By the end of this chapter, the student should be able to:
Display data graphically and interpret graphs: stemplots, histograms and boxplots.
Recognize, describe, and calculate the measures of location of data: quartiles and percentiles.
Recognize, describe, and calculate the measures of the center of data: mean, median, and mode.
Recognize, describe, and calculate the measures of the spread of data: variance, standard deviation, and range.
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 to calculate, and even more importantly, to interpret these measurements and graphs.
A statistical graph is a tool that helps you learn about the shape or distribution of a sample. The 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 in order 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 chart, the histogram, the stemandleaf plot, the frequency polygon (a type of broken line graph), pie charts, and the boxplot. In this chapter, we will briefly look at stemandleaf plots, line graphs and bar graphs. Our emphasis will be on histograms and boxplots.
One simple graph, the stemandleaf 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 one digit. For example, 23 has stem 2 and leaf 3. Four hundred thirtytwo (432) has stem 43 and leaf 2. Five thousand four hundred thirtytwo (5,432) has stem 543 and leaf 2. The decimal 9.3 has stem 9 and leaf 3. Write the stems in a vertical line from smallest the largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.
Example 2.1.
For Susan Dean’s spring precalculus 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  299 
5  355 
6  1378899 
7  2348 
8  03888 
9  0244446 
10  0 
The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% of the scores were in the 90’s or 100, a fairly high number of As.
The stemplot is a quick way to graph 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. In the example above, there were no outliers.
Example 2.2.
Create a stem plot 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
The data are the distance (in kilometers) from a home to the nearest supermarket.
Problem (Go to Solution)
Are there any outliers?
Do the data seem to have any concentration of values?
The leaves are to the right of the decimal.
Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in the example, the xaxis consists of data values and the yaxis consists of frequencies indicated by the heights of the vertical lines.
Example 2.3.
In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his/her chores. The results are shown in the table and the line graph.
Number of times teenager is reminded  Frequency 

0  2 
1  5 
2  8 
3  14 
4  7 
5  4 
Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes and they can be vertical or horizontal. The bar graph shown in Example 4 uses the data of Example 3 and is similar to the line graph. Frequencies are represented by the heights of the bars.
Example 2.4.
The bar graph shown in Example 5 has age groups represented on the xaxis and proportions on the yaxis.
Example 2.5.
By the end of March 2009, in the United States Facebook had over 56 million users. The table shows the age groups, the number of users in each age group and the proportion (%) of users in each age group. Source: http://www.insidefacebook.com/2009/03/25/numberofusfacebookusersover35nearlydoublesinlast60days/
Age groups  Number of Facebook users  Proportion (%) of Facebook users 

13  25  25,510,040  46% 
26  44  23,123,900  41% 
45  65  7,431,020  13% 
Example 2.6.
The columns in the table below contain the race/ethnicity of U.S. Public Schools: High School Class of 2009, percentages for the Advanced Placement Examinee Population for that class and percentages for the Overall Student Population. The 3dimensional graph shows the Race/Ethnicity of U.S. Public Schools on the xaxis and Advanced Placement Examinee Population percentages on the yaxis. (Source: http://www.collegeboard.com)
Race/Ethnicity  AP Examinee Population  Overall Student Population 

Asian, Asian American or Pacific Islander  10.2%  5.4% 
Black or African American  8.2%  14.5% 
Hispanic or Latino  15.5%  15.9% 
American Indian or Alaska Native  0.6%  1.2% 
White  59.4%  61.6% 
Not reported/other  6.1%  1.4% 
This book contains instructions for constructing a histogram and a box plot for the TI83+ and TI84 calculators. You can find additional instructions for using these calculators on the Texas Instruments (TI) website.
Solution to Exercise (Return to Problem)
The value 12.3 may be an outlier. Values appear to concentrate at 3 and 4 miles.
Stem  Leaf 

1  1 5 
2  3 5 7 
3  3 3 3 5 8 
4  0 2 5 5 7 8 
5  5 6 6 
6  5 7 
7  
8  
9  
10  
11  
12  3 
An observation that does not fit the rest of the data.
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 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”. The graph will have the same shape with either label. Frequency is commonly used when the data set is small and relative frequency is used when the data set is large or when we want to compare several distributions. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data. (The next section tells you how to calculate the center and the spread.)
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. (In the chapter on Sampling and Data, we defined frequency 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,
then:
For example, if 3 students in Mr. Ahab’s English class of 40 students received an A, then,
f = 3 , n = 40 , and
Seven and a half percent of the students received an A.
To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of from 5 to 15 bars or classes for clarity. 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  .0005 = 0.9995). If all the data happen to be integers and the smallest value is 2, 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.
Example 2.7.
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; 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. 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 8 bars.
We will round up to 2 and make each bar or class interval 2 units wide. Rounding up to 2 is one way to prevent a value from falling on a boundary. For this example, using 1.76 as the width would also work.
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
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 xaxis and relative frequency on the yaxis.
Example 2.8.
The following data are the number of books bought by 50 parttime 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 1 book. Ten students buy 2 books. Sixteen students buy 3 books. Six students buy 4 books. Five students buy 5 books. Two students buy 6 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 (Go to Solution)
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 _______ .
Calculate the number of bars as follows:
where 1 is the width of a bar. Therefore, bars = 6 .
The following histogram displays the number of books on the xaxis and the frequency on the yaxis.
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. Discuss, also, the shape of the histogram.
Record the data, in dollars (for example, 1.25 dollars).
Construct a histogram.
The number of times a value of the data occurs.
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.
Box plots or boxwhisker plots give a good graphical image of the concentration of the data. They also show how far from most of the data the extreme values are. The box plot is constructed from five values: the smallest value, the first quartile, the median, the third quartile, and the largest value. The median, the first quartile, and the third quartile will be discussed here, and then again in the section on measuring data in this chapter. We use these values to compare how close other data values are to them.
The median, a number, is a way of measuring the “center” of the data. You can think of the median as the “middle value,” although 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
The median is between the 7th value, 6.8, and the 8th value 7.2. To find the median, add the two values together and divide by 2.
The median is 7. Half of the values are smaller than 7 and half of the values are larger than 7.
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 is the middle value of the lower half of the data and the third quartile is the middle value of the upper half of the data. To get the idea, consider the same data set shown above:
1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5
The median or second quartile is 7. The lower half of the data is 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is 2.
1; 1; 2; 2; 4; 6; 6.8
The number 2, which is part of the data, is the first quartile. Onefourth of the values are the same or less than 2 and threefourths of the values are more than 2.
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 9.
7.2; 8; 8.3; 9; 10; 10; 11.5
The number 9, which is part of the data, is the third quartile. Threefourths of the values are less than 9 and onefourth of the values are more than 9.
To construct a box plot, use a horizontal 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. The middle fifty 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 box plot gives a good quick picture of the data.
Consider the following data:
1; 1; 2; 2; 4; 6; 6.8 ; 7.2; 8; 8.3; 9; 10; 10; 11.5
The first quartile is 2, the median is 7, and the third quartile is 9. The smallest value is 1 and the largest value is 11.5. The box plot is constructed as follows (see calculator instructions in the back of this book or on the TI web site):
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.
Example 2.9.
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:
Smallest value = 59
Largest value = 77
Q1: First quartile = 64.5
Q2: Second quartile or median= 66
Q3: Third quartile = 70
a. Each quarter has 25% of the data. 
b. The spreads of the four quarters are 64.5  59 = 5.5 (first quarter), 66  64.5 = 1.5 (second quarter), 70  66 = 4 (3rd quarter), and 77  70 = 7 (fourth quarter). So, the second quarter has the smallest spread and the fourth quarter has the largest spread. 
c. Interquartile Range: IQR = Q3 – Q1 = 70 – 64.5 = 5.5. 
d. The interval 59 through 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. 
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 1, the median and the third quartile were both 5, and the largest value was 7, the box plot would look as follows:
Example 2.10.
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 (Go to Solution)
What are the smallest and largest data values for each data set?
What is the median, the first quartile, and the third quartile for each data set?
Create a boxplot for each set of data.
Which boxplot 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?
For each data set, what percent of the data is between the smallest value and the first quartile? (Answer: 25%) the first quartile and the median? (Answer: 25%) the median and the third quartile? the third quartile and the largest value? What percent of the data is between the first quartile and the largest value? (Answer: 75%)
The first data set (the top box plot) has the widest spread for the middle 50% of the data. IQR = Q3 – Q1 is 82.5 – 56 = 26.5 for the first data set and 89 – 78 = 11 for the second data set. So, the first set of data has its middle 50% of scores more spread out.
25% of the data is between M and Q3 and 25% is between Q3 and Xmax.
Solution to Exercise (Return to Problem)
First Data Set
Xmin = 32
Q1 = 56
M = 74.5
Q3 = 82.5
Xmax = 99
Second Data Set
Xmin = 25.5
Q1 = 78
M = 81
Q3 = 89
Xmax = 98
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.
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.
The common measures of location are quartiles and percentiles (%iles). Quartiles are special percentiles. The first quartile, Q _{ 1 } is the same as the 25th percentile (25th %ile) and the third quartile, Q _{ 3 } , is the same as the 75th percentile (75th %ile). The median, M , is called both the second quartile and the 50th percentile (50th %ile).
To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Recall that quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that your score was higher than 90% of the people who took the test and lower than the scores of the remaining 10% of the people who took the test. Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively.
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 need further investigation.
Example 2.11.
Problem
For the following 13 real estate prices, calculate the IQR and determine if any prices are outliers. Prices are in dollars. (Source: San Jose Mercury News)
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
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
IQR = 649000 – 308750 = 340250
( 1.5 ) ( IQR ) = ( 1.5 ) ( 340250 ) = 510375
No house price is less than 201625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier.
Example 2.12.
Problem (Go to Solution)
For the two data sets in the test scores example, find the following:
a. The interquartile range. Compare the two interquartile ranges. 
b. Any outliers in either set. 
c. The 30th percentile and the 80th percentile for each set. How much data falls below the 30th percentile? Above the 80th percentile? 
Example 2.13. Finding Quartiles and Percentiles Using a Table
Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were (student data):
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 
Find the 28th percentile: Notice the 0.28 in the “cumulative relative frequency” column. 28% of 50 data values = 14. There are 14 values less than the 28th %ile. They include the two 4s, the five 5s, and the seven 6s. The 28th %ile is between the last 6 and the first 7. The 28th %ile is 6.5.
Find the median: Look again at the “cumulative relative frequency ” column and find 0.52. The median is the 50th %ile or the second quartile. 50% of 50 = 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 50th %ile is between the 25th (7) and 26th (7) values. The median is 7.
Find the third quartile: The third quartile is the same as the 75th 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 4s, 5s, 6s and 7s, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75th %ile, then, must be an 8 . Another way to look at the problem is to find 75% of 50 (= 37.5) and round up to 38. The third quartile, Q _{ 3 } , is the 38th value which is an 8. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.)
Example 2.14.
Problem (Go to Solution)
Using the table:
Find the 80th percentile.
Find the 90th percentile.
Find the first quartile. What is another name for the first quartile?
Construct a box plot of the data.
Collaborative Classroom Exercise: Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions.
How many students were surveyed?
What kind of sampling did you do?
Find the mean and standard deviation.
Find the mode.
Construct 2 different histograms. For each, starting value = _____ ending value = ____.
Find the median, first quartile, and third quartile.
Construct a box plot.
Construct a table of the data to find the following:
The 10th percentile
The 70th percentile
The percent of students who own less than 4 sweaters
A percentile indicates the relative standing of a data value when data are sorted into numerical order, from smallest to largest. p% 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 15th percentile.
Low percentiles always correspond to lower data values.
High percentiles always correspond to higher data values.
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 properly interpret percentiles is important not only when describing data, but is also important in later chapters of this textbook when calculating probabilities. Guideline: 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.
Additionally, you may also choose to state the percent of individuals or items with data values above the percentile.
Example 2.15.
On a timed math test, the first quartile for times for finishing the exam was 35 minutes. Interpret the first quartile in the context of this situation.
25% of students finished the exam in 35 minutes or less.
75% of students finished the exam in 35 minutes or more.
A low percentile would 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.)
Example 2.16.
On a 20 question math test, the 70th percentile for number of correct answers was 16. Interpret the 70th percentile in the context of this situation.
70% of students answered 16 or fewer questions correctly.
30% of students answered 16 or more questions correctly.
Note: A high percentile would be considered good, as answering more questions correctly is desirable.
Example 2.17.
At a certain community college, it was found that the 30th percentile of credit units that students are enrolled for is 7 units. Interpret the 30th percentile in the context of this situation.
30% of students are enrolled in 7 or fewer credit units
70% of students are enrolled in 7 or more credit units
In this example, there is no “good” or “bad” value judgment associated with a higher or lower percentile. Students attend community college for varied reasons and needs, and their course load varies according to their needs.
Do the following Practice Problems for Interpreting Percentiles
Exercise 2.6.1. (Go to Solution)
a. For runners in a race, a low time means a faster run. The winners in a race have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race? 
b. The 20th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20th percentile in the context of the situation. 
c. A bicyclist in the 90th percentile of a bicycle race between two towns completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90th percentile in the context of the situation. 
Exercise 2.6.2. (Go to Solution)
a. For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race? 
b. The 40th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40th percentile in the context of the situation. 
Exercise 2.6.3. (Go to Solution)
On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain.
Exercise 2.6.4. (Go to Solution)
Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85th percentile in the context of this situation.
Exercise 2.6.5. (Go to Solution)
In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78th percentile. Should Li be pleased or upset by this result? Explain.
Exercise 2.6.6. (Go to Solution)
In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had $1700 in damage and was in the 90th percentile. Should the manufacturer and/or a consumer be pleased or upset by this result? Explain. Write a sentence that interprets the 90th percentile in the context of this problem.
Exercise 2.6.7. (Go to Solution)
The University of California has two criteria used to set admission standards for freshman to be admitted to a college in the UC system: 
a. Students’ GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an “admissions index” score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students in the state. In this context, what percentile does the top 12% represent? 
b. Students whose GPAs are at or above the 96th percentile of all students at their high school are eligible (called eligible in the local context), even if they are not in the top 12% of all students in the state. What percent of students from each high school are “eligible in the local context”? 
Exercise 2.6.8. (Go to Solution)
Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34th percentile. The 34th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses?
**With contributions from Roberta Bloom
Solution to Exercise (Return to Problem)
For the IQRs, see the answer to the test scores example. The first data set has the larger IQR, so the scores between Q3 and Q1 (middle 50%) for the first data set are more spread out and not clustered about the median.
Xmax  Q3 = 99  82.5 = 16.5
Q1  Xmin = 56  32 = 24
is larger than 16.5 and larger than 24, so the first set has no outliers.
Xmax – Q3 = 98 – 89 = 9
Q1 – Xmin = 78 – 25.5 = 52.5
is larger than 9 but smaller than 52.5, so for the second set 45 and 25.5 are outliers.
To find the percentiles, create a frequency, relative frequency, and cumulative relative frequency chart (see “Frequency” from the Sampling and Data Chapter). Get the percentiles from that chart.
First Data Set
Second Data Set
30th %ile (7th value) = 78
80th %ile (18th value) = 90
30% of the data falls below the 30th %ile, and 20% falls above the 80th %ile.
Solution to Exercise 2.6.1. (Return to Exercise)
a. For runners in a race it is more desirable to have a low percentile for finish time. A low percentile means a short time, which is faster. 
b. INTERPRETATION: 20% of runners finished the race in 5.2 minutes or less. 80% of runners finished the race in 5.2 minutes or longer. 
c. He is among the slowest cyclists (90% of cyclists were faster than him.) INTERPRETATION: 90% of cyclists had a finish time of 1 hour, 12 minutes or less.Only 10% of cyclists had a finish time of 1 hour, 12 minutes or longer 
Solution to Exercise 2.6.2. (Return to Exercise)
a. For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a higher speed, which is faster. 
b. INTERPRETATION: 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour or more (faster). 
Solution to Exercise 2.6.3. (Return to Exercise)
On an exam you would prefer a high percentile; higher percentiles correspond to higher grades on the exam.
Solution to Exercise 2.6.4. (Return to Exercise)
When waiting in line at the DMV, the 85th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than you did. In this context, you would prefer a wait time corresponding to a lower percentile. INTERPRETATION: 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer.
Solution to Exercise 2.6.5. (Return to Exercise)
Li should be pleased. Her salary is relatively high compared to other recent college grads. 78% of recent college graduates earn less than Li does. 22% of recent college graduates earn more than Li does.
Solution to Exercise 2.6.6. (Return to Exercise)
The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of $1700 or less; only 10% had damage repair costs of $1700 or more.
Solution to Exercise 2.6.7. (Return to Exercise)
a. The top 12% of students are those who are at or above the 88th percentile of admissions index scores. 
b. The top 4% of students’ GPAs are at or above the 96th percentile, making the top 4% of students “eligible in the local context”. 
Solution to Exercise 2.6.8. (Return to Exercise)
You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of houses cost $240,000 or more.
The distance between the third quartile (Q3) and the first quartile (Q1). IQR = Q3  Q1.
An observation that does not fit the rest of the data.
A number that divides ordered data into hundredths.
Example .
Let a data set contain 200 ordered observations starting with {2.3,2.7,2.8,2.9,2.9,3.0…}. Then the first percentile is , because 1% of the data is to the left of this point on the number line and 99% of the data is on its right. The second percentile is . Percentiles may or may not be part of the data. In this example, the first percentile is not in the data, but the second percentile is. The median of the data is the second quartile and the 50th percentile. The first and third quartiles are the 25th and the 75th percentiles, respectively.
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.
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 (previously discussed under box plots in this chapter). 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 mean can also 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”): .
The Greek letter μ (pronounced “mew”) represents the population mean. If you take a truly random sample, the sample mean is a good estimate of the population mean.
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
In the second example, the frequencies are 3, 2, 1, and 5.
You can quickly find the location of the median by using the expression .
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 2 after the data has been ordered. For example, if the total number of data values is 97, then = = 49. The median is the 49th value in the ordered data. If the total number of data values is 100, then = = 50.5. The median occurs midway between the 50th and 51st values. The location of the median and the median itself 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 median itself.
Example 2.19.
Problem
AIDS data indicating the number of months an AIDS patient 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
The calculation for the mean is:
To find the median, M, first use the formula for the location. The location is:
Starting at the smallest value, the median is located between the 20th and 21st 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
The median is 24.
Example 2.20.
Problem
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
M = 30000
(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.
Another measure of the center is the mode. The mode is the most frequent value. If a data set has two values that occur the same number of times, then the set is bimodal.
Example 2.21. Statistics exam scores for 20 students are as follows
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
Find the mode.
Solution
The most frequent score is 72, which occurs five times. Mode = 72.
Example 2.22.
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 an average 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.
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.
The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean of the sample gets closer and closer to µ . This is discussed in more detail in The Central Limit Theorem.
The formula for the mean is located in the Summary of Formulas section course.
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.
# of movies  Relative Frequency 

0  5/30 
1  15/30 
2  6/30 
3  4/30 
4  1/30 
If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.
statistic of a sampling distribution is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean is an example of a statistic which estimates the population mean μ .
A number that measures the central tendency. A common name for mean is ‘average.’ The term ‘mean’ is a shortened form of ‘arithmetic mean.’ By definition, the mean for a sample (denoted by ) is , and the mean for a population (denoted by μ ) 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 than the median. The median may or may not be part of the data.
The value that appears most frequently in a set of data.
Consider the following data set:
4 ; 5 ; 6 ; 6 ; 6 ; 7 ; 7 ; 7 ; 7 ; 7 ; 7 ; 8 ; 8 ; 8 ; 9 ; 10
This data produces the histogram shown below. 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 7 for these data. In a perfectly symmetrical distribution, the mean, the median, and the mode are often the same.
The histogram for the data:
4 ; 5 ; 6 ; 6 ; 6 ; 7 ; 7 ; 7 ; 7 ; 8
is not symmetrical. The righthand side seems “chopped off” compared to the left side. The shape distribution 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 7. 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 more so.
The histogram for the data:
6 ; 7 ; 7 ; 7 ; 7 ; 8 ; 8 ; 8 ; 9 ; 10
is also not symmetrical. It is skewed to the right.
The mean is 7.7, the median is 7.5, and the mode is 7. Notice that the mean is the largest statistic, while the mode is the smallest. Again, the mean reflects the skewing the most.
To summarize, generally if the distribution of data is skewed to the left, the mean is less than the median, which is less than the mode. If the distribution of data is skewed to the right, the mode is less than the median, which is less than the mean.
Skewness and symmetry become important when we discuss probability distributions in later chapters.
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.
provides a numerical measure of the overall amount of variation in a data set
can be used to determine whether a particular data value is close to or far from the mean
The standard deviation is always positive or 0. 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 waiting times at the checkout line for customers at supermarket A and supermarket B; the average wait time at both markets is 5 minutes. At market A, the standard deviation for the waiting time is 2 minutes; at market B the standard deviation for the waiting time is 4 minutes. Because market B has a higher standard deviation, we know that there is more variation in the waiting times at market B. Overall, wait times at market B are more spread out from the average; wait times at market A are more concentrated near the average.
Suppose that Rosa and Binh both shop at Market A. Rosa waits for 7 minutes and Binh waits for 1 minute at the checkout counter. At market A, the mean wait time is 5 minutes and the standard deviation is 2 minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.
7 is 2 minutes longer than the average of 5; 2 minutes is equal to one standard deviation.
Rosa’s wait time of 7 minutes is 2 minutes longer than the average of 5 minutes.
Rosa’s wait time of 7 minutes is one standard deviation above the average of 5 minutes.
A wait time that is only one standard deviation from the average is considered close to the average.
1 is 4 minutes less than the average of 5; 4 minutes is equal to two standard deviations.
Binh’s wait time of 1 minute is 4 minutes less than the average of 5 minutes.
Binh’s wait time of 1 minute is two standard deviations below the average of 5 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 2 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 2 standard deviations. (We will learn more about this in later chapters.)
The number line may help you understand standard deviation. If we were to put 5 and 7 on a number line, 7 is to the right of 5. We say, then, that 7 is one standard deviation to the right of 5 because 5 + (1)(2) = 7. If 1 were also part of the data set, then 1 is two standard deviations to the left of 5 because 5 + (2)(2) = 1.
In general, a value = mean + (#ofSTDEV)(standard deviation)
where #ofSTDEVs = the number of standard deviations
7 is one standard deviation more than the mean of 5 because: 7=5+(1)(2)
1 is two standard deviations less than the mean of 5 because: 1=5+(−2)(2)
The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population:
sample:
Population: x = μ + (#ofSTDEV)(σ)
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 is the sample mean and the Greek symbol μ is the population mean.
If x is a data value, 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 data is for a population, in symbols a deviation is x − μ . For sample data, in symbols a deviation is x – .
The procedure to calculate the standard deviation depends on whether the data is for the entire population or comes from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is 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 an average of the squares of the deviations (the 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 data is from a population, 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 is from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n1, one less than the number of items in the sample. You can see that in the formulas below.
s = or s =
For the sample standard deviation, the denominator is n1, that is the sample size MINUS 1.
Formulas for the Population Standard Deviation
σ = or σ =
For the population standard deviation, the denominator is N, the number of items in the population.
In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is 1. If a value appears three times in the data set or population, f is 3.
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 Central Limit Theorem (not now). The notation for the standard error of the mean is 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 TI83,83+,84+ calculator, you need to select the appropriate standard deviation σ or s 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 stepbystep example to help you understand how the standard deviation measures variation from the mean.
Example 2.23.
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
The average age is 10.53 years, rounded to 2 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  
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 × .275625 = 1.1025 
10.5  4  10.5 – 10.525 = – 0.025  ( – 0.025 ) ^{2} = 0.000625  4 × .000625 = .0025 
11  6  11 – 10.525 = 0.475  ( 0.475 ) ^{2} = 0.225625  6 × .225625 = 1.35375 
11.5  3  11.5 – 10.525 = 0.975  ( 0.975 ) ^{2} = 0.950625  3 × .950625 = 2.851875 
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):
The sample standard deviation s is equal to the square root of the sample variance:
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 1.
Verify the mean and standard deviation calculated above on your calculator or computer.
Solution
For the TI83,83+,84+, enter data into the list editor. 
Put the data values in list L1 and the frequencies in list L2. 
STAT CALC 1VarStats L1, L2 
=10.525 
Use Sx because this is sample data (not a population): Sx=.715891 
For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation)
For a sample: x = + (#ofSTDEVs)(s)
For a population: x = μ + (#ofSTDEVs)( σ )
For this example, use x = + (#ofSTDEVs)(s) because the data is from a sample
Problem 2.
Find the value that is 1 standard deviation above the mean. Find .
Solution
Problem 3.
Find the value that is two standard deviations below the mean. Find .
Solution
Problem 4.
Find the values that are 1.5 standard deviations from (below and above) the mean.
Solution
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. The deviations 0.97 and 0.47 indicate that. A positive deviation occurs when the data value is greater than the mean. A negative deviation occurs when the data value is less than the mean; the deviation is 1.525 for the data value 9. If you add the deviations, the sum is always zero. (For this example, 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 n1=201=19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n1). 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 (n1) 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. When the standard deviation is 0, 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.
The formula for the standard deviation is at the end of the chapter.
Example 2.24.
Problem
Use the following data (first exam scores) from Susan Dean’s spring precalculus 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
a. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.  
b. Calculate the following to one decimal place using a TI83+ or TI84 calculator:
 
c. 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
a.
 
b.
 
c. The xaxis goes from 32.5 to 100.5; yaxis goes from 2.4 to 15 for the histogram; number of intervals is 5 for the histogram so the width of an interval is (100.5  32.5) divided by 5 which is equal to 13.6. Endpoints of the intervals: 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.
Figure 2.1. 
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 standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, it can be misleading to compare the data values directly.
For each data value, calculate how many standard deviations the value is away from its mean.
Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.
Compare the results of this calculation.
#ofSTDEVs is often called a “zscore”; we can use the symbol z. In symbols, the formulas become:
Sample  x = + z s  
Population  x = μ + z σ 
Example 2.25.
Problem
Two students, John and Ali, from different high schools, wanted to find out who had the highest G.P.A. when compared to his school. Which student had the highest G.P.A. when compared to his school?
Student  GPA  School Mean GPA  School Standard Deviation 

John  2.85  3.0  0.7 
Ali  77  80  10 
Solution
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.
;
For John,
For Ali,
John has the better G.P.A. when compared to his school because his G.P.A. is 0.21 standard deviations below his mean while Ali’s G.P.A. is 0.3 standard deviations below his mean. John’s zscore of −0.21 is higher than Ali’s zscore of −0.3 . For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.
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 2 standard deviations of the mean.
At least 89% of the data is within 3 standard deviations of the mean.
At least 95% of the data is within 4 1/2 standard deviations of the mean.
This is known as Chebyshev’s Rule.
For data having a distribution that is MOUNDSHAPED and SYMMETRIC:
Approximately 68% of the data is within 1 standard deviation of the mean.
Approximately 95% of the data is within 2 standard deviations of the mean.
More than 99% of the data is within 3 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 moundshaped and symmetric. We will learn more about this when studying the “Normal” or “Gaussian” probability distribution in later chapters.
**With contributions from Roberta Bloom
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.
Mean of the squared deviations from the mean. Square of the standard deviation. For a set of data, a deviation can be represented as where x is a value of the data and 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 1.
Commonly Used Symbols
The symbol Σ means to add or to find the sum.
n = the number of data values in a sample
N = the number of people, things, etc. in the population
= the sample mean
s = the sample standard deviation
μ = the population mean
σ = the population standard deviation
f = frequency
x = numerical value
Commonly Used Expressions
x * f = A value multiplied by its respective frequency
∑x = The sum of the values
∑x * f = The sum of values multiplied by their respective frequencies
or ( x − μ ) = Deviations from the mean (how far a value is from the mean)
or ( x − μ ) ^{2} = Deviations squared
or f ( x − μ ) ^{2} = The deviations squared and multiplied by their frequencies
Mean Formulas:
or
μ = or μ =
Standard Deviation Formulas:
s = or s =
σ = or σ =
Formulas Relating a Value, the Mean, and the Standard Deviation:
value = mean + (#ofSTDEVs)(standard deviation), where #ofSTDEVs = the number of standard deviations
x = + (#ofSTDEVs)( s )
x = μ + (#ofSTDEVs)( σ )
The student will calculate and interpret the center, spread, and location of the data.
The student will construct and interpret histograms an box plots.
Sixtyfive 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.
Data Value (# cars)  Frequency  Relative Frequency  Cumulative Relative Frequency 

Exercise 2.11.3.
What is the difference between relative frequency and frequency for each data value?
Exercise 2.11.4.
What is the difference between cumulative relative frequency and relative frequency for each data value?
Enter your data into your calculator or computer.
Determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram below. Label the horizontal and vertical axes with words. Include numerical scaling.
Calculate the following values:
Use the table in section 2.11.3 to calculate the following values:
Exercise 2.11.16. (Go to Solution)
Find the value that is 3 standard deviations:
a. Above the mean 
b. Below the mean 
Construct a box plot below. Use a ruler to measure and scale accurately.
Looking at your box plot, does it appear that the data are concentrated together, spread out evenly, or concentrated in some areas, but not in others? How can you tell?
The student will calculate measures of the center of the data.
The student will calculate the spread of the data.
The population parameters below describe the fulltime equivalent number of students (FTES) each year at Lake Tahoe Community College from 197677 through 20042005. (Source: Graphically Speaking by Bill King, LTCC Institutional Research, December 2005).
Use these values to answer the following questions:
μ = 1000 FTES
Median  1014 FTES
σ = 474 FTES
First quartile = 528.5 FTES
Third quartile = 1447.5 FTES
n = 29 years
Exercise 2.12.1. (Go to Solution)
A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer.
Exercise 2.12.4. (Go to Solution)
What percent of the FTES were from 528.5 to 1447.5? How do you know?
Exercise 2.13.1. (Go to Solution)
Twentyfive randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
# of movies  Frequency  Relative Frequency  Cumulative Relative Frequency 

0  5  
1  9  
2  6  
3  4  
4  1 
a. Find the sample mean 
b. Find the sample standard deviation, s 
c. Construct a histogram of the data. 
d. Complete the columns of the chart. 
e. Find the first quartile. 
f. Find the median. 
g. Find the third quartile. 
h. Construct a box plot of the data. 
i. What percent of the students saw fewer than three movies? 
j. Find the 40th percentile. 
k. Find the 90th percentile. 
l. Construct a line graph of the data. 
m. Construct a stem plot of the data. 
Exercise 2.13.2.
The median age for U.S. blacks currently is 30.1 years; for U.S. whites it is 36.6 years. (Source: U.S. Census)
a. Based upon this information, give two reasons why the black median age could be lower than the white median age. 
b. Does the lower median age for blacks necessarily mean that blacks die younger than whites? Why or why not? 
c. How might it be possible for blacks and whites to die at approximately the same age, but for the median age for whites to be higher? 
Exercise 2.13.3. (Go to Solution)
Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows:
X  Frequency  Relative Frequency  Cumulative Relative Frequency 

1  2  
2  5  
3  8  
4  12  
5  12  
7  1 
a. Find the sample mean 
b. Find the sample standard deviation, s 
c. Construct a histogram of the data. 
d. Complete the columns of the chart. 
e. Find the first quartile. 
f. Find the median. 
g. Find the third quartile. 
h. Construct a box plot of the data. 
i. What percent of the students owned at least five pairs? 
j. Find the 40th percentile. 
k. Find the 90th percentile. 
l. Construct a line graph of the data 
m. Construct a stem plot of the data 
Exercise 2.13.4.
600 adult Americans were asked by telephone poll, What do you think constitutes a middleclass income? The results are below. Also, include left endpoint, but not the right endpoint. (Source: Time magazine; survey by Yankelovich Partners, Inc.)
“Not sure” answers were omitted from the results.
Salary ($)  Relative Frequency 

< 20,000  0.02 
20,000  25,000  0.09 
25,000  30,000  0.19 
30,000  40,000  0.26 
40,000  50,000  0.18 
50,000  75,000  0.17 
75,000  99,999  0.02 
100,000+  0.01 
a. What percent of the survey answered “not sure” ? 
b. What percent think that middleclass is from $25,000  $50,000 ? 
c. Construct a histogram of the data

d. Find the 40th and 80th percentiles 
e. Construct a bar graph of the data 
Exercise 2.13.5. (Go to Solution)
Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year (Source: San Jose Mercury News)
177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265
a. Organize the data from smallest to largest value.  
b. Find the median.  
c. Find the first quartile.  
d. Find the third quartile.  
e. Construct a box plot of the data.  
f. The middle 50% of the weights are from _______ to _______.  
g. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why?  
h. If our population were the San Francisco 49ers, would the above data be a sample of weights or the population of weights? Why?  
i. Assume the population was the San Francisco 49ers. Find:
 
j. That same year, the average weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer? 
Exercise 2.13.6.
An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes. A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes.
a. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he? 
b. Who is the fastest runner with respect to his or her class? Explain why. 
Exercise 2.13.7.
In a survey of 20 year olds in China, Germany and America, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results.
a. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected. 
b. Explain how it is possible that more Americans than Germans surveyed have been to over eight foreign countries. 
c. Compare the three box plots. What do they imply about the foreign travel of twenty year old residents of the three countries when compared to each other? 
Exercise 2.13.8.
One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher’s attitude toward math became more positive. The twelve change scores are as follows:
3; 8; 1; 2; 0; 5; 3; 1; 1; 6; 5; 2
a. What is the average change score? 
b. What is the standard deviation for this population? 
c. What is the median change score? 
d. Find the change score that is 2.2 standard deviations below the mean. 
Exercise 2.13.9. (Go to Solution)
Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best G.P.A. when compared to his school? Explain how you determined your answer.
Student  G.P.A.  School Ave. G.P.A.  School Standard Deviation 

Thuy  2.7  3.2  0.8 
Vichet  87  75  20 
Kamala  8.6  8  0.4 
Exercise 2.13.10.
Given the following box plot:
a. Which quarter has the smallest spread of data? What is that spread?  
b. Which quarter has the largest spread of data? What is that spread?  
c. Find the Inter Quartile Range (IQR).  
d. Are there more data in the interval 5  10 or in the interval 10  13? How do you know this?  
e. Which interval has the fewest data in it? How do you know this?

Exercise 2.13.11.
Given the following box plot:
a. Think of an example (in words) where the data might fit into the above box plot. In 25 sentences, write down the example. 
b. What does it mean to have the first and second quartiles so close together, while the second to fourth quartiles are far apart? 
Exercise 2.13.12.
Santa Clara County, CA, has approximately 27,873 JapaneseAmericans. Their ages are as follows. (Source: West magazine)
Age Group  Percent of Community 

017  18.9 
1824  8.0 
2534  22.8 
3544  15.0 
4554  13.1 
5564  11.9 
65+  10.3 
a. Construct a histogram of the JapaneseAmerican community in Santa Clara County, CA. The bars will not be the same width for this example. Why not? 
b. What percent of the community is under age 35? 
c. Which box plot most resembles the information above? 
Exercise 2.13.13.
Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, each asked adult consumers the number of fiction paperbacks they had purchased the previous month. The results are below.
# of books  Freq.  Rel. Freq. 

0  10  
1  12  
2  16  
3  12  
4  8  
5  6  
6  2  
8  2 
# of books  Freq.  Rel. Freq. 

0  18  
1  24  
2  24  
3  22  
4  15  
5  10  
7  5  
9  1 
# of books  Freq.  Rel. Freq. 

01  20  
23  35  
45  12  
67  2  
89  1 
a. Find the relative frequencies for each survey. Write them in the charts. 
b. Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher’s survey. For Publishers A and B, make bar widths of 1. For Publisher C, make bar widths of 2. 
c. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical. 
d. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not? 
e. Make new histograms for Publisher A and Publisher B. This time, make bar widths of 2. 
f. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer. 
Exercise 2.13.14.
Often, cruise ships conduct all onboard transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their onboard bills for a sevenday cruise from Los Angeles to the Mexican Riviera. Below is a summary of the bills for each group.
Amount($)  Frequency  Rel. Frequency 

51100  5  
101150  10  
151200  15  
201250  15  
251300  10  
301350  5 
Amount($)  Frequency  Rel. Frequency 

100150  5  
201250  5  
251300  5  
301350  5  
351400  10  
401450  10  
451500  10  
501550  10  
551600  5  
601650  5 
a. Fill in the relative frequency for each group.  
b. Construct a histogram for the Singles group. Scale the xaxis by $50. widths. Use relative frequency on the yaxis.  
c. Construct a histogram for the Couples group. Scale the xaxis by $50. Use relative frequency on the yaxis.  
d. Compare the two graphs:
 
e. Construct a new graph for the Couples by hand. Since each couple is paying for two individuals, instead of scaling the xaxis by $50, scale it by $100. Use relative frequency on the yaxis.  
f. Compare the graph for the Singles with the new graph for the Couples:
 
i. By scaling the Couples graph differently, how did it change the way you compared it to the Singles?  
j. Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person in a couple? Explain why in one or two complete sentences. 
Exercise 2.13.15. (Go to Solution)
Refer to the following histograms and box plot. Determine which of the following are true and which are false. Explain your solution to each part in complete sentences.
a. The medians for all three graphs are the same. 
b. We cannot determine if any of the means for the three graphs is different. 
c. The standard deviation for (b) is larger than the standard deviation for (a). 
d. We cannot determine if any of the third quartiles for the three graphs is different. 
Exercise 2.13.16.
Refer to the following box plots.
a. In complete sentences, explain why each statement is false.
 
b. For which group, Data 1 or Data 2, is the value of “7” more likely to be an outlier? Explain why in complete sentences 
Exercise 2.13.17. (Go to Solution)
In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirtysix lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference.
a. Organize the data in a chart. 
b. Find the median, the first quartile, and the third quartile. 
c. Find the 65th percentile. 
d. Find the 10th percentile. 
e. Construct a box plot of the data. 
f. The middle 50% of the conferences last from _______ days to _______ days. 
g. Calculate the sample mean of days of engineering conferences. 
h. Calculate the sample standard deviation of days of engineering conferences. 
i. Find the mode. 
j. If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice. 
k. Give two reasons why you think that 3  5 days seem to be popular lengths of engineering conferences. 
Exercise 2.13.18.
A survey of enrollment at 35 community colleges across the United States yielded the following figures (source: Microsoft Bookshelf):
6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622
a. Organize the data into a chart with five intervals of equal width. Label the two columns “Enrollment” and “Frequency.” 
b. Construct a histogram of the data. 
c. If you were to build a new community college, which piece of information would be more valuable: the mode or the average size? 
d. Calculate the sample average. 
e. Calculate the sample standard deviation. 
f. A school with an enrollment of 8000 would be how many standard deviations away from the mean? 
Exercise 2.13.19. (Go to Solution)
The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years. (Source: Bureau of the Census)
a. What does it mean for the median age to rise? 
b. Give two reasons why the median age could rise. 
c. For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not? 
Exercise 2.13.20.
A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results.
a. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected for that car series.  
b. Which group is most likely to have an outlier? Explain how you determined that.  
c. Compare the three box plots. What do they imply about the age of purchasing a BMW from the series when compared to each other?  
d. Look at the BMW 5 series. Which quarter has the smallest spread of data? What is that spread?  
e. Look at the BMW 5 series. Which quarter has the largest spread of data? What is that spread?  
f. Look at the BMW 5 series. Find the Inter Quartile Range (IQR).  
g. Look at the BMW 5 series. Are there more data in the interval 3138 or in the interval 4555? How do you know this?  
h. Look at the BMW 5 series. Which interval has the fewest data in it? How do you know this?

Exercise 2.13.21. (Go to Solution)
The following box plot shows the U.S. population for 1990, the latest available year. (Source: Bureau of the Census, 1990 Census)
a. Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)? How do you know? 
b. 12.6% are age 65 and over. Approximately what percent of the population are of working age adults (above age 17 to age 65)? 
Exercise 2.13.22.
Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information:
Javier  Ercilla  

6.0 miles  6.0 miles  
s  4.0 miles  7.0 miles 
a. How can you determine which survey was correct ? 
b. Explain what the difference in the results of the surveys implies about the data. 
c. If the two histograms depict the distribution of values for each supervisor, which one depicts Ercilia’s sample? How do you know?
Figure 2.2. 
d. If the two box plots depict the distribution of values for each supervisor, which one depicts Ercilia’s sample? How do you know? Figure 2.3. 
Exercise 2.13.23. (Go to Solution)
Student grades on a chemistry exam were:
77, 78, 76, 81, 86, 51, 79, 82, 84, 99
a. Construct a stemandleaf plot of the data. 
b. Are there any potential outliers? If so, which scores are they? Why do you consider them outliers? 
The next three questions refer to the following information. We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section.
Number of years  Frequency 

7  1 
14  3 
15  1 
18  1 
19  4 
20  3 
22  1 
23  1 
26  1 
40  2 
42  2 
Total = 20 
Exercise 2.13.26. (Go to Solution)
Is this a sample or the entire population?
A. sample 
B. entire population 
C. neither 
The next two questions refer to the following table. X = the number of days per week that 100 clients use a particular exercise facility.
X  Frequency 

0  3 
1  12 
2  33 
3  28 
4  11 
5  9 
6  4 
Exercise 2.13.28. (Go to Solution)
The number that is 1.5 standard deviations BELOW the mean is approximately:
A. 0.7 
B. 4.8 
C. 2.8 
D. Cannot be determined 
The next two questions refer to the following histogram. Suppose one hundred eleven people who shopped in a special Tshirt store were asked the number of Tshirts they own costing more than $19 each.
Exercise 2.13.29. (Go to Solution)
The percent of people that own at most three (3) Tshirts costing more than $19 each is approximately:
A. 21 
B. 59 
C. 41 
D. Cannot be determined 
Exercise 2.13.30. (Go to Solution)
If the data were collected by asking the first 111 people who entered the store, then the type of sampling is:
A. cluster 
B. simple random 
C. stratified 
D. convenience 
Exercise 2.13.31. (Go to Solution)
Below are the 2008 obesity rates by U.S. states and Washington, DC. (Source: http://www.cdc.gov/obesity/data/trends.html#State)
State  Percent (%)  State  Percent (%) 

Alabama  31.4  Montana  23.9 
Alaska  26.1  Nebraska  26.6 
Arizona  24.8  Nevada  25 
Arkansas  28.7  New Hampshire  24 
California  23.7  New Jersey  22.9 
Colorado  18.5  New Mexico  25.2 
Connecticut  21  New York  24.4 
Delaware  27  North Carolina  29 
Washington, DC  21.8  North Dakota  27.1 
Florida  24.4  Ohio  28.7 
Georgia  27.3  Oklahoma  30.3 
Hawaii  22.6  Oregon  24.2 
Idaho  24.5  Pennsylvania  27.7 
Illinois  26.4  Rhode Island  21.5 
Indiana  26.3  South Carolina  30.1 
Iowa  26  South Dakota  27.5 
Kansas  27.4  Tennessee  30.6 
Kentucky  29.8  Texas  28.3 
Louisiana  28.3  Utah  22.5 
Maine  25.2  Vermont  22.7 
Maryland  26  Virginia  25 
Massachusetts  20.9  Washington  25.4 
Michigan  28.9  West Virginia  31.2 
Minnesota  24.3  Wisconsin  25.4 
Mississippi  32.8  Wyoming  24.6 
Missouri  28.5 
a.. Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the xaxis with the states. 
b.. Use a random number generator to randomly pick 8 states. Construct a bar graph of the obesity rates of those 8 states. 
c.. Construct a bar graph for all the states beginning with the letter “A.” 
d.. Construct a bar graph for all the states beginning with the letter “M.” 
Exercise 2.13.32. (Go to Solution)
A music school has budgeted to purchase 3 musical instruments. They plan to purchase a piano costing $3000, a guitar costing $550, and a drum set costing $600. The average cost for a piano is $4,000 with a standard deviation of $2,500. The average cost for a guitar is $500 with a standard deviation of $200. The average cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer numerically.
Exercise 2.13.33. (Go to Solution)
Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the table below. (Note that this is the data presented for publisher B in homework exercise 13).
# of books  Freq.  Rel. Freq. 

0  18  
1  24  
2  24  
3  22  
4  15  
5  10  
7  5  
9  1 
Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion.
If a data value is identified as an outlier, what should be done about it?
Are any data values further than 2 standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is moundshaped and symmetric, rather than for skewed data.)
Do parts (a) and (c) of this problem give the same answer?
Examine the shape of the data. Which part, (a) or (c), of this question gives a more appropriate result for this data?
Based on the shape of the data which is the most appropriate measure of center for this data: mean, median or mode?
**Exercises 32 and 33 contributed by Roberta Bloom
Solution to Exercise 2.13.1. (Return to Exercise)
a. 1.48 
b. 1.12 
e. 1 
f. 1 
g. 2 
h.

i. 80% 
j. 1 
k. 3 
Solution to Exercise 2.13.3. (Return to Exercise)
a. 3.78 
b. 1.29 
e. 3 
f. 4 
g. 5 
h.

i. 32.5% 
j. 4 
k. 5 
Solution to Exercise 2.13.5. (Return to Exercise)
b. 241  
c. 205.5  
d. 272.5  
e.
 
f. 205.5, 272.5  
g. sample  
h. population  
i.
 
j. Young 
Solution to Exercise 2.13.17. (Return to Exercise)
b. 4,3,5 
c. 4 
d. 3 
e.

f. 3,5 
g. 3.94 
h. 1.28 
i. 3 
j. mode 
Solution to Exercise 2.13.31. (Return to Exercise)
Example solution for b using the random number generator for the Ti84 Plus to generate a simple random sample of 8 states. Instructions are below.
Number the entries in the table 1  51 (Includes Washington, DC; Numbered vertically) 
Press MATH 
Arrow over to PRB 
Press 5:randInt( 
Enter 51,1,8) 
Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}. Corresponding percents are {28.7 21.8 24.5 26 28.9 32.8 25 24.6}.
Solution to Exercise 2.13.32. (Return to Exercise)
For pianos, the cost of the piano is 0.4 standard deviations BELOW average. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE average. For drums, the cost of the drum set is 1.0 standard deviations BELOW average. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar cost the most in comparison to the cost of other instruments of the same type.
Solution to Exercise 2.13.33. (Return to Exercise)
IQR = 4 – 1 = 3 ; Q1 – 1.5*IQR = 1 – 1.5(3) = 3.5 ; Q3 + 1.5*IQR = 4 + 1.5(3) = 8.5 ;The data value of 9 is larger than 8.5. The purchase of 9 books in one month is an outlier.
The outlier should be investigated to see if there is an error or some other problem in the data; then a decision whether to include or exclude it should be made based on the particular situation. If it was a correct value then the data value should remain in the data set. If there is a problem with this data value, then it should be corrected or removed from the data. For example: If the data was recorded incorrectly (perhaps a 9 was miscoded and the correct value was 6) then the data should be corrected. If it was an error but the correct value is not known it should be removed from the data set.
xbar – 2s = 2.45 – 2*1.88 = 1.31 ; xbar + 2s = 2.45 + 2*1.88 = 6.21 ; Using this method, the five data values of 7 books purchased and the one data value of 9 books purchased would be considered unusual.
No: part (a) identifies only the value of 9 to be an outlier but part (c) identifies both 7 and 9.
The data is skewed (to the right). It would be more appropriate to use the method involving the IQR in part (a), identifying only the one value of 9 books purchased as an outlier. Note that part (c) remarks that identifying unusual data values by using the criteria of being further than 2 standard deviations away from the mean is most appropriate when the data are moundshaped and symmetric.
The data are skewed to the right. For skewed data it is more appropriate to use the median as a measure of center.
Class Time:
Names:
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.
Record the number of pairs of shoes you own:
Randomly survey 30 classmates. Record their values.
_____  _____  _____  _____  _____ 
_____  _____  _____  _____  _____ 
_____  _____  _____  _____  _____ 
_____  _____  _____  _____  _____ 
_____  _____  _____  _____  _____ 
_____  _____  _____  _____  _____ 
Construct a histogram. Make 56 intervals. Sketch the graph using a ruler and pencil. Scale the axes.
Figure 2.4.
Calculate the following:
=
s =
Are the data discrete or continuous? How do you know?
Describe the shape of the histogram. Use complete sentences.
Are there any potential outliers? Which value(s) is (are) it (they)? Use a formula to check the end values to determine if they are potential outliers.
Determine the following:
Minimum value =
Median =
Maximum value =
First quartile =
Third quartile =
IQR =
Construct a box plot of data
What does the shape of the box plot imply about the concentration of data? Use complete sentences.
Using the box plot, how can you determine if there are potential outliers?
How does the standard deviation help you to determine concentration of the data and whether or not there are potential outliers?
What does the IQR represent in this problem?
Show your work to find the value that is 1.5 standard deviations:
a. Above the mean: 
b. Below the mean: 