考试，但他没有复习，每题都随机猜测答案。每道题有三个可能的选项。求该学生猜对超过 75% 题目的概率。

94\.

94.

Six different colored dice are rolled. Of interest is the number of dice that show a one.

同时掷六枚不同颜色的骰子。关注的是掷出一点（即显示数字 1）的骰子数量。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. On average, how many dice would you expect to show a one?

4. 平均而言，你预计有多少枚骰子显示一点？

5. Find the probability that all six dice show a one.

5. 求六枚骰子全部显示一点的概率。

6. Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically.

6. 三枚还是四枚骰子显示一点的可能性更大？用数值对你的回答进行量化说明。

95\.

95.

More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses.

在规模最大的高等院校（总注册人数超过 15,000）中，超过 96% 提供某种在线课程。假设随机选取 13 所这样的院校。我们关注提供远程学习课程的院校数量。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. On average, how many schools would you expect to offer such courses?

4. 平均而言，你预计有多少所院校提供此类课程？

5. Find the probability that at most ten offer such courses.

5. 求至多 10 所提供此类课程的概率。

6. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence.

6. 12 所还是 13 所更可能提供此类课程？用数值对你的回答进行量化说明，并用完整的句子作答。

96\.

96.

Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen.

假设约 85% 的毕业班学生参加自己的毕业典礼。随机选取一组 22 名毕业班学生。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many are expected to attend their graduation?

4. 预计有多少人参加自己的毕业典礼？

5. Find the probability that 17 or 18 attend.

5. 求 17 人或 18 人参加的概率。

6. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically.

6. 根据数值，如果全部 22 人都参加了毕业典礼，你会感到意外吗？用数值对你的回答进行说明。

97\.

97.

At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey 25 fencers at The Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon.

在击剑中心，60% 的击剑手以花剑作为主要武器。我们在该击剑中心随机调查 25 名击剑手。我们关注不以花剑作为主要武器的击剑手人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many are expected to not to use the foil as their main weapon?

4. 预计有多少人不以花剑作为主要武器？

5. Find the probability that six do not use the foil as their main weapon.

5. 求恰有 6 人不以花剑作为主要武器的概率。

6. Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically.

6. 根据数值，如果全部 25 人都不以花剑作为主要武器，你会感到意外吗？用数值对你的回答进行说明。

98\.

98.

Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school.

当地一所高中的学生中，约有 8% 在整个高中四年都参加课后体育运动。随机选取一组 60 名毕业班学生。关注的是在整个高中四年都参加课后体育运动的人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many seniors are expected to have participated in after-school sports all four years of high school?

4. 预计有多少名毕业班学生整个高中四年都参加了课后体育运动？

5. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

5. 根据数值，如果没有一名毕业班学生整个高中四年都参加课后体育运动，你会感到意外吗？用数值对你的回答进行说明。

6. Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

6. 根据数值，这四名还是五名毕业班学生整个高中四年都参加了课后体育运动的可能性更大？用数值对你的回答进行说明。

99\.

99.

The chance of an IRS audit for a tax return with over \$25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent.

对于收入超过 \$25,000 的纳税申报表，每年被国税局（IRS）审计的几率约为 2%。我们关注具有该收入水平的人在 20 年期间预期的审计次数。假设各年相互独立。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many audits are expected in a 20-year period?

4. 在 20 年期间预计会被审计多少次？

5. Find the probability that a person is not audited at all.

5. 求某人完全不被审计的概率。

6. Find the probability that a person is audited more than twice.

6. 求某人被审计超过两次的概率。

100\.

100.

It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies.

据估计，只有约 30% 的加利福尼亚州居民备有充足的地震应急物资。假设随机调查 11 名加州居民。我们关注备有充足地震应急物资的人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. What is the probability that at least eight have adequate earthquake supplies?

4. 至少有 8 人备有充足地震应急物资的概率是多少？

5. Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why?

5. 被调查的居民中，谁的可能性更大：无人备有还是全部备有充足的地震应急物资？为什么？

6. How many residents do you expect will have adequate earthquake supplies?

6. 你预计有多少名居民备有充足的地震应急物资？

101\.

101.

There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being \$1. The player places a bet on a number or object. The "house" rolls three dice. If none of the dice show the number or object that was bet, the house keeps the \$1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back his or her \$1 bet, plus \$1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back his or her \$1 bet, plus \$2 profit. If all three dice show the number or object bet, the player gets back his or her \$1 bet, plus \$3 profit. Let *X* = number of matches and *Y* = profit per game.

中国新年和越南新年都有两种类似的游戏。在中国版本中，使用标有数字 1、2、3、4、5、6 的均匀骰子，以及标有这些数字的棋盘。在越南版本中，使用绘有葫芦、鱼、公鸡、螃蟹、小龙虾和鹿的均匀骰子，棋盘上也绘有这六种物件。我们以每注 \$1 进行游戏。玩家对某个数字或物件下注。"庄家"掷三枚骰子。如果没有任何一枚骰子显示所下注的数字或物件，庄家收下这 \$1 赌注。如果其中一枚骰子显示所下注的数字或物件（而另外两枚没有显示），玩家拿回自己的 \$1 赌注，外加 \$1 利润。如果有两枚骰子显示所下注的数字或物件（而第三枚没有显示），玩家拿回自己的 \$1 赌注，外加 \$2 利润。如果三枚骰子都显示所下注的数字或物件，玩家拿回自己的 \$1 赌注，外加 \$3 利润。令 *X* = 匹配的数量，*Y* = 每局的利润。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. List the values that *Y* may take on. Then, construct one PDF table that includes both *X* and *Y* and their probabilities.

4. 列出 *Y* 可能取的值。然后，构造一张同时包含 *X* 与 *Y* 及其概率的概率分布函数（PDF）表。

5. Calculate the average expected matches over the long run of playing this game for the player.

5. 计算玩家长期玩此游戏时每局预期的匹配数量的平均值。

6. Calculate the average expected earnings over the long run of playing this game for the player.

6. 计算玩家长期玩此游戏时每局预期的收益平均值。

7. Determine who has the advantage, the player or the house.

7. 判断谁占有优势，是玩家还是庄家。

102\.

102.

According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let *X* = the number of people who have access to electricity.

根据世界银行的数据，截至 2009 年，乌干达只有 9% 的人口能用上电。假设我们在乌干达随机抽样调查 150 人。令 *X* = 能用上电的人数。

1. What is the probability distribution for *X*?

1. *X* 的概率分布是什么？

2. Using the formulas, calculate the mean and standard deviation of *X*.

2. 利用公式，计算 *X* 的均值与标准差。

3. Use your calculator to find the probability that 15 people in the sample have access to electricity.

3. 用计算器求出样本中有 15 人能用上电的概率。

4. Find the probability that at most ten people in the sample have access to electricity.

4. 求样本中至多 10 人能用上电的概率。

5. Find the probability that more than 25 people in the sample have access to electricity.

5. 求样本中超过 25 人能用上电的概率。

103\.

103.

The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in Afghanistan is 28.1%. Suppose you choose 15 people in Afghanistan at random. Let *X* = the number of people who are literate.

一个国家的识字率衡量的是 15 岁及以上人口中能够读写的比例。阿富汗的识字率为 28.1%。假设你在阿富汗随机选取 15 人。令 *X* = 识字的人数。

1. Sketch a graph of the probability distribution of *X*.

1. 画出 *X* 的概率分布图。

2. Using the formulas, calculate the (i) mean and (ii) standard deviation of *X*.

2. 利用公式，计算 *X* 的（i）均值与（ii）标准差。

3. Find the probability that more than five people in the sample are literate. Is it is more likely that three people or four people are literate.

3. 求样本中超过 5 人识字的概率。究竟是 3 人识字还是 4 人识字的可能性更大。

4.4 Geometric Distribution 4.4 几何分布

104\.

104.

A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call.

一位想买二手红色 Miata 汽车的消费者会不停地给经销商打电话，直到找到一家有这款车的经销商。她估计任意一家独立的经销商有这款车的概率为 28%。我们关注的是她必须打电话的经销商数量。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. On average, how many dealerships would we expect her to have to call until she finds one that has the car?

4. 平均而言，我们预期她需要打多少家经销商的电话才能找到一家有这款车的？

5. Find the probability that she must call at most four dealerships.

5. 求她最多需要打四家经销商电话的概率。

6. Find the probability that she must call three or four dealerships.

6. 求她需要打三家或四家经销商电话的概率。

105\.

105.

Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl.

假设美国成年人观看超级碗的概率为 40%。每个人被视为相互独立。我们关注的是必须调查多少名美国成年人才能找到一名会观看超级碗的人。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many adults in America do you expect to survey until you find one who will watch the Super Bowl?

4. 你预期需要调查多少名美国成年人才能找到一名会观看超级碗的？

5. Find the probability that you must ask seven people.

5. 求你必须询问七个人的概率。

6. Find the probability that you must ask three or four people.

6. 求你必须询问三个人或四个人的概率。

106\.

106.

It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies.

据估计，只有约 30% 的加利福尼亚州居民备有充足的地震应急物资。假设我们关注的是必须调查多少名加州居民才能找到一名没有充足地震应急物资的居民。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies?

4. 在找到一名没有充足地震应急物资的加州居民之前，我们只需调查一名或两名居民的概率是多少？

5. What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies?

5. 在找到一名没有充足地震应急物资的加州居民之前，我们必须至少调查三名加州居民的概率是多少？

6. How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies?

6. 你预期需要调查多少名加州居民才能找到一名没有充足地震应急物资的居民？

7. How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies?

7. 你预期需要调查多少名加州居民才能找到一名有充足地震应急物资的居民？

107\.

107.

In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once.

在 L.L. Bean® 的一本春季商品目录中，192 页里有 29 页刊登了鞋类广告。假设我们随机调查 20 页。我们关注的是刊登鞋类广告的页数。每一页可以被重复抽取。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many pages do you expect to advertise footwear on them?

4. 你预期有多少页会刊登鞋类广告？

5. Is it probable that all twenty will advertise footwear on them? Why or why not?

5. 全部二十页都刊登鞋类广告是否可能？为什么可能或不可能？

6. What is the probability that fewer than ten will advertise footwear on them?

6. 少于十页刊登鞋类广告的概率是多少？

7. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable *X* and give its distribution.

7. 提醒：一页可以被重复抽取。我们关注的是必须随机调查多少页才能找到一页刊登了鞋类广告的页面。定义随机变量 *X* 并给出它的分布。

8. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it?

8. 你只需调查至多三页就能找到一页刊登鞋类广告的页面的概率是多少？

9. How many pages do you expect to need to survey in order to find one that advertises footwear?

9. 你预期需要调查多少页才能找到一页刊登鞋类广告的？

108\.

108.

Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let *F* be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome.

假设你正在进行掷一枚均匀六面骰子的概率试验。令 *F* 为掷出四点或五点的事件。你关注的是需要掷多少次骰子才能得到第一次四点或五点的结果。

*p* = probability of success (event *F* occurs)
*q* = probability of failure (event *F* does not occur)

*p* = 成功的概率（事件 *F* 发生）
*q* = 失败的概率（事件 *F* 不发生）

1. Write the description of the random variable *X*.

1. 写出随机变量 *X* 的描述。

2. What are the values that *X* can take on?

2. *X* 可以取哪些值？

3. Find the values of *p* and *q*.

3. 求 *p* 和 *q* 的值。

4. Find the probability that the first occurrence of event *F* (rolling a four or five) is on the second trial.

4. 求事件 *F*（掷出四点或五点）首次发生在第二次试验的概率。

109\.

109.

Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does *X* take on?

Ellen 每周有三天练习音乐。她在 85% 的情况下三天都练习，8% 的情况下练习两天，4% 的情况下练习一天，3% 的情况下一天也不练习。随机选取一周。*X* 取哪些值？

110\.

110.

The World Bank records the prevalence of HIV in countries around the world. According to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who are infected with HIV.”¹ In South Africa, the prevalence of HIV is 17.3%. Let *X* = the number of people you test until you find a person infected with HIV.

世界银行记录世界各国 HIV 的患病率。根据其数据，"HIV 患病率指 15 至 49 岁人群中感染 HIV 的百分比。"¹ 在南非，HIV 患病率为 17.3%。令 *X* = 你检测的人数，直到找到一名感染 HIV 的人。

1. Sketch a graph of the distribution of the discrete random variable *X*.

1. 画出离散随机变量 *X* 分布的草图。

2. What is the probability that you must test 30 people to find one with HIV?

2. 你必须检测 30 个人才能找到一名 HIV 感染者的概率是多少？

3. What is the probability that you must ask ten people?

3. 你必须询问十个人的概率是多少？

4. Find the (i) mean and (ii) standard deviation of the distribution of *X*.

4. 求 *X* 分布的（i）均值和（ii）标准差。

111\.

111.

According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let *X* = the number of millenials you ask until you find a person without a profile on a social networking site.

根据皮尤研究中心最近的一项调查，75% 的千禧一代（1981 至 1995 年出生的人）在社交网站上有个人主页。令 *X* = 你询问的千禧一代人数，直到找到一名没有社交网站个人主页的人。

1. Describe the distribution of *X*.

1. 描述 *X* 的分布。

2. Find the (i) mean and (ii) standard deviation of *X*.

2. 求 *X* 的（i）均值和（ii）标准差。

3. What is the probability that you must ask ten people to find one person without a social networking site?

3. 你必须询问十个人才能找到一名没有社交网站个人主页的人的概率是多少？

4. What is the probability that you must ask 20 people to find one person without a social networking site?

4. 你必须询问 20 个人才能找到一名没有社交网站个人主页的人的概率是多少？

5. What is the probability that you must ask *at most* five people?

5. 你必须询问*至多*五个人的概率是多少？

4.5 Hypergeometric Distribution 4.5 超几何分布

112\.

112.

A group of Martial Arts students is planning on participating in an upcoming demonstration. Six are students of Tae Kwon Do; seven are students of Shotokan Karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of Shotokan Karate students in that first demonstration.

一群武术学员计划参加即将举行的一场演示。其中六人是跆拳道学员，七人是松涛馆空手道学员。假设随机挑选八名学员参加第一场演示。我们关注的是第一场演示中松涛馆空手道学员的人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many Shotokan Karate students do we expect to be in that first demonstration?

4. 我们预期第一场演示中有多少名松涛馆空手道学员？

113\.

113.

在 L.L. Bean® 的一本春季商品目录中，192 页里有 29 页刊登了鞋类广告。假设我们随机调查 20 页。我们关注的是刊登鞋类广告的页数。每一页最多只能被抽取一次。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many pages do you expect to advertise footwear on them?

4. 你预期有多少页会刊登鞋类广告？

5. Calculate the standard deviation.

5. 计算标准差。

114\.

114.

Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that ten people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient.

假设正在组建一个技术工作组来研究教师中的技术意识情况。假设将从 28 名志愿者中随机选出十人进入委员会，其中 20 人具备技术能力，8 人不具备。我们关注的是委员会中不具备技术能力的人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many instructors do you expect on the committee who are not technically proficient?

4. 你预期委员会中有多少名不具备技术能力的教师？

5. Find the probability that at least five on the committee are not technically proficient.

5. 求委员会中至少有五人不具备技术能力的概率。

6. Find the probability that at most three on the committee are not technically proficient.

6. 求委员会中至多有三人不具备技术能力的概率。

115\.

115.

Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the Boston Celtics and four volunteers from the New England Patriots. We are interested in the number of Patriots picked.

假设九名马萨诸塞州运动员将出席一场慈善活动。这九人是从波士顿凯尔特人队的八名志愿者和新英格兰爱国者队的四名志愿者中随机选出的。我们关注的是被选中的爱国者队队员人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. Are you choosing the nine athletes with or without replacement?

4. 你是在有放回还是无放回的情况下选出这九名运动员的？

116\.

116.

A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart?

一手桥牌定义为从一副 52 张的牌中随机无放回地选出 13 张牌。在一副标准牌中，每种花色各有 13 张牌：红桃、黑桃、梅花和方块。发到一手不含红桃的牌的概率是多少？

1. What is the group of interest?

1. 关注的群体是什么？

2. How many are in the group of interest?

2. 关注群体中有多少个？

3. How many are in the other group?

3. 另一群体中有多少个？

4. Let *X* = \_\_\_\_\_\_\_\_\_. What values does *X* take on?

4. 令 *X* = \_\_\_\_\_\_\_\_\_。*X* 取哪些值？

5. The probability question is *P*(\_\_\_\_\_\_\_).

5. 概率问题是 *P*(\_\_\_\_\_\_\_)。

6. Find the probability in question.

6. 求问题中的概率。

7. Find the (i) mean and (ii) standard deviation of *X*.

7. 求 *X* 的（i）均值和（ii）标准差。

4.6 Poisson Distribution 4.6 泊松分布

117.

The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let *X* = the number of calls received at noon.

明尼阿波利斯一家律师事务所的交换台在周一中午时段平均接到 5.5 个打入电话。经验表明，现有员工每小时最多可处理六个电话。令 *X* = 中午接到的电话数。

1. Find the mean and standard deviation of *X*.

1. 求 *X* 的均值与标准差。

2. What is the probability that the office receives at most six calls at noon on Monday?

2. 该事务所在周一中午接到至多六个电话的概率是多少？

3. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon?

3. 求该律师事务所中午接到六个电话的概率。对于平均在中午接到 5.5 个打入电话的事务所员工来说，这意味着什么？

4. What is the probability that the office receives more than eight calls at noon?

4. 该事务所中午接到超过八个电话的概率是多少？

118\.

118.

The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let *X* = the number of births in an hour.

位于菲律宾马尼拉的 Dr. Jose Fabella Memorial Hospital 产科病房是世界上最繁忙的病房之一，平均每天有 60 例分娩。令 *X* = 一小时内分娩的例数。

1. Find the mean and standard deviation of *X*.

1. 求 *X* 的均值与标准差。

2. Sketch a graph of the probability distribution of *X*.

2. 画出 *X* 的概率分布图。

3. What is the probability that the maternity ward will deliver three babies in one hour?

3. 该产科病房在一小时内接生三名婴儿的概率是多少？

4. What is the probability that the maternity ward will deliver at most three babies in one hour?

4. 该产科病房在一小时内至多接生三名婴儿的概率是多少？

5. What is the probability that the maternity ward will deliver more than five babies in one hour?

5. 该产科病房在一小时内接生超过五名婴儿的概率是多少？

119.

A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.

一家圣诞树灯泡制造商知道其 3% 的灯泡有缺陷。分别用二项分布与泊松分布，求一串 100 只灯泡中至多有四只缺陷灯泡的概率。

120\.

120.

The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.

日本女性一生平均生育 1.37 个孩子。假设随机抽取一名日本女性。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. Find the probability that she has no children.

4. 求她没有孩子的概率。

5. Find the probability that she has fewer children than the Japanese average.

5. 求她生育的孩子数少于日本平均值的概率。

6. Find the probability that she has more children than the Japanese average.

6. 求她生育的孩子数多于日本平均值的概率。

121.

The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen.

西班牙女性一生平均生育 1.47 个孩子。假设随机抽取一名西班牙女性。

1. In words, define the Random Variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. Find the probability that she has no children.

4. 求她没有孩子的概率。

5. Find the probability that she has fewer children than the Spanish average.

5. 求她生育的孩子数少于西班牙平均值的概率。

6. Find the probability that she has more children than the Spanish average .

6. 求她生育的孩子数多于西班牙平均值的概率。

122\.

122.

Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces:

育龄母猫平均每年产三窝。假设随机抽取一只育龄母猫。求她在一年内产下以下窝数的概率：

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_\_\_

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_\_\_

4. Find the probability that she has no litters in one year.

4. 求她在一年内没有产窝的概率。

5. Find the probability that she has at least two litters in one year.

5. 求她在一年内至少产两窝的概率。

6. Find the probability that she has exactly three litters in one year.

6. 求她在一年内恰好产三窝的概率。

123.

The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.

幸运饼干中多出一张签文的概率约为 3%。给定一袋 144 块幸运饼干，我们关注带有额外签文的饼干数量。本题可用两种分布求解，但只使用其中一种分布来解。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many cookies do we expect to have an extra fortune?

4. 我们预计有多少块饼干带有额外签文？

5. Find the probability that none of the cookies have an extra fortune.

5. 求没有饼干带有额外签文的概率。

6. Find the probability that more than three have an extra fortune.

6. 求超过三块饼干带有额外签文的概率。

7. As *n* increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.

7. 当 *n* 增大时，使用两种分布得到的概率会发生什么变化？用完整的句子说明。

124\.

124.

According to the South Carolina Department of Mental Health web site, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following.

据南卡罗来纳州心理健康部网站，每 200 名美国女性中平均有 1 名患有厌食症。从随机选取的 600 名美国女性群体中，确定以下各项。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of*X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many are expected to suffer from anorexia?

4. 预计有多少人患厌食症？

5. Find the probability that no one suffers from anorexia.

5. 求没有人患厌食症的概率。

6. Find the probability that more than four suffer from anorexia.

6. 求超过四人患厌食症的概率。

125.

The chance of an IRS audit for a tax return with over \$25,000 in income is about 2% per year. Suppose that 100 people with tax returns over \$25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions.

对于收入超过 \$25,000 的纳税申报表，每年被 IRS 审计的概率约为 2%。假设随机选取 100 名收入超过 \$25,000 的纳税人。我们关注一年内被审计的人数。使用泊松分布回答以下问题。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many are expected to be audited?

4. 预计有多少人被审计？

5. Find the probability that no one was audited.

5. 求没有人被审计的概率。

6. Find the probability that at least three were audited.

6. 求至少三人被审计的概率。

126\.

126.

Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.

当地一所高中约有 8% 的学生在高中四年全程参加课后体育运动。随机选取一组 60 名毕业班学生，关注的是在高中四年全程参加课后体育运动的人数。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. How many seniors are expected to have participated in after-school sports all four years of high school?

4. 预计有多少名毕业班学生曾在高中四年全程参加课后体育运动？

5. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

5. 基于数值，若所有毕业班学生都未在高中四年全程参加课后体育运动，你会感到意外吗？用数值说明你的理由。

6. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

6. 基于数值，四名还是五名毕业班学生在高中四年全程参加课后体育运动更有可能？用数值说明你的理由。

127.

On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes.

平均而言，业余厨师 Pierre 每做两盘蛋糕面糊会掉进三片蛋壳。假设你就买了他的一个蛋糕。

1. In words, define the random variable *X*.

1. 用文字定义随机变量 *X*。

2. List the values that *X* may take on.

2. 列出 *X* 可能取的值。

3. Give the distribution of *X*. *X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

3. 给出 *X* 的分布。*X* ~ \_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)

4. On average, how many pieces of egg shell do you expect to be in the cake?

4. 平均而言，你预计蛋糕里会有多少片蛋壳？

5. What is the probability that there will not be any pieces of egg shell in the cake?

5. 蛋糕里没有任何蛋壳片的概率是多少？

6. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes?

6. 假设你连续六周每周都买一个 Pierre 的蛋糕，这些蛋糕中没有任何蛋壳的概率是多少？

7. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?

7. 基于给出的 Pierre 的平均值，蛋糕里可能出现七片蛋壳吗？为什么？

*Use the following information to answer the next two exercises:* The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week.

使用以下信息回答接下来的两道习题：Plum 太太的猫因想玩耍而每晚把她弄醒的平均次数为十次。我们关注她每周被猫弄醒的次数。

128\.

128.

In words, the random variable *X* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

用文字表述，随机变量 *X* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

1. the number of times Mrs. Plum’s cats wake her up each week.

1. Plum 太太的猫每周把她弄醒的次数。

2. the number of times Mrs. Plum’s cats wake her up each hour.

2. Plum 太太的猫每小时把她弄醒的次数。

3. the number of times Mrs. Plum’s cats wake her up each night.

3. Plum 太太的猫每晚把她弄醒的次数。

4. the number of times Mrs. Plum’s cats wake her up.

4. Plum 太太的猫把她弄醒的次数。

129.

Find the probability that her cats will wake her up no more than five times next week.

求她的猫在下周把她弄醒不超过五次的概率。

1. 0.5000

2. 0.9329

3. 0.0378

4. 0.0671

Footnotes 脚注

1"Prevalence of HIV, total (% of populations ages 15-49)," The World Bank, 2013. (accessed May 15, 2013).

1"艾滋病患病率，总计（占 15–49 岁人口百分比）"，世界银行，2013 年。（2013 年 5 月 15 日访问）

4 Discrete Random Variables 离散随机变量

Introduction 引言

Random Variable Notation 随机变量记号

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量的概率分布函数（PDF）

Problem 问题

Solution 解答

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

Problem 问题

Solution 解答

Problem 问题

Solution 解答

Problem 问题

Solution 解答

4.3 Binomial Distribution 4.3 二项分布

Problem 问题

Problem 问题

Solution 解答

Notation for the Binomial: *B* = Binomial Probability Distribution Function 二项分布的记号：*B* = 二项概率分布函数

Problem 问题

Solution 解答

Problem 问题

4.4 Geometric Distribution 4.4 几何分布

Problem 问题

Solution 解答

Notation for the Geometric: G = Geometric Probability Distribution Function 几何分布记号：G = 几何概率分布函数

Problem 问题

Solution 解答

4.5 Hypergeometric Distribution 4.5 超几何分布

Problem 问题

Solution 解答

Problem 问题

Solution 解答

Problem 问题

Solution 解答

Notation for the Hypergeometric: H = Hypergeometric Probability Distribution Function 超几何分布的记号：H = 超几何概率分布函数

4.6 Poisson Distribution 4.6 泊松分布

Problem 问题

Solution 解答

Problem 问题

Solution 解答

Notation for the Poisson: P = Poisson Probability Distribution Function 泊松分布的记号：P = 泊松概率分布函数

Problem 问题

Solution 解答

Problem 问题

Solution 解答

Problem 问题

Solution 解答

4.7 Discrete Distribution (Playing Card Experiment) 4.7 离散分布（扑克牌实验）

Discrete Distribution (Playing Card Experiment) 离散分布（扑克牌实验）

4.8 Discrete Distribution (Lucky Dice Experiment) 4.8 离散分布（幸运骰子实验）

Discrete Distribution (Lucky Dice Experiment) 离散分布（幸运骰子实验）

Key Terms 关键术语

Chapter Review 本章回顾

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量的概率分布函数（PDF）

4.2 Mean or Expected Value and Standard Deviation 4.2 均值（期望值）与标准差

4.3 Binomial Distribution 4.3 二项分布

4.4 Geometric Distribution 4.4 几何分布

4.5 Hypergeometric Distribution 4.5 超几何分布

4.6 Poisson Distribution 4.6 泊松分布

Formula Review 公式回顾

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

4.3 Binomial Distribution 4.3 二项分布

4.4 Geometric Distribution 4.4 几何分布

4.5 Hypergeometric Distribution 4.5 超几何分布

4.6 Poisson Distribution 4.6 泊松分布

Practice 练习

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量概率分布函数（PDF）

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

4.3 Binomial Distribution 4.3 二项分布

4.4 Geometric Distribution 4.4 几何分布

4.5 Hypergeometric Distribution 4.5 超几何分布

4.6 Poisson Distribution 4.6 泊松分布

Homework 作业

4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 4.1 离散随机变量的概率分布函数（PDF）

4.2 Mean or Expected Value and Standard Deviation 4.2 均值或期望值与标准差

4.3 Binomial Distribution 4.3 二项分布

4.4 Geometric Distribution 4.4 几何分布

4.5 Hypergeometric Distribution 4.5 超几何分布

4.6 Poisson Distribution 4.6 泊松分布

Footnotes 脚注

Notation for the Binomial: B = Binomial Probability Distribution Function 二项分布的记号：B = 二项概率分布函数