2 Descriptive Statistics 描述统计学

本页译自 OpenStax《Introductory Statistics》第 2 章 Descriptive Statistics。公式经本地 MathJax 渲染，自定义宏已注入。

Introduction 引言

Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data.

In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called "Descriptive Statistics." You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs.

A statistical graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied.

Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs. Our emphasis will be on histograms and box plots.

2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 2.1 茎叶图（Stemplots）、线图与条形图

One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.

For Susan Dean's spring pre-calculus class, scores for the first exam were as follows (smallest to largest):

33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100

| Stem | Leaf |

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

| 3 | 3 |

| 4 | 2 9 9 |

| 5 | 3 5 5 |

| 6 | 1 3 7 8 8 9 9 |

| 7 | 2 3 4 8 |

| 8 | 0 3 8 8 8 |

| 9 | 0 2 4 4 4 4 6 |

| 10 | 0 |

Table 2.1 Stem-and-Leaf Graph

The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% $\left( \frac{8}{31} \right)$ were in the 90s or 100, a fairly high number of As.

For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):

32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61

Construct a stem plot for the data.

The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later.

The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data:

1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3

Problem 问题

Do the data seem to have any concentration of values?

The leaves are to the right of the decimal.

Solution 解答

The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers.

| Stem | Leaf |

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

| 1 | 1 5 |

| 2 | 3 5 7 |

| 3 | 2 3 3 5 8 |

| 4 | 0 2 5 5 7 8 |

| 5 | 5 6 |

| 6 | 5 7 |

| 7 | |

| 8 | |

| 9 | |

| 10 | |

| 11 | |

| 12 | 3 |

Table 2.2

The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers:

0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0

Problem 问题

A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. Table 2.4 and Table 2.5 show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data.

Solution 解答

| Ages at Inauguration | | Ages at Death |

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

| 9 9 8 7 7 7 6 3 2 | 4 | 6 9 |

| 8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 2 1 1 1 1 1 0 | 5 | 3 6 6 7 7 8 |

| 9 8 5 4 4 2 1 1 1 0 | 6 | 0 0 3 3 4 4 5 6 7 7 7 8 |

| | 7 | 0 1 1 1 2 3 4 7 8 8 9 |

| | 8 | 0 1 3 5 8 |

| | 9 | 0 0 3 3 |

Table 2.3

| President | Age | President | Age | President | Age |

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

| Washington | 57 | Lincoln | 52 | Hoover | 54 |

| J. Adams | 61 | A. Johnson | 56 | F. Roosevelt | 51 |

| Jefferson | 57 | Grant | 46 | Truman | 60 |

| Madison | 57 | Hayes | 54 | Eisenhower | 62 |

| Monroe | 58 | Garfield | 49 | Kennedy | 43 |

| J. Q. Adams | 57 | Arthur | 51 | L. Johnson | 55 |

| Jackson | 61 | Cleveland | 47 | Nixon | 56 |

| Van Buren | 54 | B. Harrison | 55 | Ford | 61 |

| W. H. Harrison | 68 | Cleveland | 55 | Carter | 52 |

| Tyler | 51 | McKinley | 54 | Reagan | 69 |

| Polk | 49 | T. Roosevelt | 42 | G.H.W. Bush | 64 |

| Taylor | 64 | Taft | 51 | Clinton | 47 |

| Fillmore | 50 | Wilson | 56 | G. W. Bush | 54 |

| Pierce | 48 | Harding | 55 | Obama | 47 |

| Buchanan | 65 | Coolidge | 51 | | |

Table 2.4 Presidential Ages at Inauguration

| President | Age | President | Age | President | Age |

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

| Washington | 67 | Lincoln | 56 | Hoover | 90 |

| J. Adams | 90 | A. Johnson | 66 | F. Roosevelt | 63 |

| Jefferson | 83 | Grant | 63 | Truman | 88 |

| Madison | 85 | Hayes | 70 | Eisenhower | 78 |

| Monroe | 73 | Garfield | 49 | Kennedy | 46 |

| J. Q. Adams | 80 | Arthur | 56 | L. Johnson | 64 |

| Jackson | 78 | Cleveland | 71 | Nixon | 81 |

| Van Buren | 79 | B. Harrison | 67 | Ford | 93 |

| W. H. Harrison | 68 | Cleveland | 71 | Reagan | 93 |

| Tyler | 71 | McKinley | 58 | | |

| Polk | 53 | T. Roosevelt | 60 | | |

| Taylor | 65 | Taft | 72 | | |

| Fillmore | 74 | Wilson | 67 | | |

| Pierce | 64 | Harding | 57 | | |

| Buchanan | 77 | Coolidge | 60 | | |

Table 2.5 Presidential Age at Death

The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stem-and-leaf plot of these wins and losses.

| Losses | Wins | Year | Losses | Wins | Year |

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

| 34 | 48 | 1968–1969 | 41 | 41 | 1989–1990 |

| 34 | 48 | 1969–1970 | 39 | 43 | 1990–1991 |

| 46 | 36 | 1970–1971 | 44 | 38 | 1991–1992 |

| 46 | 36 | 1971–1972 | 39 | 43 | 1992–1993 |

| 36 | 46 | 1972–1973 | 25 | 57 | 1993–1994 |

| 47 | 35 | 1973–1974 | 40 | 42 | 1994–1995 |

| 51 | 31 | 1974–1975 | 36 | 46 | 1995–1996 |

| 53 | 29 | 1975–1976 | 26 | 56 | 1996–1997 |

| 51 | 31 | 1976–1977 | 32 | 50 | 1997–1998 |

| 41 | 41 | 1977–1978 | 19 | 31 | 1998–1999 |

| 36 | 46 | 1978–1979 | 54 | 28 | 1999–2000 |

| 32 | 50 | 1979–1980 | 57 | 25 | 2000–2001 |

| 51 | 31 | 1980–1981 | 49 | 33 | 2001–2002 |

| 40 | 42 | 1981–1982 | 47 | 35 | 2002–2003 |

| 39 | 43 | 1982–1983 | 54 | 28 | 2003–2004 |

| 42 | 40 | 1983–1984 | 69 | 13 | 2004–2005 |

| 48 | 34 | 1984–1985 | 56 | 26 | 2005–2006 |

| 32 | 50 | 1985–1986 | 52 | 30 | 2006–2007 |

| 25 | 57 | 1986–1987 | 45 | 37 | 2007–2008 |

| 32 | 50 | 1987–1988 | 35 | 47 | 2008–2009 |

| 30 | 52 | 1988–1989 | 29 | 53 | 2009–2010 |

Table 2.6

Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in Example 2.4, the ***x*-axis (horizontal axis) consists of data values and the *y*-axis (vertical axis) consists of frequency points**. The frequency points are connected using line segments.

In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in Table 2.7 and in Figure 2.2.

| Number of times teenager is reminded | Frequency |

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

| 0 | 2 |

| 1 | 5 |

| 2 | 8 |

| 3 | 14 |

| 4 | 7 |

| 5 | 4 |

Table 2.7

In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in Table 2.8. Construct a line graph.

| Number of times in shop | Frequency |

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

| 0 | 7 |

| 1 | 10 |

| 2 | 14 |

| 3 | 9 |

Table 2.8

Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in Example 2.5 has age groups represented on the ***x*-axis and proportions on the *y*-axis**.

Problem 问题

By the end of 2011, Facebook had over 146 million users in the United States. Table 2.9 shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph using this data.

| Age groups | Number of Facebook users | Proportion (%) of Facebook users |

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

| 13–25 | 65,082,280 | 45% |

| 26–44 | 53,300,200 | 36% |

| 45–64 | 27,885,100 | 19% |

Table 2.9

Solution 解答

The population in Park City is made up of children, working-age adults, and retirees. Table 2.10 shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions.

| Age groups | Number of people | Proportion of population |

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

| Children | 67,059 | 19% |

| Working-age adults | 152,198 | 43% |

| Retirees | 131,662 | 38% |

Table 2.10

Problem 问题

The columns in Table 2.11 contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the *x*-axis, and the Advanced Placement examinee population percentages on the *y*-axis.

| Race/Ethnicity | AP Examinee Population | Overall Student Population |

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

| 1 = Asian, Asian American or Pacific Islander | 10.3% | 5.7% |

| 2 = Black or African American | 9.0% | 14.7% |

| 3 = Hispanic or Latino | 17.0% | 17.6% |

| 4 = American Indian or Alaska Native | 0.6% | 1.1% |

| 5 = White | 57.1% | 59.2% |

| 6 = Not reported/other | 6.0% | 1.7% |

Table 2.11

Solution 解答

Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district.

| District | Registered voter population | Overall city population |

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

| 1 | 15.5% | 19.4% |

| 2 | 12.2% | 15.6% |

| 3 | 9.8% | 9.0% |

| 4 | 17.4% | 18.5% |

| 5 | 22.8% | 20.7% |

| 6 | 22.3% | 16.8% |

Table 2.12

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2.2 Histograms, Frequency Polygons, and Time Series Graphs 2.2 直方图、频数多边形与时间序列图

For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more.

A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data.

The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample. (Remember, frequency is defined as the number of times an answer occurs.) If:

• *f* = frequency
• *n* = total number of data values (or the sum of the individual frequencies), and
• *RF* = relative frequency,

• *f* = 频数
• *n* = 数据值的总数（或各频数之和），且
• *RF* = 相对频数，

then:

$${\text{RF} = \frac{f}{n}}{}$$

For example, if three students in Mr. Ahab's English class of 40 students received from 90% to 100%, then, *f* = 3, *n* = 40, and *RF* = $\frac{f}{n}$ = $\frac{3}{40}$ = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures.

To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data.

The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured.

60; 60.5; 61; 61; 61.5

63.5; 63.5; 63.5

64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5

66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5

68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5

70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71

72; 72; 72; 72.5; 72.5; 73; 73.5

74

The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.

60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95.

The largest value is 74, so 74 + 0.05 = 74.05 is the ending value.

Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars.

$$\begin{matrix}{74.05 - 59.95 = 14.1} \\{14.1 \div 8 = 1.76}\end{matrix}$$

We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the number of bars or class intervals is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals.

The boundaries are:

• 59.95
• 59.95 + 2 = 61.95
• 61.95 + 2 = 63.95
• 63.95 + 2 = 65.95
• 65.95 + 2 = 67.95
• 67.95 + 2 = 69.95
• 69.95 + 2 = 71.95
• 71.95 + 2 = 73.95
• 73.95 + 2 = 75.95

• 59.95
• 59.95 + 2 = 61.95
• 61.95 + 2 = 63.95
• 63.95 + 2 = 65.95
• 65.95 + 2 = 67.95
• 67.95 + 2 = 69.95
• 69.95 + 2 = 71.95
• 71.95 + 2 = 73.95
• 73.95 + 2 = 75.95

The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.

The following histogram displays the heights on the *x*-axis and relative frequency on the *y*-axis.

The following data are the shoe sizes of 50 male students. The sizes are discrete data since shoe size is measured in whole and half units only. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.

9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5

11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5

12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14

Create a histogram for the following data: the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data, since books are counted.

1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1

2; 2; 2; 2; 2; 2; 2; 2; 2; 2

3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3

4; 4; 4; 4; 4; 4

5; 5; 5; 5; 5

6; 6

Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books.

Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.

Problem 问题

Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from \_\_\_\_\_\_\_ to \_\_\_\_\_\_\_, the 5 in the middle of the interval from \_\_\_\_\_\_\_ to \_\_\_\_\_\_\_, and the \_\_\_\_\_\_\_ in the middle of the interval from \_\_\_\_\_\_\_ to \_\_\_\_\_\_\_ .

Solution 解答

• 3.5 to 4.5
• 4.5 to 5.5
• 6
• 5.5 to 6.5

• 3.5 至 4.5
• 4.5 至 5.5
• 6
• 5.5 至 6.5

Calculate the number of bars as follows:

$$\begin{matrix}{6.5 - 0.5 = 6} \\{6 \div 1 = 6}\end{matrix}$$

where 1 is the width of a bar. Therefore, bars = 6.

The following histogram displays the number of books on the *x*-axis and the frequency on the *y*-axis.

Go to Appendix G Notes for the TI-83, 83+, 84, 84+ Calculators. There are calculator instructions for entering data and for creating a customized histogram. Create the histogram for Example 2.8.

• Press Y=. Press CLEAR to delete any equations.
• Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1, press CLEAR and then arrow down. If necessary, do the same for L2.
• Into L1, enter 1, 2, 3, 4, 5, 6.
• Into L2, enter 11, 10, 16, 6, 5, 2.
• Press WINDOW. Set Xmin = .5, Xmax = 6.5, Xscl = (6.5 – .5)/6, Ymin = –1, Ymax = 20, Yscl = 1, Xres = 1.
• Press 2^nd Y=. Start by pressing 4:Plotsoff ENTER.
• Press 2^nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE. Arrow to the 3^rd picture (histogram). Press ENTER.
• Arrow down to Xlist: Enter L1 (2^nd 1). Arrow down to Freq. Enter L2 (2^nd 2).
• Press GRAPH.
• Use the TRACE key and the arrow keys to examine the histogram.

• 按 Y=。按 CLEAR 删除任何方程。
• 按 STAT 1:EDIT。若 L1 中已有数据，将光标上移到 L1 的名称处，按 CLEAR 再下移。必要时对 L2 做同样操作。
• 在 L1 中输入 1、2、3、4、5、6。
• 在 L2 中输入 11、10、16、6、5、2。
• 按 WINDOW。设置 Xmin = .5，Xmax = 6.5，Xscl = (6.5 – .5)/6，Ymin = –1，Ymax = 20，Yscl = 1，Xres = 1。
• 按 2^nd Y=。首先按 4:Plotsoff ENTER。
• 按 2^nd Y=。按 1:Plot1。按 ENTER。下移光标至 TYPE。右移光标至第 3^rd 个图（直方图）。按 ENTER。
• 下移光标至 Xlist：输入 L1（2^nd 1）。下移光标至 Freq。输入 L2（2^nd 2）。
• 按 GRAPH。
• 使用 TRACE 键和箭头键查看直方图。

The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted.

1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1

2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2

3; 3; 3; 3; 3; 3; 3; 3

20 student athletes play one sport. 22 student athletes play two sports. Eight student athletes play three sports.

*Fill in the blanks for the following sentence.* Since the data consist of the numbers 1, 2, 3, and the starting point is 0.5, a width of one places the 1 in the middle of the interval 0.5 to \_\_\_\_\_, the 2 in the middle of the interval from \_\_\_\_\_ to \_\_\_\_\_, and the 3 in the middle of the interval from \_\_\_\_\_ to \_\_\_\_\_.

Problem 问题

Using this data set, construct a histogram.

| Number of Hours My Classmates Spent Playing Video Games on Weekends | | | | |

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

| 9.95 | 10 | 2.25 | 16.75 | 0 |

| 19.5 | 22.5 | 7.5 | 15 | 12.75 |

| 5.5 | 11 | 10 | 20.75 | 17.5 |

| 23 | 21.9 | 24 | 23.75 | 18 |

| 20 | 15 | 22.9 | 18.8 | 20.5 |

Table 2.13

Solution 解答

Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram.

The following data represent the number of employees at various restaurants in New York City. Using this data, create a histogram.

22; 35; 15; 26; 40; 28; 18; 20; 25; 34; 39; 42; 24; 22; 19; 27; 22; 34; 40; 20; 38; and 28

Use 10–19 as the first interval.

Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals.

Frequency Polygons 频数多边形

Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons.

To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the *x*-axis and *y*-axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them.

A frequency polygon was constructed from the frequency table below.

| Frequency Distribution for Calculus Final Test Scores | | | |

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

| Lower Bound | Upper Bound | Frequency | Cumulative Frequency |

| 49.5 | 59.5 | 5 | 5 |

| 59.5 | 69.5 | 10 | 15 |

| 69.5 | 79.5 | 30 | 45 |

| 79.5 | 89.5 | 40 | 85 |

| 89.5 | 99.5 | 15 | 100 |

Table 2.14

The first label on the *x*-axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the *x*-axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the *x*-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.

Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in Table 2.15.

| Age at Inauguration | Frequency |

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

| 41.5–46.5 | 4 |

| 46.5–51.5 | 11 |

| 51.5–56.5 | 14 |

| 56.5–61.5 | 9 |

| 61.5–66.5 | 4 |

| 66.5–71.5 | 2 |

Table 2.15

Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets.

We will construct an overlay frequency polygon comparing the scores from Example 2.10 with the students’ final numeric grade.

| Frequency Distribution for Calculus Final Test Scores | | | |

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

| Lower Bound | Upper Bound | Frequency | Cumulative Frequency |

| 49.5 | 59.5 | 5 | 5 |

| 59.5 | 69.5 | 10 | 15 |

| 69.5 | 79.5 | 30 | 45 |

| 79.5 | 89.5 | 40 | 85 |

| 89.5 | 99.5 | 15 | 100 |

Table 2.16

| Frequency Distribution for Calculus Final Grades | | | |

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

| Lower Bound | Upper Bound | Frequency | Cumulative Frequency |

| 49.5 | 59.5 | 10 | 10 |

| 59.5 | 69.5 | 10 | 20 |

| 69.5 | 79.5 | 30 | 50 |

| 79.5 | 89.5 | 45 | 95 |

| 89.5 | 99.5 | 5 | 100 |

Table 2.17

Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with this data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected.

One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph.

Constructing a Time Series Graph 构建时间序列图

To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur.

Problem 问题

The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only.

| Year | Jan | Feb | Mar | Apr | May | Jun | Jul |

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

| 2003 | 181.7 | 183.1 | 184.2 | 183.8 | 183.5 | 183.7 | 183.9 |

| 2004 | 185.2 | 186.2 | 187.4 | 188.0 | 189.1 | 189.7 | 189.4 |

| 2005 | 190.7 | 191.8 | 193.3 | 194.6 | 194.4 | 194.5 | 195.4 |

| 2006 | 198.3 | 198.7 | 199.8 | 201.5 | 202.5 | 202.9 | 203.5 |

| 2007 | 202.416 | 203.499 | 205.352 | 206.686 | 207.949 | 208.352 | 208.299 |

| 2008 | 211.080 | 211.693 | 213.528 | 214.823 | 216.632 | 218.815 | 219.964 |

| 2009 | 211.143 | 212.193 | 212.709 | 213.240 | 213.856 | 215.693 | 215.351 |

| 2010 | 216.687 | 216.741 | 217.631 | 218.009 | 218.178 | 217.965 | 218.011 |

| 2011 | 220.223 | 221.309 | 223.467 | 224.906 | 225.964 | 225.722 | 225.922 |

| 2012 | 226.665 | 227.663 | 229.392 | 230.085 | 229.815 | 229.478 | 229.104 |

Table 2.18

| Year | Aug | Sep | Oct | Nov | Dec | Annual |

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

| 2003 | 184.6 | 185.2 | 185.0 | 184.5 | 184.3 | 184.0 |

| 2004 | 189.5 | 189.9 | 190.9 | 191.0 | 190.3 | 188.9 |

| 2005 | 196.4 | 198.8 | 199.2 | 197.6 | 196.8 | 195.3 |

| 2006 | 203.9 | 202.9 | 201.8 | 201.5 | 201.8 | 201.6 |

| 2007 | 207.917 | 208.490 | 208.936 | 210.177 | 210.036 | 207.342 |

| 2008 | 219.086 | 218.783 | 216.573 | 212.425 | 210.228 | 215.303 |

| 2009 | 215.834 | 215.969 | 216.177 | 216.330 | 215.949 | 214.537 |

| 2010 | 218.312 | 218.439 | 218.711 | 218.803 | 219.179 | 218.056 |

| 2011 | 226.545 | 226.889 | 226.421 | 226.230 | 225.672 | 224.939 |

| 2012 | 230.379 | 231.407 | 231.317 | 230.221 | 229.601 | 229.594 |

Table 2.19

Solution 解答

The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO₂ emissions for the United States.

| CO2 Emissions | | | |

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

| | Ukraine | United Kingdom | United States |

| 2003 | 352,259 | 540,640 | 5,681,664 |

| 2004 | 343,121 | 540,409 | 5,790,761 |

| 2005 | 339,029 | 541,990 | 5,826,394 |

| 2006 | 327,797 | 542,045 | 5,737,615 |

| 2007 | 328,357 | 528,631 | 5,828,697 |

| 2008 | 323,657 | 522,247 | 5,656,839 |

| 2009 | 272,176 | 474,579 | 5,299,563 |

Table 2.20

Uses of a Time Series Graph 时间序列图的用途

Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot.

---

2.3 Measures of the Location of the Data 2.3 数据的位置度量

The common measures of location are quartiles and percentiles

Quartiles are special percentiles. The first quartile, *Q*₁, is the same as the 25^th percentile, and the third quartile, *Q*₃, is the same as the 75^th percentile. The median, *M*, is called both the second quartile and the 50^th percentile.

To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90^th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.

Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75^th percentile. That translates into a score of at least 1220.

Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.

The median is a number that measures the "center" of the data. You can think of the median as the "middle value," but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data.

1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1

Ordered from smallest to largest:

1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5

Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.

$$\begin{matrix}{6.8 + 7.2 = 14} \\{14 \div 2 = 7}\end{matrix}$$

The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.

Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, *Q*₁, is the middle value of the lower half of the data, and the third quartile, *Q*₃, is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set:

1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5

The median or second quartile is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two.

1; 1; 2; 2; 4; 6; 6.8

The number two, which is part of the data, is the first quartile. One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two.

The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine.

The third quartile, *Q*3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example.

The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile (*Q*₃) and the first quartile (*Q*₁).

*IQR* = *Q*₃ – *Q*₁

The *IQR* can help to determine potential outliers. **A value is suspected to be a potential outlier if it is less than (1.5)(*IQR*) below the first quartile or more than (1.5)(*IQR*) above the third quartile**. Potential outliers always require further investigation.

A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data.

Problem 问题

For the following 13 real estate prices, calculate the *IQR* and determine if any prices are potential outliers. Prices are in dollars.

389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000

Solution 解答

Order the data from smallest to largest.

114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000

*M* = 488,800

*Q*₁ = $\frac{\text{230,500~+~387,000}}{2}$ = 308,750

*Q*₃ = $\frac{\text{639,000~+~659,000}}{2}$ = 649,000

*IQR* = 649,000 – 308,750 = 340,250

(1.5)(*IQR*) = (1.5)(340,250) = 510,375

*Q*₁ – (1.5)(*IQR*) = 308,750 – 510,375 = –201,625

*Q*₃ + (1.5)(*IQR*) = 649,000 + 510,375 = 1,159,375

No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier.

For the following 11 salaries, calculate the *IQR* and determine if any salaries are outliers. The salaries are in dollars.

\$33,000; \$64,500; \$28,000; \$54,000; \$72,000; \$68,500; \$69,000; \$42,000; \$54,000; \$120,000; \$40,500

Problem 问题

For the two data sets in the test scores example, find the following:

1. The interquartile range. Compare the two interquartile ranges.

2. Any outliers in either set.

Solution 解答

The five number summary for the day and night classes is

| | Minimum | *Q*₁ | Median | *Q*₃ | Maximum |

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

| Day | 32 | 56 | 74.5 | 82.5 | 99 |

| Night | 25.5 | 78 | 81 | 89 | 98 |

Table 2.21

1. The IQR for the day group is *Q*₃ – *Q*₁ = 82.5 – 56 = 26.5

The IQR for the night group is *Q*₃ – *Q*₁ = 89 – 78 = 11

The interquartile range (the spread or variability) for the day class is larger than the night class *IQR*. This suggests more variation will be found in the day class’s class test scores.

2. Day class outliers are found using the IQR times 1.5 rule. So,

• *Q*₁ - *IQR*(1.5) = 56 – 26.5(1.5) = 16.25
• *Q*₃ + *IQR*(1.5) = 82.5 + 26.5(1.5) = 122.25

• *Q*₁ – IQR(1.5) = 56 – 26.5(1.5) = 16.25
• *Q*₃ + IQR(1.5) = 82.5 + 26.5(1.5) = 122.25

Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers.

Night class outliers are calculated as:

• *Q*₁ – *IQR* (1.5) = 78 – 11(1.5) = 61.5
• *Q*₃ + IQR(1.5) = 89 + 11(1.5) = 105.5

• *Q*₁ – IQR(1.5) = 78 – 11(1.5) = 61.5
• *Q*₃ + IQR(1.5) = 89 + 11(1.5) = 105.5

For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier.

Find the interquartile range for the following two data sets and compare them.

Test Scores for Class *A*

69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94

Test Scores for Class *B*

90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100

Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were:

| AMOUNT OF SLEEP PER SCHOOL NIGHT (HOURS) | FREQUENCY | RELATIVE FREQUENCY | CUMULATIVE RELATIVE FREQUENCY |

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

| 4 | 2 | 0.04 | 0.04 |

| 5 | 5 | 0.10 | 0.14 |

| 6 | 7 | 0.14 | 0.28 |

| 7 | 12 | 0.24 | 0.52 |

| 8 | 14 | 0.28 | 0.80 |

| 9 | 7 | 0.14 | 0.94 |

| 10 | 3 | 0.06 | 1.00 |

Table 2.22

Find the 28^th percentile. Notice the 0.28 in the "cumulative relative frequency" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28^th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28^th percentile is between the last six and the first seven. The 28^th percentile is 6.5.

Find the median. Look again at the "cumulative relative frequency" column and find 0.52. The median is the 50^th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50^th percentile is between the 25^th, or seven, and 26^th, or seven, values. The median is seven.

Find the third quartile. The third quartile is the same as the 75^th percentile. You can "eyeball" this answer. If you look at the "cumulative relative frequency" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75^th percentile, then, must be an eight. Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, *Q*₃, is the 38^th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.)

Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65^th percentile.

| Amount of time spent on route (hours) | Frequency | Relative Frequency | Cumulative Relative Frequency |

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

| 2 | 12 | 0.30 | 0.30 |

| 3 | 14 | 0.35 | 0.65 |

| 4 | 10 | 0.25 | 0.90 |

| 5 | 4 | 0.10 | 1.00 |

Table 2.23

Problem 问题

Using Table 2.22:

1. Find the 80^th percentile.

2. Find the 90^th percentile.

3. Find the first quartile. What is another name for the first quartile?

Solution 解答

Using the data from the frequency table, we have:

1. The 80^th percentile is between the last eight and the first nine in the table (between the 40^th and 41^st values). Therefore, we need to take the mean of the 40^th an 41^st values. The 80^th percentile $= \frac{8 + 9}{2} = 8.5$

2. The 90^th percentile will be the 45^th data value (location is 0.90(50) = 45) and the 45^th data value is nine.

3. *Q*₁ is also the 25^th percentile. The 25^th percentile location calculation: *P*₂₅ = 0.25(50) = 12.5 ≈ 13 the 13^th data value. Thus, the 25th percentile is six.

Refer to the Table 2.23. Find the third quartile. What is another name for the third quartile?

Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions:

1. How many students were surveyed?

2. What kind of sampling did you do?

3. Construct two different histograms. For each, starting value = \_\_\_\_\_ ending value = \_\_\_\_.

4. Find the median, first quartile, and third quartile.

5. Construct a table of the data to find the following:

1. the 10^th percentile

2. the 70^th percentile

3. the percent of students who own less than four sweaters

A Formula for Finding the *k*th Percentile 求第 k 百分位数的公式

If you were to do a little research, you would find several formulas for calculating the *k*^th percentile. Here is one of them.

*k* = the *k^th* percentile. It may or may not be part of the data.

*i* = the index (ranking or position of a data value)

*n* = the total number of data

• Order the data from smallest to largest.
• Calculate $i = \frac{k}{100}(n + 1)$
• If *i* is an integer, then the *k^th* percentile is the data value in the *i^th* position in the ordered set of data.
• If *i* is not an integer, then round *i* up and round *i* down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example.

• 将数据从小到大排序。
• 计算 $i = \frac{k}{100}(n + 1)$
• 若 *i* 为整数，则第 *k*^th 百分位数就是有序数据集中位于第 *i*^th 个位置的数据值。
• 若 *i* 不是整数，则将 *i* 向上和向下取整到最接近的整数。取有序数据集中这两个位置上的两个数据值的平均数。通过例子更容易理解。

Problem 问题

Listed are 29 ages for Academy Award winning best actors *in order from smallest to largest.*

18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

1. Find the 70^th percentile.

2. Find the 83^rd percentile.

Solution 解答

1. - *k* = 70

• *i* = the index
• *n* = 29

• *i* = 索引
• *n* = 29

*i* = $\frac{k}{100}$ (*n* + 1) = ($\frac{70}{100}$)(29 + 1) = 21. Twenty-one is an integer, and the data value in the 21^st position in the ordered data set is 64. The 70^th percentile is 64 years.

2. - *k* = 83^rd percentile

• *i* = the index
• *n* = 29

• *i* = 索引
• *n* = 29

*i* = $\frac{k}{100}$ (*n* + 1) = ($\frac{83}{100}$)(29 + 1) = 24.9, which is NOT an integer. Round it down to 24 and up to 25. The age in the 24^th position is 71 and the age in the 25^th position is 72. Average 71 and 72. The 83^rd percentile is 71.5 years.

Listed are 29 ages for Academy Award winning best actors *in order from smallest to largest.*

18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

Calculate the 20^th percentile and the 55^th percentile.

You can calculate percentiles using calculators and computers. There are a variety of online calculators.

A Formula for Finding the Percentile of a Value in a Data Set 求数据集中某一数值百分位数的公式

• Order the data from smallest to largest.
• *x* = the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile.
• *y* = the number of data values equal to the data value for which you want to find the percentile.
• *n* = the total number of data.
• Calculate $\frac{x + 0.5y}{n}$(100). Then round to the nearest integer.

• 将数据从小到大排序。
• *x* = 从数据列表底部往上数、直到（但不包含）你想求百分位数的那个数据值的数据个数。
• *y* = 等于你想求百分位数的那个数据值的数据个数。
• *n* = 数据的总个数。
• 计算 $\frac{x + 0.5y}{n}$(100)。然后四舍五入到最接近的整数。

Problem 问题

Listed are 29 ages for Academy Award winning best actors *in order from smallest to largest.*

18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

1. Find the percentile for 58.

2. Find the percentile for 25.

Solution 解答

1. Counting from the bottom of the list, there are 18 data values less than 58. There is one value of 58.

*x* = 18 and *y* = 1.$\frac{x + 0.5y}{n}$(100) = $\frac{18 + 0.5(1)}{29}$(100) = 63.80. 58 is the 64^th percentile.

2. Counting from the bottom of the list, there are three data values less than 25. There is one value of 25.

*x* = 3 and *y* = 1.$\frac{x + 0.5y}{n}$(100) = $\frac{3 + 0.5(1)}{29}$(100) = 12.07. Twenty-five is the 12^th percentile.

Listed are 30 ages for Academy Award winning best actors in order from smallest to largest.

18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

Find the percentiles for 47 and 31.

Interpreting Percentiles, Quartiles, and Median 解释百分位数、四分位数与中位数

A percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the pth percentile. For example, 15% of data values are less than or equal to the 15^th percentile.

• Low percentiles always correspond to lower data values.
• High percentiles always correspond to higher data values.

• 较低的百分位数总是对应于较小的数据值。
• 较高的百分位数总是对应于较大的数据值。

A percentile may or may not correspond to a value judgment about whether it is "good" or "bad." The interpretation of whether a certain percentile is "good" or "bad" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered "good;" in other contexts a high percentile might be considered "good". In many situations, there is no value judgment that applies.

Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text.

When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information.

• information about the context of the situation being considered
• the data value (value of the variable) that represents the percentile
• the percent of individuals or items with data values below the percentile
• the percent of individuals or items with data values above the percentile.

• 所考虑情境的背景信息
• 代表该百分位数的数据值（即变量的值）
• 数据值低于该百分位数的个体或物品所占的百分比
• 数据值高于该百分位数的个体或物品所占的百分比

Problem 问题

On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation.

Solution 解答

• Twenty-five percent of students finished the exam in 35 minutes or less.
• Seventy-five percent of students finished the exam in 35 minutes or more.
• A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)

• 25% 的学生在 35 分钟或更短时间内完成了考试。
• 75% 的学生在 35 分钟或更长时间内完成了考试。
• 较低的百分位数可被认为是好的，因为在限时考试中完成得更快是理想的（如果花太久，你可能答不完）。

For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation.

Problem 问题

On a 20 question math test, the 70^th percentile for number of correct answers was 16. Interpret the 70^th percentile in the context of this situation.

On a 60 point written assignment, the 80^th percentile for the number of points earned was 49. Interpret the 80^th percentile in the context of this situation.

Problem 问题

At a community college, it was found that the 30^th percentile of credit units that students are enrolled for is seven units. Interpret the 30^th percentile in the context of this situation.

During a season, the 40^th percentile for points scored per player in a game is eight. Interpret the 40^th percentile in the context of this situation.

Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown.

0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes

10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes;

30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes

Determine the following five values.

• Min = 0
• *Q*₁ = 20
• Med = 40
• *Q*₃ = 60
• Max = 300

• Min = 0
• *Q*₁ = 20
• Med = 40
• *Q*₃ = 60
• Max = 300

If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the *IQR* is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.

However, the principal needs to be careful. The value 300 appears to be a potential outlier.

*Q*₃ + 1.5(*IQR*) = 60 + (1.5)(40) = 120.

The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values:

• Min = 0
• *Q*₁ = 20
• *Q*₃ = 60
• Max = 120

• Min = 0
• *Q*₁ = 20
• *Q*₃ = 60
• Max = 120

We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results.

2.4 Box Plots 2.4 箱线图

Box plots (also called box-and-whisker plots or box-whisker plots) give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. A box plot is constructed from five values: the minimum value, the first quartile, the median, the third quartile, and the maximum value. We use these values to compare how close other data values are to them.

To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The "whiskers" extend from the ends of the box to the smallest and largest data values. The median or second quartile can be between the first and third quartiles, or it can be one, or the other, or both. The box plot gives a good, quick picture of the data.

You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values.

Consider, again, this dataset.

1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5

The first quartile is two, the median is seven, and the third quartile is nine. The smallest value is one, and the largest value is 11.5. The following image shows the constructed box plot.

See the calculator instructions on the TI web site or in the appendix.

The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line.

It is important to start a box plot with a scaled number line. Otherwise the box plot may not be useful.

The following data are the heights of 40 students in a statistics class.

59; 60; 61; 62; 62; 63; 63; 64; 64; 64; 65; 65; 65; 65; 65; 65; 65; 65; 65; 66; 66; 67; 67; 68; 68; 69; 70; 70; 70; 70; 70; 71; 71; 72; 72; 73; 74; 74; 75; 77

Construct a box plot with the following properties; the calculator intructions for the minimum and maximum values as well as the quartiles follow the example.

• Minimum value = 59
• Maximum value = 77
• *Q*1: First quartile = 64.5
• *Q*2: Second quartile or median= 66
• *Q*3: Third quartile = 70

• 最小值 = 59
• 最大值 = 77
• *Q*1：第一四分位数 = 64.5
• *Q*2：第二四分位数或中位数 = 66
• *Q*3：第三四分位数 = 70

1. Each quarter has approximately 25% of the data.

2. The spreads of the four quarters are 64.5 – 59 = 5.5 (first quarter), 66 – 64.5 = 1.5 (second quarter), 70 – 66 = 4 (third quarter), and 77 – 70 = 7 (fourth quarter). So, the second quarter has the smallest spread and the fourth quarter has the largest spread.

3. Range = maximum value – the minimum value = 77 – 59 = 18

4. Interquartile Range: *IQR* = *Q*3 – *Q*1 = 70 – 64.5 = 5.5.

5. The interval 59–65 has more than 25% of the data so it has more data in it than the interval 66 through 70 which has 25% of the data.

6. The middle 50% (middle half) of the data has a range of 5.5 inches.

To find the minimum, maximum, and quartiles:

Enter data into the list editor (Pres STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, and then arrow down.

Put the data values into the list L1.

Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.

Press ENTER.

Use the down and up arrow keys to scroll.

Smallest value = 59.

Largest value = 77.

*Q*₁: First quartile = 64.5.

*Q*₂: Second quartile or median = 66.

*Q*₃: Third quartile = 70.

To construct the box plot:

Press 4:Plotsoff. Press ENTER.

Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER.

Arrow down to Xlist: Press 2nd 1 for L1

Arrow down to Freq: Press ALPHA. Press 1.

Press Zoom. Press 9: ZoomStat.

Press TRACE, and use the arrow keys to examine the box plot.

The following data are the number of pages in 40 books on a shelf. Construct a box plot using a graphing calculator, and state the interquartile range.

136; 140; 178; 190; 205; 215; 217; 218; 232; 234; 240; 255; 270; 275; 290; 301; 303; 315; 317; 318; 326; 333; 343; 349; 360; 369; 377; 388; 391; 392; 398; 400; 402; 405; 408; 422; 429; 450; 475; 512

For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like:

In this case, at least 25% of the values are equal to one. Twenty-five percent of the values are between one and five, inclusive. At least 25% of the values are equal to five. The top 25% of the values fall between five and seven, inclusive.

Test scores for a college statistics class held during the day are:

99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90

Test scores for a college statistics class held during the evening are:

98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5

Problem 问题

1. Find the smallest and largest values, the median, and the first and third quartile for the day class.

2. Find the smallest and largest values, the median, and the first and third quartile for the night class.

3. For each data set, what percentage of the data is between the smallest value and the first quartile? the first quartile and the median? the median and the third quartile? the third quartile and the largest value? What percentage of the data is between the first quartile and the largest value?

4. Create a box plot for each set of data. Use one number line for both box plots.

5. Which box plot has the widest spread for the middle 50% of the data (the data between the first and third quartiles)? What does this mean for that set of data in comparison to the other set of data?

Solution 解答

1. - Min = 32

• *Q*₁ = 56
• *M* = 74.5
• *Q*₃ = 82.5
• Max = 99

• *Q*₁ = 56
• *M* = 74.5
• *Q*₃ = 82.5
• 最大值 = 99

2. - Min = 25.5

• *Q*₁ = 78
• *M* = 81
• *Q*₃ = 89
• Max = 98

• *Q*₁ = 78
• *M* = 81
• *Q*₃ = 89
• 最大值 = 98

3. Day class: There are six data values ranging from 32 to 56: 30%. There are six data values ranging from 56 to 74.5: 30%. There are five data values ranging from 74.5 to 82.5: 25%. There are five data values ranging from 82.5 to 99: 25%. There are 16 data values between the first quartile, 56, and the largest value, 99: 75%. Night class:

4.

5. The first data set has the wider spread for the middle 50% of the data. The *IQR* for the first data set is greater than the *IQR* for the second set. This means that there is more variability in the middle 50% of the first data set.

The following data set shows the heights in inches for the boys in a class of 40 students.

66; 66; 67; 67; 68; 68; 68; 68; 68; 69; 69; 69; 70; 71; 72; 72; 72; 73; 73; 74

The following data set shows the heights in inches for the girls in a class of 40 students.

61; 61; 62; 62; 63; 63; 63; 65; 65; 65; 66; 66; 66; 67; 68; 68; 68; 69; 69; 69

Construct a box plot using a graphing calculator for each data set, and state which box plot has the wider spread for the middle 50% of the data.

Graph a box-and-whisker plot for the data values shown.

10; 10; 10; 15; 35; 75; 90; 95; 100; 175; 420; 490; 515; 515; 790

The five numbers used to create a box-and-whisker plot are:

• Min: 10
• *Q*₁: 15
• Med: 95
• *Q*₃: 490
• Max: 790

• 最小值：10
• *Q*₁：15
• 中位数：95
• *Q*₃：490
• 最大值：790

The following graph shows the box-and-whisker plot.

Follow the steps you used to graph a box-and-whisker plot for the data values shown.

0; 5; 5; 15; 30; 30; 45; 50; 50; 60; 75; 110; 140; 240; 330

---

2.5 Measures of the Center of the Data 2.5 数据中心位置的度量

The "center" of a data set is also a way of describing location. The two most widely used measures of the "center" of the data are the mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center.

The words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. However, in practice among non-statisticians, “average" is commonly accepted for “arithmetic mean.”

When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an *x* with a bar over it (pronounced “*x* bar”): $\overset{–}{x}$.

The Greek letter *μ* (pronounced "mew") represents the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random.

To see that both ways of calculating the mean are the same, consider the sample:

1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4

$$\overline{x} = \frac{1 + 1 + 1 + 2 + 2 + 3 + 4 + 4 + 4 + 4 + 4}{11} = 2.7$$ $$\overline{x} = \frac{3(1) + 2(2) + 1(3) + 5(4)}{11} = 2.7$$

In the second calculation, the frequencies are 3, 2, 1, and 5.

You can quickly find the location of the median by using the expression $\frac{n + 1}{2}$.

The letter *n* is the total number of data values in the sample. If *n* is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If *n* is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered. For example, if the total number of data values is 97, then $\frac{n + 1}{2}$= $\frac{97 + 1}{2}$ = 49. The median is the 49^th value in the ordered data. If the total number of data values is 100, then $\frac{n + 1}{2}$= $\frac{100 + 1}{2}$ = 50.5. The median occurs midway between the 50^th and 51^st values. The location of the median and the value of the median are not the same. The upper case letter *M* is often used to represent the median. The next example illustrates the location of the median and the value of the median.

Problem 问题

AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest):

3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47;

Calculate the mean and the median.

Solution 解答

The calculation for the mean is:

$\overline{x} = \frac{\left\lbrack 3 + 4 + (8)(2) + 10 + 11 + 12 + 13 + 14 + (15)(2) + (16)(2) + \text{...} + 35 + 37 + 40 + (44)(2) + 47 \right\rbrack}{40} = {23.6}$

To find the median, *M*, first use the formula for the location. The location is:

$\frac{n + 1}{2} = \frac{40 + 1}{2} = 20.5$

Starting at the smallest value, the median is located between the 20^th and 21^st values (the two 24s):

3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47;

$M = \frac{24 + 24}{2} = 24$

To find the mean and the median:

Clear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER.

Enter data into the list editor. Press STAT 1:EDIT.

Put the data values into list L1.

Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.

Press the down and up arrow keys to scroll.

$\overline{x}$ = 23.6, *M* = 24

The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and median.

3; 4; 5; 7; 7; 7; 7; 8; 8; 9; 9; 10; 10; 10; 10; 10; 11; 12; 12; 13; 14; 14; 15; 15; 17; 17; 18; 19; 19; 19; 21; 21; 22; 22; 23; 24; 24; 24; 24

Problem 问题

Suppose that in a small town of 50 people, one person earns \$5,000,000 per year and the other 49 each earn \$30,000. Which is the better measure of the "center": the mean or the median?

Solution 解答

$\overline{x} = \frac{5,000,000 + 49(30,000)}{50} = 129,400$

*M* = 30,000

(There are 49 people who earn \$30,000 and one person who earns \$5,000,000.)

The median is a better measure of the "center" than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data.

In a sample of 60 households, one house is worth \$2,500,000. Twenty-nine houses are worth \$280,000, and all the others are worth \$315,000. Which is the better measure of the "center": the mean or the median?

Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal.

Statistics exam scores for 20 students are as follows:

50; 53; 59; 59; 63; 63; 72; 72; 72; 72; 72; 76; 78; 81; 83; 84; 84; 84; 90; 93

Problem 问题

Find the mode.

Solution 解答

The most frequent score is 72, which occurs five times. Mode = 72.

The number of books checked out from the library from 25 students are as follows:

0; 0; 0; 1; 2; 3; 3; 4; 4; 5; 5; 7; 7; 7; 7; 8; 8; 8; 9; 10; 10; 11; 11; 12; 12

Find the mode.

Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice.

When is the mode the best measure of the "center"? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing.

The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red.

Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software.

Five credit scores are 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice. Consider the annual earnings of workers at a factory. The mode is \$25,000 and occurs 150 times out of 301. The median is \$50,000 and the mean is \$47,500. What would be the best measure of the "center"?

The Law of Large Numbers and the Mean 大数定律与均值

The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean $\overline{x}$ of the sample is very likely to get closer and closer to *µ*. This is discussed in more detail later in the text.

Sampling Distributions and Statistic of a Sampling Distribution 抽样分布与抽样分布的统计量

You can think of a sampling distribution as a relative frequency distribution with a great many samples. (See Sampling and Data for a review of relative frequency). Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below.

Table 2.24

If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.

A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean $\overline{x}$ is an example of a statistic which estimates the population mean *μ*.

Calculating the Mean of Grouped Frequency Tables 计算分组频数表的均值

When only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean: *mean* = $\frac{data\ sum}{number\ of\ data\ values}$ We simply need to modify the definition to fit within the restrictions of a frequency table.

Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is $\frac{lower\ boundary + upper\ boundary}{2}$. We can now modify the mean definition to be $Mean\ of\ Frequency\ Table = \frac{\sum{fm}}{\sum f}$ where *f* = the frequency of the interval and *m* = the midpoint of the interval.

Problem 问题

A frequency table displaying professor Blount’s last statistic test is shown. Find the best estimate of the class mean.

| Grade Interval | Number of Students |

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

| 50–56.5 | 1 |

| 56.5–62.5 | 0 |

| 62.5–68.5 | 4 |

| 68.5–74.5 | 4 |

| 74.5–80.5 | 2 |

| 80.5–86.5 | 3 |

| 86.5–92.5 | 4 |

| 92.5–98.5 | 1 |

Table 2.25

Solution 解答

• Find the midpoints for all intervals

• 求出所有区间的中点

| Grade Interval | Midpoint |

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

| 50–56.5 | 53.25 |

| 56.5–62.5 | 59.5 |

| 62.5–68.5 | 65.5 |

| 68.5–74.5 | 71.5 |

| 74.5–80.5 | 77.5 |

| 80.5–86.5 | 83.5 |

| 86.5–92.5 | 89.5 |

| 92.5–98.5 | 95.5 |

Table 2.26

• Calculate the sum of the product of each interval frequency and midpoint.$\sum^{}fm$

• 计算每个区间的频数与其中点之乘积的和。$\sum^{}fm$

$53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.25$

• $\mu = \frac{\sum{fm}}{\sum f} = \frac{1460.25}{19} = 76.86$

• $\mu = \frac{\sum{fm}}{\sum f} = \frac{1460.25}{19} = 76.86$

Maris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data:

| Hours Teenagers Spend on Video Games | Number of Teenagers |

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

| 0–3.5 | 3 |

| 3.5–7.5 | 7 |

| 7.5–11.5 | 12 |

| 11.5–15.5 | 7 |

| 15.5–19.5 | 9 |

Table 2.27

What is the best estimate for the mean number of hours spent playing video games?

2.6 Skewness and the Mean, Median, and Mode 2.6 偏度与均值、中位数和众数

Consider the following data set.

4; 5; 6; 6; 6; 7; 7; 7; 7; 7; 7; 8; 8; 8; 9; 10

This data set can be represented by following histogram. Each interval has width one, and each value is located in the middle of an interval.

The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median.

The histogram for the data: 4; 5; 6; 6; 6; 7; 7; 7; 7; 8 (shown in Figure 2.17) is not symmetrical. The right-hand side seems "chopped off" compared to the left side. A distribution of this type is called skewed to the left because it is pulled out to the left.

The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so.

The histogram for the data: 6; 7; 7; 7; 7; 8; 8; 8; 9; 10 Figure 2.18, is also not symmetrical. It is skewed to the right.

The mean is 7.7, the median is 7.5, and the mode is seven. Of the three statistics, the mean is the largest, while the mode is the smallest. Again, the mean reflects the skewing the most.

The mean is affected by outliers that do not influence the mean. Therefore, when the distribution of data is skewed to the left, the mean is often less than the median. When the distribution is skewed to the right, the mean is often greater than the median. In symmetric distributions, we expect the mean and median to be approximately equal in value. This is an important connection between the shape of the distribution and the relationship of the mean and median. It is not, however, true for every data set. The most common exceptions occur in sets of discrete data.

Skewness and symmetry become important when we discuss probability distributions in later chapters.

Problem 问题

Statistics are used to compare and sometimes identify authors. The following lists shows a simple random sample that compares the letter counts for three authors.

Terry: 7; 9; 3; 3; 3; 4; 1; 3; 2; 2

Davis: 3; 3; 3; 4; 1; 4; 3; 2; 3; 1

Maris: 2; 3; 4; 4; 4; 6; 6; 6; 8; 3

1. Make a dot plot for the three authors and compare the shapes.

2. Calculate the mean for each.

3. Calculate the median for each.

4. Describe any pattern you notice between the shape and the measures of center.

Solution 解答

1.

2. Terry’s mean is 3.7, Davis’ mean is 2.7, Maris’ mean is 4.6.

3. Terry’s median is three, Davis’ median is three. Maris’ median is four.

4. It appears that the median is always closest to the high point (the mode), while the mean tends to be farther out on the tail. In a symmetrical distribution, the mean and the median are both centrally located close to the high point of the distribution.

Discuss the mean, median, and mode for each of the following problems. Is there a pattern between the shape and measure of the center?

a\.

b\.

| The Ages Former U.S Presidents Died | |

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

| 4 | 6 9 |

| 5 | 3 6 7 7 7 8 |

| 6 | 0 0 3 3 4 4 5 6 7 7 7 8 |

| 7 | 0 1 1 2 3 4 7 8 8 9 |

| 8 | 0 1 3 5 8 |

| 9 | 0 0 3 3 |

| Key: 8\|0 means 80. | |

Table 2.28

c\.

2.7 Measures of the Spread of the Data 2.7 数据的离散程度度量

An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.

The standard deviation 标准差

• provides a numerical measure of the overall amount of variation in a data set, and
• can be used to determine whether a particular data value is close to or far from the mean.

• 提供了对数据集中总体变异数量的数值度量，并且
• 可用于判断某个特定数据值是接近还是远离均值。

The standard deviation provides a measure of the overall variation in a data set 标准差提供了对数据集中总体变异程度的度量

The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.

Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket *A* and supermarket *B*. the average wait time at both supermarkets is five minutes. At supermarket *A*, the standard deviation for the wait time is two minutes; at supermarket *B* the standard deviation for the wait time is four minutes.

Because supermarket *B* has a higher standard deviation, we know that there is more variation in the wait times at supermarket *B*. Overall, wait times at supermarket *B* are more spread out from the average; wait times at supermarket *A* are more concentrated near the average.

The standard deviation can be used to determine whether a data value is close to or far from the mean. 标准差可用于判断某个数据值是接近还是远离均值。

Suppose that Rosa and Binh both shop at supermarket *A*. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket *A*, the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.

Rosa waits for seven minutes:

• Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.
• Rosa's wait time of seven minutes is two minutes longer than the average of five minutes.
• Rosa's wait time of seven minutes is one standard deviation above the average of five minutes.

• 7 比平均值 5 多 2 分钟；2 分钟等于一个标准差。
• Rosa 等候 7 分钟，比平均值 5 分钟多 2 分钟。
• Rosa 等候 7 分钟，比平均值 5 分钟高出 1 个标准差。

Binh waits for one minute.

• One is four minutes less than the average of five; four minutes is equal to two standard deviations.
• Binh's wait time of one minute is four minutes less than the average of five minutes.
• Binh's wait time of one minute is two standard deviations below the average of five minutes.
• A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)

• 1 比平均值 5 少 4 分钟；4 分钟等于 2 个标准差。
• Binh 等候 1 分钟，比平均值 5 分钟少 4 分钟。
• Binh 等候 1 分钟，比平均值 5 分钟低 2 个标准差。
• 一个偏离平均值 2 个标准差的数据值，恰好处于许多统计学家认为"远离平均值"的临界线上。把偏离超过 2 个标准差的数据视为远离均值，更多是一种近似的"经验法则"，而非严格规则。一般而言，数据分布的形态会影响有多少数据落在超过 2 个标准差之外。（你将在后面的章节中学到更多相关内容。）

The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because 5 + (1)(2) = 7.

If one were also part of the data set, then one is two standard deviations to the left of five because 5 + (–2)(2) = 1.

• In general, a value = mean + (#ofSTDEV)(standard deviation)
• where \#ofSTDEVs = the number of standard deviations
• \#ofSTDEV does not need to be an integer
• One is two standard deviations less than the mean of five because: 1 = 5 + (–2)(2).

• 一般而言，某数值 = 均值 +（标准差个数）（标准差）
• 其中 #ofSTDEVs = 标准差的个数
• #ofSTDEV 不一定是整数
• 1 比均值 5 少 2 个标准差，因为：1 = 5 + (–2)(2)。

The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population.

• sample: $x\text{~=~}\overline{x}\text{~+~}(\# ofSTDEV)(s)$
• Population: $x = \mu + (\# ofSTDEV)(\sigma)$

• 样本： $x\text{~=~}\overline{x}\text{~+~}(\# ofSTDEV)(s)$
• 总体： $x = \mu + (\# ofSTDEV)(\sigma)$

The lower case letter *s* represents the sample standard deviation and the Greek letter *σ* (sigma, lower case) represents the population standard deviation.

The symbol $\overline{x}$ is the sample mean and the Greek symbol $\mu$ is the population mean.

Calculating the Standard Deviation 计算标准差

If *x* is a number, then the difference "*x* – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is *x* – *μ*. For sample data, in symbols a deviation is *x* – $\overline{x}$.

The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter *σ* (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of *σ*.

To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the *x* – $\overline{x}$ values for a sample, or the *x* – *μ* values for a population). The symbol *σ*² represents the population variance; the population standard deviation *σ* is the square root of the population variance. The symbol *s*² represents the sample variance; the sample standard deviation *s* is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.

If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by *N*, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by ***n* – 1**, one less than the number of items in the sample.

Formulas for the Sample Standard Deviation 样本标准差公式

• $s = \sqrt{\frac{\Sigma{(x - \overline{x})}^{2}}{n - 1}}$ or $s = \sqrt{\frac{\Sigma f{(x - \overline{x})}^{2}}{n - 1}}$
• For the sample standard deviation, the denominator is ***n* - 1**, that is the sample size MINUS 1.

• $s = \sqrt{\frac{\Sigma{(x - \overline{x})}^{2}}{n - 1}}$ 或 $s = \sqrt{\frac{\Sigma f{(x - \overline{x})}^{2}}{n - 1}}$
• 对于样本标准差，分母为 n - 1，即样本量减 1。

Formulas for the Population Standard Deviation 总体标准差公式

• $\sigma~ = ~\sqrt{\frac{\Sigma{(x - \mu)}^{2}}{N}}$ or $\sigma~ = ~\sqrt{\frac{\Sigma f{(x–\mu)}^{2}}{N}}$
• For the population standard deviation, the denominator is *N*, the number of items in the population.

• $\sigma~ = ~\sqrt{\frac{\Sigma{(x - \mu)}^{2}}{N}}$ 或 $\sigma~ = ~\sqrt{\frac{\Sigma f{(x–\mu)}^{2}}{N}}$
• 对于总体标准差，分母为 N，即总体中的项数。

In these formulas, *f* represents the frequency with which a value appears. For example, if a value appears once, *f* is one. If a value appears three times in the data set or population, *f* is three.

Sampling Variability of a Statistic 统计量的抽样变异性

The statistic of a sampling distribution was discussed in Descriptive Statistics: Measuring the Center of the Data. How much the statistic varies from one sample to another is known as the sampling variability of a statistic. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean in the chapter The Central Limit Theorem (not now). The notation for the standard error of the mean is $\frac{\sigma}{\sqrt{n}}$ where *σ* is the standard deviation of the population and n is the size of the sample.

**In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation *σ_x* or *s_x* from the summary statistics.** We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)

In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of *n* = 20 fifth grade students. The ages are rounded to the nearest half year:

9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5;

$$\overline{x} = \frac{\text{9~+~9}\text{.5(2)~+~10(4)~+~10}\text{.5(4)~+~11(6)~+~11}\text{.5(3)}}{20} = 10.525$$

The average age is 10.53 years, rounded to two places.

The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating *s*.

| Data | Freq. | Deviations | *Deviations*² | (Freq.)(*Deviations*²) |

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

| *x* | *f* | (*x* – $\overline{x}$) | (*x* – $\overline{x}$)² | (*f*)(*x* – $\overline{x}$)² |

| 9 | 1 | 9 – 10.525 = –1.525 | (–1.525)² = 2.325625 | 1 × 2.325625 = 2.325625 |

| 9.5 | 2 | 9.5 – 10.525 = –1.025 | (–1.025)² = 1.050625 | 2 × 1.050625 = 2.101250 |

| 10 | 4 | 10 – 10.525 = –0.525 | (–0.525)² = 0.275625 | 4 × 0.275625 = 1.1025 |

| 10.5 | 4 | 10.5 – 10.525 = –0.025 | (–0.025)² = 0.000625 | 4 × 0.000625 = 0.0025 |

| 11 | 6 | 11 – 10.525 = 0.475 | (0.475)² = 0.225625 | 6 × 0.225625 = 1.35375 |

| 11.5 | 3 | 11.5 – 10.525 = 0.975 | (0.975)² = 0.950625 | 3 × 0.950625 = 2.851875 |

| | | | | The total is 9.7375 |

Table 2.29

The sample variance, *s*², is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1):

$s^{2} = \frac{9.7375}{20 - 1} = 0.5125$

The sample standard deviation *s* is equal to the square root of the sample variance:

$s = \sqrt{0.5125} = 0.715891,$ which is rounded to two decimal places, *s* = 0.72.

Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy.

Problem 问题

• For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation). Verify the mean and standard deviation or a calculator or computer.
• For a sample: *x* = $\overline{x}$ + (#ofSTDEVs)(*s*)
• For a population: *x* = *μ* + (#ofSTDEVs)(*σ*)
• For this example, use *x* = $\overline{x}$ + (#ofSTDEVs)(*s*) because the data is from a sample

• 对于以下各题，请记住数值 = 均值 +（标准差个数）（标准差）。用计算器或计算机验证均值和标准差。
• 对于样本：*x* = $\overline{x}$ +（标准差个数）(*s*)
• 对于总体：*x* = *μ* +（标准差个数）(*σ*)
• 本例中使用 *x* = $\overline{x}$ +（标准差个数）(*s*)，因为数据来自一个样本

1. Verify the mean and standard deviation on your calculator or computer.

2. Find the value that is one standard deviation above the mean. Find ($\overline{x}$ + 1s).

3. Find the value that is two standard deviations below the mean. Find ($\overline{x}$ – 2s).

4. Find the values that are 1.5 standard deviations from (below and above) the mean.

Solution 解答

1. - Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.

• Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.
• Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.
• Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.
• $\overline{x}$ = 10.525
• Use Sx because this is sample data (not a population): Sx=0.715891

• 将数据输入列表编辑器。按 STAT 1:EDIT。如有必要，向上移到名称处清除列表。按 CLEAR 再向下移。
• 把数据值（9、9.5、10、10.5、11、11.5）放入列表 L1，频数（1、2、4、4、6、3）放入列表 L2。用方向键移动。
• 按 STAT 并移到 CALC。按 1:1-VarStats，输入 L1（2nd 1）、L2（2nd 2）。别忘了逗号。按 ENTER。
• $\overline{x}$ = 10.525
• 使用 Sx，因为这是样本数据（不是总体）：Sx=0.715891

2. ($\overline{x}$ + 1s) = 10.53 + (1)(0.72) = 11.25

3. ($\overline{x}$ – 2*s*) = 10.53 – (2)(0.72) = 9.09

4. - ($\overline{x}$ – 1.5*s*) = 10.53 – (1.5)(0.72) = 9.45

• ($\overline{x}$ + 1.5*s*) = 10.53 + (1.5)(0.72) = 11.61

• ($\overline{x}$ + 1.5*s*) = 10.53 + (1.5)(0.72) = 11.61

On a baseball team, the ages of each of the players are as follows:

21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40

Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean.

Explanation of the standard deviation calculation shown in the table 表中标准差计算方法的解释

The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. (For Example 2.32, there are *n* = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.

The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.

Notice that instead of dividing by *n* = 20, the calculation divided by *n* – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one (*n* – 1). Why not divide by *n*? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (*n* – 1) gives a better estimate of the population variance.

Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.

The standard deviation, *s* or *σ*, is either zero or larger than zero. Describing the data with reference to the spread is called "variability". The variability in data depends upon the method by which the outcomes are obtained; for example, by measuring or by random sampling. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make *s* or *σ* very large.

The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot.

Problem 问题

Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:

33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100

1. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.

2. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:

1. The sample mean

2. The sample standard deviation

3. The median

4. The first quartile

5. The third quartile

6. *IQR*

3. Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.

Solution 解答

1. See Table 2.30

2. 1. The sample mean = 73.5

2. The sample standard deviation = 17.9

3. The median = 73

4. The first quartile = 61

5. The third quartile = 90

6. *IQR* = 90 – 61 = 29

3. The *x*-axis goes from 32.5 to 100.5; *y*-axis goes from –2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is (100.5 – 32.5) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, 32.5 + 13.6 = 46.1, 46.1 + 13.6 = 59.7, 59.7 + 13.6 = 73.3, 73.3 + 13.6 = 86.9, 86.9 + 13.6 = 100.5 = the ending value; No data values fall on an interval boundary.

| Data | Frequency | Relative Frequency | Cumulative Relative Frequency |

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

| 33 | 1 | 0.032 | 0.032 |

| 42 | 1 | 0.032 | 0.064 |

| 49 | 2 | 0.065 | 0.129 |

| 53 | 1 | 0.032 | 0.161 |

| 55 | 2 | 0.065 | 0.226 |

| 61 | 1 | 0.032 | 0.258 |

| 63 | 1 | 0.032 | 0.29 |

| 67 | 1 | 0.032 | 0.322 |

| 68 | 2 | 0.065 | 0.387 |

| 69 | 2 | 0.065 | 0.452 |

| 72 | 1 | 0.032 | 0.484 |

| 73 | 1 | 0.032 | 0.516 |

| 74 | 1 | 0.032 | 0.548 |

| 78 | 1 | 0.032 | 0.580 |

| 80 | 1 | 0.032 | 0.612 |

| 83 | 1 | 0.032 | 0.644 |

| 88 | 3 | 0.097 | 0.741 |

| 90 | 1 | 0.032 | 0.773 |

| 92 | 1 | 0.032 | 0.805 |

| 94 | 4 | 0.129 | 0.934 |

| 96 | 1 | 0.032 | 0.966 |

| 100 | 1 | 0.032 | 0.998 (Why isn't this value 1?) |

Table 2.30

The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73 – 33 = 40) than the spread in the upper 50% (100 – 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (*IQR* = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.

The following data show the different types of pet food stores in the area carry.

6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;

Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.

Standard deviation of Grouped Frequency Tables 分组频数表的标准差

Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula: $Mean\ of\ Frequency\ Table = \frac{\sum{fm}}{\sum f}$

where $f =$ interval frequencies and *m* = interval midpoints.

Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how "unusual" individual data is compared to the mean.

Find the standard deviation for the data in Table 2.31.

| Class | Frequency, $f$ | Midpoint, $m$ | $f \cdot m$ | $\overline{x}$ | $m - \overline{x}$ | $\left( m - \overline{x} \right)^{2}$ | $f\left( m - \overline{x} \right)^{2}$ |

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

| 0–2 | 1 | 1 | 1 | 7.58 | -6.58 | 43.2964 | 43.2964 |

| 3–5 | 6 | 4 | 24 | 7.58 | -3.58 | 12.8164 | 76.8984 |

| 6–8 | 10 | 7 | 70 | 7.58 | -0.58 | 0.3364 | 3.364 |

| 9–11 | 7 | 10 | 70 | 7.58 | 2.42 | 5.8564 | 40.9948 |

| 12–14 | 0 | 13 | 0 | 7.58 | 5.42 | 29.3764 | 0 |

| 15–17 | 2 | 16 | 32 | 7.58 | 8.42 | 70.8964 | 141.7928 |

| SUM ($\mathbf{\Sigma}$) | 26 | | 197 | | | | 43.2964 |

Table 2.31

The values in the second, third, and fourth columns of Table 2.31 are used to calculate the mean of the grouped frequency table, the value in the fifth column.

$$\overline{x} = \frac{\Sigma fm}{\Sigma f} = \frac{197}{26} \approx 7.58.$$ 2.1

After calculating $\overline{x}$, find the difference, $m - \overline{x}$, for each midpoint, $m$. Next, square each difference. In the final column, calculate the product of the frequency and the squared difference for each class.

The table makes it easy to use the formula for calculating the standard deviation of a grouped frequency table:

$$s_{x} = \sqrt{\frac{\Sigma f\left( m - \overline{x} \right)^{2}}{n - 1}} = \sqrt{\frac{43.2964}{26 - 1}} \approx 3.50.$$ 2.2

Although the formula is not complicated, these calculations are typically performed using technology.

Find the standard deviation for the data from the previous example

| Class | Frequency, *f* |

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

| 0–2 | 1 |

| 3–5 | 6 |

| 6–8 | 10 |

| 9–11 | 7 |

| 12–14 | 0 |

| 15–17 | 2 |

Table 2.32

First, press the STAT key and select 1:Edit

Input the midpoint values into L1 and the frequencies into L2

Select STAT, CALC, and 1: 1-Var Stats

Select 2^nd then 1 then , 2^nd then 2 Enter

You will see displayed both a population standard deviation, *σ_x*, and the sample standard deviation, *s_x*.

Comparing Values from Different Data Sets 比较来自不同数据集的数值

The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.

• For each data value, calculate how many standard deviations away from its mean the value is.
• Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for \#ofSTDEVs.
• $\# ofSTDEVs = \frac{\text{value~–~mean}}{\text{standard~deviation}}$
• Compare the results of this calculation.

• 对于每一个数据值，计算它偏离其均值多少个标准差。
• 使用公式：value = mean + (#ofSTDEVs)(standard deviation)；解出 \#ofSTDEVs。
• $\# ofSTDEVs = \frac{\text{value~–~mean}}{\text{standard~deviation}}$
• 比较这一计算的结果。

\#ofSTDEVs is often called a "*z*-score"; we can use the symbol *z*. In symbols, the formulas become:

| | | |

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

| Sample | $x$ = $\overline{x}$ + *zs* | $z = \frac{x\ - \ \overline{x}}{s}$ |

| Population | $x$ = $\mu$ + *zσ* | $z = \frac{x\ - \ \mu}{\sigma}$ |

Table 2.33

Problem 问题

Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?

| Student | GPA | School Mean GPA | School Standard Deviation |

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

| John | 2.85 | 3.0 | 0.7 |

| Ali | 77 | 80 | 10 |

Table 2.34

Solution 解答

For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.

$z = \operatorname{\#\ of\ STDEVs} = \frac{\text{value~}–\text{mean}}{\text{standard~deviation}} = \frac{x–\mu}{\sigma}$

For John, $z = \# ofSTDEVs = \frac{2.85–3.0}{0.7} = –0.21$

For Ali, $z = \# ofSTDEVs = \frac{77 - 80}{10} = - 0.3$

John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.

John's *z*-score of –0.21 is higher than Ali's *z*-score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.

Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?

| Swimmer | Time (seconds) | Team Mean Time | Team Standard Deviation |

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

| Angie | 26.2 | 27.2 | 0.8 |

| Beth | 27.3 | 30.1 | 1.4 |

Table 2.35

The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.

For ANY data set, no matter what the distribution of the data is:

• At least 75% of the data is within two standard deviations of the mean.
• At least 89% of the data is within three standard deviations of the mean.
• At least 95% of the data is within 4.5 standard deviations of the mean.
• This is known as Chebyshev's Rule.

• 至少有 75% 的数据落在距均值两个标准差之内。
• 至少有 89% 的数据落在距均值三个标准差之内。
• 至少有 95% 的数据落在距均值 4.5 个标准差之内。
• 这称为切比雪夫法则（Chebyshev's Rule）。

For data having a distribution that is BELL-SHAPED and SYMMETRIC:

• Approximately 68% of the data is within one standard deviation of the mean.
• Approximately 95% of the data is within two standard deviations of the mean.
• More than 99% of the data is within three standard deviations of the mean.
• This is known as the Empirical Rule.
• It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.

• 大约有 68% 的数据落在距均值一个标准差之内。
• 大约有 95% 的数据落在距均值两个标准差之内。
• 超过 99% 的数据落在距均值三个标准差之内。
• 这称为经验法则（Empirical Rule）。
• 需注意，该法则仅当数据分布呈钟形且对称时才适用。在后面章节学习"正态"或"高斯"概率分布时，我们会了解更多相关内容。

2.8 Descriptive Statistics 2.8 描述统计学

Descriptive Statistics 描述统计学

Class Time:

Names:

Student Learning Outcomes

• The student will construct a histogram and a box plot.
• The student will calculate univariate statistics.
• The student will examine the graphs to interpret what the data implies.

• 学生将绘制直方图和箱线图。
• 学生将计算单变量统计量。
• 学生将考察图形，以解释数据所隐含的信息。

Collect the Data Record the number of pairs of shoes you own.

1. Randomly survey 30 classmates about the number of pairs of shoes they own. Record their values.

| | | | | |

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

| | | | | |

| | | | | |

| | | | | |

| | | | | |

| | | | | |

| | | | | |

Table 2.36 Survey Results

2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil and scale the axes.

3. Calculate the following values.

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

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

4. Are the data discrete or continuous? How do you know?

5. In complete sentences, describe the shape of the histogram.

6. Are there any potential outliers? List the value(s) that could be outliers. Use a formula to check the end values to determine if they are potential outliers.

Analyze the Data

1. Determine the following values.

1. Min = \_\_\_\_\_

2. *M* = \_\_\_\_\_

3. Max = \_\_\_\_\_

4. *Q*₁ = \_\_\_\_\_

5. *Q*₃ = \_\_\_\_\_

6. *IQR* = \_\_\_\_\_

2. Construct a box plot of data

3. What does the shape of the box plot imply about the concentration of data? Use complete sentences.

4. Using the box plot, how can you determine if there are potential outliers?

5. How does the standard deviation help you to determine concentration of the data and whether or not there are potential outliers?

6. What does the *IQR* represent in this problem?

7. Show your work to find the value that is 1.5 standard deviations:

1. above the mean.

2. below the mean.

Key Terms 关键术语

Box plot

a graph that gives a quick picture of the middle 50% of the data

First Quartile

the value that is the median of the of the lower half of the ordered data set

Frequency

the number of times a value of the data occurs

Frequency Polygon

looks like a line graph but uses intervals to display ranges of large amounts of data

Frequency Table

a data representation in which grouped data is displayed along with the corresponding frequencies

Histogram

a graphical representation in *x*-*y* form of the distribution of data in a data set; *x* represents the data and *y* represents the frequency, or relative frequency. The graph consists of contiguous rectangles.

Interquartile Range

or *IQR*, is the range of the middle 50 percent of the data values; the *IQR* is found by subtracting the first quartile from the third quartile.

Interval

also called a class interval; an interval represents a range of data and is used when displaying large data sets

Mean

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

Median

a number that separates ordered data into halves; half the values are the same number or smaller than the median and half the values are the same number or larger than the median. The median may or may not be part of the data.

Midpoint

the mean of an interval in a frequency table

Mode

the value that appears most frequently in a set of data

Outlier

an observation that does not fit the rest of the data

Paired Data Set

two data sets that have a one to one relationship so that:

• both data sets are the same size, and
• each data point in one data set is matched with exactly one point from the other set.

• 两组数据的规模相同；且
• 一组数据中的每个数据点都恰好与另一组数据中的一个点相匹配。

Percentile

a number that divides ordered data into hundredths; percentiles may or may not be part of the data. The median of the data is the second quartile and the 50^th percentile. The first and third quartiles are the 25^th and the 75^th percentiles, respectively.

Quartiles

the numbers that separate the data into quarters; quartiles may or may not be part of the data. The second quartile is the median of the data.

Relative Frequency

the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes

Skewed

used to describe data that is not symmetrical; when the right side of a graph looks "chopped off" compared the left side, we say it is "skewed to the left." When the left side of the graph looks "chopped off" compared to the right side, we say the data is "skewed to the right." Alternatively: when the lower values of the data are more spread out, we say the data are skewed to the left. When the greater values are more spread out, the data are skewed to the right.

Standard Deviation

a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: *s* for sample standard deviation and σ for population standard deviation.

Variance

mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as *x* – $\overline{x}$ where *x* is a value of the data and $\overline{x}$ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.

Chapter Review 章节回顾

2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 2.1 茎叶图（Stemplots）、线图与条形图

A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used.

2.2 Histograms, Frequency Polygons, and Time Series Graphs 2.2 直方图、频数多边形与时间序列图

A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on *y*-axis with the frequency being graphed on the *x*-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time.

2.3 Measures of the Location of the Data 2.3 数据的位置度量

The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50^th percentile would be greater than 50 percent of the other observations in the set. Quartiles divide data into quarters. The first quartile (*Q*₁) is the 25^th percentile,the second quartile (*Q*₂ or median) is 50^th percentile, and the third quartile (*Q*₃) is the 75^th percentile. The interquartile range, or *IQR*, is the range of the middle 50 percent of the data values. The *IQR* is found by subtracting *Q*₁ from *Q*₃, and can help determine outliers by using the following two expressions.

• *Q*₃ + *IQR*(1.5)
• *Q*₁ – *IQR*(1.5)

• *Q*₃ + *IQR*(1.5)
• *Q*₁ – *IQR*(1.5)

2.4 Box Plots 2.4 箱线图

Box plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data.

2.5 Measures of the Center of the Data 2.5 数据中心位置的度量

The mean and the median can be calculated to help you find the "center" of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set.

2.6 Skewness and the Mean, Median, and Mode 2.6 偏度与均值、中位数、众数

Looking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A left (or negative) skewed distribution has a shape like Figure 2.17. A right (or positive) skewed distribution has a shape like Figure 2.18. A symmetrical distribution looks like Figure 2.16.

2.7 Measures of the Spread of the Data 2.7 数据离散程度的度量

The standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population.

• The Standard Deviation allows us to compare individual data or classes to the data set mean numerically.
• *s* = $\sqrt{\frac{\sum^{}{(x - \overline{x})}^{2}}{n - 1}}$ or *s* = $\sqrt{\frac{\sum^{}f{(x - \overline{x})}^{2}}{n - 1}}$ is the formula for calculating the standard deviation of a sample. To calculate the standard deviation of a population, we would use the population mean, *μ*, and the formula *σ* = $\sqrt{\frac{\sum^{}{(x - \mu)}^{2}}{N}}$ or *σ* = $\sqrt{\frac{\sum^{}f{(x - \mu)}^{2}}{N}}$.

• 标准差使我们可以从数值上把单个数据或各个组与数据集的均值进行比较。
• *s* = $\sqrt{\frac{\sum^{}{(x - \overline{x})}^{2}}{n - 1}}$ 或 *s* = $\sqrt{\frac{\sum^{}f{(x - \overline{x})}^{2}}{n - 1}}$ 是计算样本标准差的公式。要计算总体的标准差，我们会使用总体均值 *μ*，以及公式 *σ* = $\sqrt{\frac{\sum^{}{(x - \mu)}^{2}}{N}}$ 或 *σ* = $\sqrt{\frac{\sum^{}f{(x - \mu)}^{2}}{N}}$。

Formula Review 公式回顾

2.3 Measures of the Location of the Data 2.3 数据的位置度量

$i = \left( \frac{k}{100} \right)\left( {n + 1} \right)$

where *i* = the ranking or position of a data value,

*k* = the kth percentile,

*n* = total number of data.

Expression for finding the percentile of a data value: $\left( \frac{x\text{~+~}0.5y}{n} \right)$(100)

where *x* = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,

*y* = the number of data values equal to the data value for which you want to find the percentile,

*n* = total number of data

2.5 Measures of the Center of the Data 2.5 数据中心位置的度量

$\mu = \frac{\sum{fm}}{\sum f}$ Where *f* = interval frequencies and *m* = interval midpoints.

2.7 Measures of the Spread of the Data 2.7 数据离散程度的度量

$s_{x} = \sqrt{\frac{\sum{fm^{2}}}{n} - {\overline{x}}^{2}}$ where $\begin{array}{l}{s_{x} = \text{~sample~standard~deviation}} \\{\overline{x}\text{~=~sample~mean}}\end{array}$

Practice 练习

2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 2.1 茎叶图（Stemplot）、线图与条形图

*For each of the following data sets, create a stem plot and identify any outliers.*

1. The miles per gallon rating for 30 cars are shown below (lowest to highest).

19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43

2\. The height in feet of 25 trees is shown below (lowest to highest).

25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54

3. The data are the prices of different laptops at an electronics store. Round each value to the nearest ten.

249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610

4\. The data are daily high temperatures in a town for one month.

61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95

*For the next three exercises, use the data to construct a line graph.*

5. In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in Table 2.37.

| Number of times in store | Frequency |

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

| 1 | 4 |

| 2 | 10 |

| 3 | 16 |

| 4 | 6 |

| 5 | 4 |

Table 2.37

6. In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table 2.38.

| Years since last purchase | Frequency |

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

| 0 | 2 |

| 1 | 8 |

| 2 | 13 |

| 3 | 22 |

| 4 | 16 |

| 5 | 9 |

Table 2.38

7. Several children were asked how many TV shows they watch each day. The results of the survey are shown in Table 2.39.

| Number of TV Shows | Frequency |

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

| 0 | 12 |

| 1 | 18 |

| 2 | 36 |

| 3 | 7 |

| 4 | 2 |

Table 2.39

8. The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. Table 2.40 shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.

| Seasons | Number of students | Proportion of population |

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

| Spring | 8 | 24% |

| Summer | 9 | 26% |

| Autumn | 11 | 32% |

| Winter | 6 | 18% |

Table 2.40

9. Using the data from Mrs. Ramirez’s math class supplied in Table 2.40, construct a bar graph showing the percentages.

10\. David County has six high schools. Each school sent students to participate in a county-wide science competition. Table 2.41 shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.

| High School | Science competition population | Overall student population |

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

| Alabaster | 28.9% | 8.6% |

| Concordia | 7.6% | 23.2% |

| Genoa | 12.1% | 15.0% |

| Mocksville | 18.5% | 14.3% |

| Tynneson | 24.2% | 10.1% |

| West End | 8.7% | 28.8% |

Table 2.41

11. Use the data from the David County science competition supplied in Table 2.41. Construct a bar graph that shows the county-wide population percentage of students at each school.

2.2 Histograms, Frequency Polygons, and Time Series Graphs 2.2 直方图、频数多边形与时间序列图

12\. Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.

| Data Value (# cars) | Frequency | Relative Frequency | Cumulative Relative Frequency |

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

| | | | |

| | | | |

| | | | |

| | | | |

Table 2.42

13. What does the frequency column in Table 2.42 sum to? Why?

14\. What does the relative frequency column in Table 2.42 sum to? Why?

15. What is the difference between relative frequency and frequency for each data value in Table 2.42?

16\. What is the difference between cumulative relative frequency and relative frequency for each data value?

17. To construct the histogram for the data in Table 2.42, determine appropriate minimum and maximum *x* and *y* values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.

18\. Construct a frequency polygon for the following:

1. | Pulse Rates for Women | Frequency |

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

| 60–69 | 12 |

| 70–79 | 14 |

| 80–89 | 11 |

| 90–99 | 1 |

| 100–109 | 1 |

| 110–119 | 0 |

| 120–129 | 1 |

Table 2.43

2. | Actual Speed in a 30 MPH Zone | Frequency |

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

| 42–45 | 25 |

| 46–49 | 14 |

| 50–53 | 7 |

| 54–57 | 3 |

| 58–61 | 1 |

Table 2.44

3. | Tar (mg) in Nonfiltered Cigarettes | Frequency |

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

| 10–13 | 1 |

| 14–17 | 0 |

| 18–21 | 15 |

| 22–25 | 7 |

| 26–29 | 2 |

Table 2.45

19. Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger.

| Depth of Hunger | Frequency |

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

| 230–259 | 21 |

| 260–289 | 13 |

| 290–319 | 5 |

| 320–349 | 7 |

| 350–379 | 1 |

| 380–409 | 1 |

| 410–439 | 1 |

Table 2.46

20. Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?

| Life Expectancy at Birth – Women | Frequency |

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

| 49–55 | 3 |

| 56–62 | 3 |

| 63–69 | 1 |

| 70–76 | 3 |

| 77–83 | 8 |

| 84–90 | 2 |

Table 2.47

| Life Expectancy at Birth – Men | Frequency |

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

| 49–55 | 3 |

| 56–62 | 3 |

| 63–69 | 1 |

| 70–76 | 1 |

| 77–83 | 7 |

| 84–90 | 5 |

Table 2.48

21. Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births.

| | | | | | | | |

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

| Sex/Year | 1855 | 1856 | 1857 | 1858 | 1859 | 1860 | 1861 |

| Female | 45,545 | 49,582 | 50,257 | 50,324 | 51,915 | 51,220 | 52,403 |

| Male | 47,804 | 52,239 | 53,158 | 53,694 | 54,628 | 54,409 | 54,606 |

| Total | 93,349 | 101,821 | 103,415 | 104,018 | 106,543 | 105,629 | 107,009 |

Table 2.49

| | | | | | | | | |

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

| Sex/Year | 1862 | 1863 | 1864 | 1865 | 1866 | 1867 | 1868 | 1869 |

| Female | 51,812 | 53,115 | 54,959 | 54,850 | 55,307 | 55,527 | 56,292 | 55,033 |

| Male | 55,257 | 56,226 | 57,374 | 58,220 | 58,360 | 58,517 | 59,222 | 58,321 |

| Total | 107,069 | 109,341 | 112,333 | 113,070 | 113,667 | 114,044 | 115,514 | 113,354 |

Table 2.50

| | | | | | | |

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