3.3: Measures of the Spread of the Data
 Page ID
 21501
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
 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.
The standard deviation provides a measure of the overall variation in a data set
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.
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.
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.)
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 \(5 + (2)(2) = 1\).
Figure \(\PageIndex{1}\)
 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: \(1 = 5 + (2)(2)\).
The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population.
 sample: \[x = \bar{x} + \text{(#ofSTDEV)(s)}\]
 Population: \[x = \mu + \text{(#ofSTDEV)(s)}\]
The lower case letter s represents the sample standard deviation and the Greek letter \(\sigma\) (sigma, lower case) represents the population standard deviation.
Calculating the Standard Deviation
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 \(x  \mu\). For sample data, in symbols a deviation is \(x  \bar{x}\).
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 \(\sigma\) (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 \(\sigma\).
To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the \(x  \bar{x}\) values for a sample, or the \(x  \mu\) values for a population). The symbol \(\sigma^{2}\) represents the population variance; the population standard deviation \(\sigma\) 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
\[s = \sqrt{\dfrac{\sum(x\bar{x})^{2}}{n1}} \label{eq1}\]
or
\[s = \sqrt{\dfrac{\sum f (x\bar{x})^{2}}{n1}} \label{eq2}\]
For the sample standard deviation, the denominator is \(n  1\), that is the sample size MINUS 1.
Formulas for the Population Standard Deviation
\[\sigma = \sqrt{\dfrac{\sum(x\mu)^{2}}{N}} \label{eq3} \]
or
\[\sigma = \sqrt{\dfrac{\sum f (x\mu)^{2}}{N}} \label{eq4}\]
For the population standard deviation, the denominator is \(N\), the number of items in the population.
In Equations \ref{eq2} and \ref{eq4}, \(f\) 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
The statistic of a sampling distribution was discussed in Section 2.6. 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 Chapter 7. The notation for the standard error of the mean is \(\dfrac{\sigma}{\sqrt{n}}\) where \(\sigma\) 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 \(\sigma_{x}\) 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 stepbystep example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)
Example \(\PageIndex{1}\)
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;
\[\bar{x} = \dfrac{9+9.5(2)+10(4)+10.5(4)+11(6)+11.5(3)}{20} = 10.525\]
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 – \(\bar{x}\))  (x – \(\bar{x}\))2  (f)(x – \(\bar{x}\))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 
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):
\[s^{2} = \dfrac{9.7375}{201} = 0.5125\]
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.
 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: \(x\) = \(\bar{x}\) + (#ofSTDEVs)(s)
 For a population: \(x\) = \(\mu\) + (#ofSTDEVs)\(\sigma\)
 For this example, use x = \(\bar{x}\) + (#ofSTDEVs)(s) because the data is from a sample
 Verify the mean and standard deviation on your calculator or computer.
 Find the value that is one standard deviation above the mean. Find (\(\bar{x}\) + 1s).
 Find the value that is two standard deviations below the mean. Find (\(\bar{x}\) – 2s).
 Find the values that are 1.5 standard deviations from (below and above) the mean.
Solution

 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 1:1VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.
 \(\bar{x}\) = 10.525
 Use Sx because this is sample data (not a population): Sx=0.715891
 (\(\bar{x} + 1s) = 10.53 + (1)(0.72) = 11.25\)
 (\(\bar{x}  2s) = 10.53 – (2)(0.72) = 9.09\)

 (\(\bar{x}  1.5s) = 10.53 – (1.5)(0.72) = 9.45\)
 (\(\bar{x} + 1.5s) = 10.53 + (1.5)(0.72) = 11.61\)
Exercise 2.8.1
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.
Answer
\(\mu\) = 30.68
\(s = 6.09\)
(\(\bar{x} + 2s = 30.68 + (2)(6.09) = 42.86\).
Explanation of the standard deviation calculation shown in the table
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 \(\PageIndex{1}\), 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 \(\sigma\), is either zero or larger than zero. 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 \(\sigma\) 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.
Example \(\PageIndex{2}\)
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
 Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
 Calculate the following to one decimal place using a TI83+ or TI84 calculator:
 The sample mean
 The sample standard deviation
 The median
 The first quartile
 The third quartile
 IQR
 Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.
Answer
 See Table

 The sample mean = 73.5
 The sample standard deviation = 17.9
 The median = 73
 The first quartile = 61
 The third quartile = 90
 IQR = 90 – 61 = 29
 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.
Figure 2.9.2.
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.
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?) 
Exercise \(\PageIndex{2}\)
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 TI83+ or TI84 calculator.
Answer
\(\mu = 9.3\) and \(s = 2.2\)
Standard deviation of Grouped Frequency Tables
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:
\[\text{Mean of Frequency Table} = \dfrac{\sum fm}{\sum f}\]
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.
Example \(\PageIndex{3}\)
Find the standard deviation for the data in Table \(\PageIndex{3}\).
Class  Frequency, f  Midpoint, m  m^{2}  \(\bar{x}\)  fm^{2}  Standard Deviation 

0–2  1  1  1  7.58  1  3.5 
3–5  6  4  16  7.58  96  3.5 
6–8  10  7  49  7.58  490  3.5 
9–11  7  10  100  7.58  700  3.5 
12–14  0  13  169  7.58  0  3.5 
15–17  2  16  256  7.58  512  3.5 
For this data set, we have the mean, \(\bar{x}\) = 7.58 and the standard deviation, \(s_{x}\) = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated, \(s_{x} = \sqrt{\dfrac{f(m  \bar{x})^{2}}{n1}}\) where \(s_{x}\) = sample standard deviation, \(\bar{x}\) = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.
Find the standard deviation for the data from the previous example
Class  02  35  68  9–11  12–14  15–17 

Frequency, f  1  6  10  7  0  2 
First, press the STAT key and select 1:Edit
Figure 2.8.3
Input the midpoint values into L1 and the frequencies into L2
Figure 2.8.4
Select STAT, CALC, and 1: 1Var Stats
Figure 2.8.5
Select 2^{nd} then 1 then , 2^{nd} then 2 Enter
Figure 2.8.6
You will see displayed both a population standard deviation, \(\sigma_{x}\), and the sample standard deviation, \(s_{x}\).
Comparing Values from Different Data Sets
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.
 \(\text{#ofSTDEVs} = \dfrac{\text{valuemean}}{\text{standard deviation}}\)
 Compare the results of this calculation.
#ofSTDEVs is often called a "zscore"; we can use the symbol \(z\). In symbols, the formulas become:
Sample  \(x = \bar{x} + zs\)  \(z = \dfrac{x  \bar{x}}{s}\) 
Population  \(x = \mu + z\sigma\)  \(z = \dfrac{x  \mu}{\sigma}\) 
Example \(\PageIndex{4}\)
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 
Answer
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.
\(z = \text{#ofSTDEVs} = \left(\dfrac{\text{valuemean}}{\text{standard deviation}}\right) = \left(\dfrac{x + \mu}{\sigma}\right)\)
For John,
\(z\) = \(\text{#ofSTDEVs} = \left(\dfrac{2.853.0}{0.7}\right) = 0.21\)
For Ali,
\(z = \text{#ofSTDEVs} = (\dfrac{7780}{10}) = 0.3\)
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 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.
Exercise \(\PageIndex{4}\)
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 
Answer
For Angie:
\[z = \left(\dfrac{26.227.2}{0.8}\right) = 1.25\]
For Beth:
\[z = \left(\dfrac{27.330.1}{1.4}\right) = 2\]
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.
For data having a distribution that is BELLSHAPED 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 bellshaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.
References
 Data from Microsoft Bookshelf.
 King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutionalresearch (accessed April 3, 2013).
Chapter Review
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.
 \(s = \sqrt{\dfrac{\sum(x\bar{x})^{2}}{n1}}\) or \(s = \sqrt{\dfrac{\sum f (x\bar{x})^{2}}{n1}}\) 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, \(\mu\), and the formula \(\sigma = \sqrt{\dfrac{\sum(x\mu)^{2}}{N}}\) or \(\sigma = \sqrt{\dfrac{\sum f (x\mu)^{2}}{N}}\).
∑f(x−μ)2N−−−−−−−−−√ .
Formula Review
\[s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n}  \bar{x}^2}\]
where \(s_{x} \text{sample standard deviation}\) and \(\bar{x} = \text{sample mean}\)
Use the following information to answer the next two exercises: The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles.
29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150
Exercise 2.8.4
Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth.
Answer
\(s\) = 34.5
Exercise 2.8.5
Find the value that is one standard deviation below the mean.
Exercise 2.8.6
Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team?
Baseball Player  Batting Average  Team Batting Average  Team Standard Deviation 

Fredo  0.158  0.166  0.012 
Karl  0.177  0.189  0.015 
Answer
For Fredo:
\(z\) = \(\dfrac{0.1580.166}{0.012}\) = –0.67
For Karl:
\(z\) = \(\dfrac{0.1770.189}{0.015}\) = –0.8
Fredo’s zscore of –0.67 is higher than Karl’s zscore of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.
Exercise 2.8.7
Use Table to find the value that is three standard deviations:
 above the mean
 below the mean
Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84.
Exercise 2.8.5
Find the standard deviation for the following frequency tables using the formula. Check the calculations with the TI 83/84.

Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 
Daily Low Temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 
Points per Game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2
Answer
 \(s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n}  \bar{x}^{2}} = \sqrt{\dfrac{193157.45}{30}  79.5^{2}} = 10.88\)
 \(s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n}  \bar{x}^{2}} = \sqrt{\dfrac{380945.3}{101}  60.94^{2}} = 7.62\)
 \(s_{x} = \sqrt{\dfrac{\sum fm^{2}}{n}  \bar{x}^{2}} = \sqrt{\dfrac{440051.5}{86}  70.66^{2}} = 11.14\)
Bringing It Together
Exercise 2.8.7
Twentyfive randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
# of movies  Frequency 

0  5 
1  9 
2  6 
3  4 
4  1 
 Find the sample mean \(\bar{x}\).
 Find the approximate sample standard deviation, \(s\).
Answer
 1.48
 1.12
Exercise 2.8.8
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 

1  2 
2  5 
3  8 
4  12 
5  12 
6  0 
7  1 
 Find the sample mean \(\bar{x}\)
 Find the sample standard deviation, s
 Construct a histogram of the data.
 Complete the columns of the chart.
 Find the first quartile.
 Find the median.
 Find the third quartile.
 Construct a box plot of the data.
 What percent of the students owned at least five pairs?
 Find the 40^{th} percentile.
 Find the 90^{th} percentile.
 Construct a line graph of the data
 Construct a stemplot of the data
Exercise 2.8.9
Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year.
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
 Organize the data from smallest to largest value.
 Find the median.
 Find the first quartile.
 Find the third quartile.
 Construct a box plot of the data.
 The middle 50% of the weights are from _______ to _______.
 If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why?
 If our population included every team member who ever played for the San Francisco 49ers, would the above data be a sample of weights or the population of weights? Why?
 Assume the population was the San Francisco 49ers. Find:
 the population mean, \(\mu\).
 the population standard deviation, \(\sigma\).
 the weight that is two standard deviations below the mean.
 When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he?
 That same year, the mean 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?
Answer
 174; 177; 178; 184; 185; 185; 185; 185; 188; 190; 200; 205; 205; 206; 210; 210; 210; 212; 212; 215; 215; 220; 223; 228; 230; 232; 241; 241; 242; 245; 247; 250; 250; 259; 260; 260; 265; 265; 270; 272; 273; 275; 276; 278; 280; 280; 285; 285; 286; 290; 290; 295; 302
 241
 205.5
 272.5
 205.5, 272.5
 sample
 population

 236.34
 37.50
 161.34
 0.84 std. dev. below the mean
 Young
Exercise 2.8.10
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 12 change scores are as follows:
3; 8; –1; 2; 0; 5; –3; 1; –1; 6; 5; –2
 What is the mean change score?
 What is the standard deviation for this population?
 What is the median change score?
 Find the change score that is 2.2 standard deviations below the mean.
Exercise 2.8.11
Refer to Figure determine which of the following are true and which are false. Explain your solution to each part in complete sentences.
<figure id="fsidm70725344">
Figure 2.8.6.
</figure>
 The medians for all three graphs are the same.
 We cannot determine if any of the means for the three graphs is different.
 The standard deviation for graph b is larger than the standard deviation for graph a.
 We cannot determine if any of the third quartiles for the three graphs is different.
Answer
 True
 True
 True
 False
Exercise 2.8.12
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.
 Organize the data in a chart.
 Find the median, the first quartile, and the third quartile.
 Find the 65^{th} percentile.
 Find the 10^{th} percentile.
 Construct a box plot of the data.
 The middle 50% of the conferences last from _______ days to _______ days.
 Calculate the sample mean of days of engineering conferences.
 Calculate the sample standard deviation of days of engineering conferences.
 Find the mode.
 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.
 Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences.
Exercise 2.8.13
A survey of enrollment at 35 community colleges across the United States yielded the following figures:
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
 Organize the data into a chart with five intervals of equal width. Label the two columns "Enrollment" and "Frequency."
 Construct a histogram of the data.
 If you were to build a new community college, which piece of information would be more valuable: the mode or the mean?
 Calculate the sample mean.
 Calculate the sample standard deviation.
 A school with an enrollment of 8000 would be how many standard deviations away from the mean?
Answer

Enrollment Frequency 10005000 10 500010000 16 1000015000 3 1500020000 3 2000025000 1 2500030000 2  Check student’s solution.
 mode
 8628.74
 6943.88
 –0.09
Use the following information to answer the next two exercises. \(X =\) 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 2.8.14
The 80^{th} percentile is _____
 5
 80
 3
 4
Exercise 2.8.15
The number that is 1.5 standard deviations BELOW the mean is approximately _____
 0.7
 4.8
 –2.8
 Cannot be determined
Answer
a
Exercise 2.8.16
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.
# 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 IQRto 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 two 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?
Glossary
 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.
Contributors
 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 \(x\) – \(\bar{x}\) where \(x\) is a value of the data and \(\bar{x}\) 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.
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