Measures of the Spread of the Data, Descriptive Statistics | Statistics

Learning objectives

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 all the data are 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 whereas 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 both Rosa and Binh 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. A z-score is a standardized score that lets us compare data sets. It tells us how many standard deviations a data value is from the mean and is calculated as the ratio of the difference in a particular score and the population mean to the population standard deviation.

We can use the given information to create the table below.

Supermarket	Population Standard Deviation, σ	Individual Score, x	Population Mean, μ
Supermarket A	2 minutes	7, 1	5
Supermarket B	4 minutes		5

Since Rosa and Binh only shop at Supermarket A, we can ignore the row for Supermarket B.

We need the values from the first row to determine the number of standard deviations above or below the mean each individual wait time is; we can do so by calculating two different z-scores.

Rosa waited for seven minutes, so the z-score representing this deviation from the population mean may be calculated as

z = \frac{x - μ}{σ} = \frac{7 - 5}{2} = 1 .

The z-score of one tells us that Rosa’s wait time is one standard deviation above the mean wait time of five minutes.

Binh waited for one minute, so the z-score representing this deviation from the population mean may be calculated as

z = \frac{x - μ}{σ} = \frac{1 - 5}{2} = - 2 .

The z-score of −2 tells us that Binh’s wait time is two standard deviations below the mean wait time 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 they are 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 farther 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.

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows a number line in intervals of 1 from 0 to 7.

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 as follows:

Sample: $x = \bar{x} + (# o f S T D E V) (s)$
Population: $x = μ + (# o f S T D E V) (σ) .$

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

The symbol

x

is the sample mean, and the Greek symbol

μ

is the population mean.

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 – μ. 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 lowercase letter s represents the sample standard deviation and the Greek letter σ (lowercase sigma) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ.

To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the x – $\bar{x}$ values for a sample or the x – μ values for a population). The symbol σ² represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s² 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

$s = \sqrt{\frac{Σ {(x - \bar{x})}^{2}}{n - 1}}$ or $s = \sqrt{\frac{Σ f {(x - \bar{x})}^{2}}{n - 1}}$
For the sample standard deviation, the denominator is n−; that is, the sample size minus 1.

Formulas for the Population Standard Deviation

$σ = \sqrt{\frac{Σ {(x - μ)}^{2}}{N}}$ or $σ = \sqrt{\frac{Σ f {(x - μ)}^{2}}{N}}$
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 one. If a value appears three times in the data set or population, f is three.

Types of Variability in Samples

When researchers study a population, they often use a sample, either for convenience or because it is not possible to access the entire population. Variability is the term used to describe the differences that may occur in these outcomes. Common types of variability include the following:

Observational or measurement variability
Natural variability
Induced variability
Sample variability

Here are some examples to describe each type of variability:

Example 1: Measurement variability

Measurement variability occurs when there are differences in the instruments used to measure or in the people using those instruments. If we are gathering data on how long it takes for a ball to drop from a height by having students measure the time of the drop with a stopwatch, we may experience measurement variability if the two stopwatches used were made by different manufacturers. For example, one stopwatch measures to the nearest second, whereas the other one measures to the nearest tenth of a second. We also may experience measurement variability because two different people are gathering the data. Their reaction times in pressing the button on the stopwatch may differ; thus, the outcomes will vary accordingly. The differences in outcomes may be affected by measurement variability.

Example 2: Natural variability

Natural variability arises from the differences that naturally occur because members of a population differ from each other. For example, if we have two identical corn plants and we expose both plants to the same amount of water and sunlight, they may still grow at different rates simply because they are two different corn plants. The difference in outcomes may be explained by natural variability.

Example 3: Induced variability

Induced variability is the counterpart to natural variability. This occurs because we have artificially induced an element of variation that, by definition, was not present naturally. For example, we assign people to two different groups to study memory, and we induce a variable in one group by limiting the amount of sleep they get. The difference in outcomes may be affected by induced variability.

Example 4: Sample variability

Sample variability occurs when multiple random samples are taken from the same population. For example, if I conduct four surveys of 50 people randomly selected from a given population, the differences in outcomes may be affected by sample variability.

Sampling Variability of a Statistic

The statistic of a sampling distribution was discussed in Descriptive Statistics: Measures of 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. The standard error is the standard deviation of the sampling distribution. In other words, it is the average standard deviation that results from repeated sampling. You will cover the standard error of the mean in the chapter The Central Limit Theorem (not now). The notation for the standard error of the mean is $\frac{σ}{\sqrt{n}}$ , where σ is the standard deviation of the population and n is the size of the sample.

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} = \frac{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	Frequency	Deviations	Deviations²	(Frequency)(Deviations²)
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 = –.525	(–.525)² = .275625	4 × .275625 = 1.1025
10.5	4	10.5 – 10.525 = –.025	(–.025)² = .000625	4 × .000625 = .0025
11	6	11 – 10.525 = .475	(.475)² = .225625	6 × .225625 = 1.35375
11.5	3	11.5 – 10.525 = .975	(.975)² = .950625	3 × .950625 = 2.851875
				The total is 9.7375.

The last column simply multiplies each squared deviation by the frequency for the corresponding data value.

The sample variance, s², is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1):

s^{2} = \frac{9.7375}{20 - 1} = .5125

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

$s = \sqrt{.5125} = .715891,$ which is rounded to two decimal places, s = .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 on a calculator or computer. Note that these formulas are derived by algebraically manipulating the z-score formulas, given either parameters or statistics.
For a sample: x = $\bar{x}$ + (#ofSTDEVs)(s)
For a population: x = μ + (#ofSTDEVs)(σ)
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.

- Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2^nd 1 for L1, the comma (,), and 2^nd 2 for L2.
- Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.
- Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.
- Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2^nd 1), L2 (2^nd 2). Do not forget the comma. Press ENTER.
- $\bar{x}$ = 10.525.
- Use Sx because this is sample data (not a population): Sx=.715891.
( $\bar{x}$ + 1s) = 10.53 + (1)(.72) = 11.25
( $\bar{x}$ – 2s) = 10.53 – (2)(.72) = 9.09
- ( $\bar{x}$ – 1.5s) = 10.53 – (1.5)(.72) = 9.45
- ( $\bar{x}$ + 1.5s) = 10.53 + (1.5)(.72) = 11.61

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 .97 and .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. We can sum the products of the frequencies and deviations to show that the sum of the deviations is always zero. $1 (- 1.525) + 2 (- 1.025) + 4 (- .525) + 4 (- .025) + 6 (.475) + 3 (.975) = 0$ For element-655, 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.

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

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

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 TI-83+ or TI-84 calculator:
1. The sample mean
2. The sample standard deviation
3. The median
4. The first quartile
5. The third quartile
6. IQR
Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.

See id6947804.
Entering the data values into a list in your graphing calculator and then selecting Stat, Calc, and 1-Var Stats will produce the one-variable statistics you need.
The x-axis goes from 32.5 to 100.5; the y-axis goes from –2.4 to 15 for the histogram. The number of intervals is 5, so the width of an interval is (100.5 – 32.5) divided by 5, 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.

View original figure (external; image not reproduced in this text edition)

Original figure description: A hybrid image displaying both a histogram and box plot described in detail in the answer solution above.

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 percent is greater (73 – 33 = 40) than the spread in the upper 50 percent (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 percent of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25 percent of the exam scores are Ds and Fs.

Data	Frequency	Relative Frequency	Cumulative Relative Frequency
33	1	.032	.032
42	1	.032	.064
49	2	.065	.129
53	1	.032	.161
55	2	.065	.226
61	1	.032	.258
63	1	.032	.290
67	1	.032	.322
68	2	.065	.387
69	2	.065	.452
72	1	.032	.484
73	1	.032	.516
74	1	.032	.548
78	1	.032	.580
80	1	.032	.612
83	1	.032	.644
88	3	.097	.741
90	1	.032	.773
92	1	.032	.805
94	4	.129	.934
96	1	.032	.966
100	1	.032	.998 (Why isn't this value 1?)

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 $M e a n o f F r e q u e n c y T a b l e = \frac{\sum f m}{\sum f},$
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 are when compared to the mean.

Find the standard deviation for the data in fs-idm67330768.

Class	Frequency, f	Midpoint, m	m²	$\bar{x}$ ²	fm²	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 = .58. While the formula for calculating the standard deviation is not complicated, $s_{x} = \sqrt{\frac{f {(m - \bar{x})}^{2}}{n - 1}},$ 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.

Try It

Find the standard deviation for the data from the previous example:

Class	Frequency, f
0–2	1
3–5	6
6–8	10
9–11	7
12–14	0
15–17	2

First, press the STAT key and select 1:Edit.

View original figure (external; image not reproduced in this text edition)

Original figure description: Figure three shows the words edit, calc, tests on the top of a calculator. 1: edit 2: sort A 3: Sort D, 4:Clr List, and 5:Set up Editor

Input the midpoint values into L1 and the frequencies into L2.

View original figure (external; image not reproduced in this text edition)

Original figure description: Figure 4 shows three columns: L1. L2.and L3. L1 shows the numbers 1, 4, 7, 10, 13 and 16. L2 shows the numbers 1, 6, 10, 7, 0, 2. L3 shows nothing. There is an unsolved equation L2 multiplied by 7 = blank.

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

View original figure (external; image not reproduced in this text edition)

Original figure description: Figure 5 shows edit, calc, and tests. Calc is highlighted. It then shows the following options. 1: 1-var stats, 2: 2-var stats, 3: med-med, 4: LinReg (ax+b), 5: QuadReg, 6: CubicReg, 7 QuartReg.

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

View original figure (external; image not reproduced in this text edition)

Original figure description: This calculator screen shows the following calculator steps. In line 1, X bar is equal to 7.576923077. In line 2, uppercase sigma x is equal to 197. In line 3, uppercase sigma x squared is equal to 1799. In line 4, capital S of x is equal to 3.500549407. In line 5, lowercase sigma of x is equal to 3.432571103. In line 6, n preceded by a downward-facing arrow is equal to 26.

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

Comparing Values from Different Data Sets

As explained before, a z-score allows us to compare statistics 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.
In symbols, the formulas for calculating z-scores become the following.

Sample $z = \frac{x - \bar{x}}{s}$

Population $z = \frac{x - μ}{σ}$

As shown in the table, when only a sample mean and sample standard deviation are given, the top formula is used. When the population mean and population standard deviation are given, the bottom formula is used.

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	.7
Ali	77	80	10

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.

$z = # of STDEVs = \frac{value - mean}{standard deviation} = \frac{x + μ}{σ}$

For John, $z = # o f S T D E V s = \frac{2.85 - 3.0}{.7} = - 0.21$

For Ali, $z = # o f S T D E V s = \frac{77 - 80}{10} = - 0.3$

John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean, while Ali's GPA is .3 standard deviations below his school's mean.

John's z-score of –.21 is higher than Ali's z-score of –.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. The z-score representing John's score does not fall as far below the mean as the z-score representing Ali's score.

Try It

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	.8
Beth	27.3	30.1	1.4

For Angie: z = $\frac{26 .2 – 27 .2}{0 .8}$ = –1.25

For Beth: z = $\frac{27 .3 - 30 .1}{1. 4}$ = –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, the following are true:

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

A bell-shaped distribution is one that is normal and symmetric, meaning the curve can be folded along a line of symmetry drawn through the median, and the left and right sides of the curve would fold on each other symmetrically.. With a bell-shaped distribution, the mean, median, and mode are all located at the same place.

For data having a distribution that is bell-shaped and symmetric, the following are true:

Approximately 68 percent of the data is within one standard deviation of the mean.
Approximately 95 percent of the data is within two standard deviations of the mean.
More than 99 percent 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 applies only when the shape of the distribution of the data is bell-shaped and symmetric; we will learn more about this when studying the Normal or Gaussian probability distribution in later chapters.

References

King, B. (2005, Dec.). Graphically Speaking. Retrieved from http://www.ltcc.edu/web/about/institutional-research

Microsoft Bookshelf. (n.d.).

Chapter Review

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

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

Formula Review

$s_{x} = \sqrt{\frac{\sum f m^{2}}{n} - {\bar{x}}^{2}}$ where $\begin{matrix} s_{x} = sample standard deviation \\ \bar{x} = sample mean \end{matrix}$ $z = \frac{(x - \bar{x})}{s}$ and $z = \frac{(x - μ)}{θ} .$

For each of the examples given below, tell whether the differences in outcomes may be explained by measurement variability, natural variability, induced variability, or sampling variability.

Scientists randomly select five groups of 10 women from a population of 1,000 women to record their body fat percentage. The scientists compute the mean body fat percentage from each group. The differences in outcomes may be attributed to which type of variability?

sampling variability

A pharmaceutical company randomly assigns participants to one of two groups: one is a control group receiving a placebo, and another is a treatment group receiving a new drug to lower blood pressure. The differences in outcomes may be attributed to which type of variability?

induced variability

Jaiqua and Harold are trying to determine how ramp steepness affects the speed of a ball rolling down the ramp. They measure the time it takes for the ball to roll down ramps of differing slopes. When Jaiqua rolls the ball and Harold works the stopwatch, they get different results than when Harold rolls the ball and Jaiqua works the stopwatch. The differences in outcomes may be attributed to which type of variability?

measurement variability

Twenty people begin the same workout program on the same day and continue for three months. During that time, all participants worked out for the same amount of time and did the same number of exercises and repetitions. Each person was weighed at both the beginning and the end of the program. The differences in outcomes regarding the amount of weight lost may be attributed to which type of variability?

natural variability

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

Use a graphing calculator or computer to find the standard deviation and round to the nearest tenth.

s = 34.5

Find the value that is one standard deviation below the mean.

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	.158	.166	.012
Karl	.177	.189	.015

For Fredo: z = $\frac{.158 – .166}{.012}$ = –0.67.

For Karl: z = $\frac{.177 – .189}{.015}$ = –.8.

Fredo’s z score of –.67 is higher than Karl’s z score of –.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.

Use fs-idm133764640 to find the value that is three standard deviations

aabove the mean, and
bbelow the mean

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

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

$s_{x} = \sqrt{\frac{\sum f m^{2}}{n} - {\bar{x}}^{2}} = \sqrt{\frac{193,157.45}{30} - {79.5}^{2}} = 10.88$
$s_{x} = \sqrt{\frac{\sum f m^{2}}{n} - {\bar{x}}^{2}} = \sqrt{\frac{380,945.3}{101} - {60.94}^{2}} = 7.62$
$s_{x} = \sqrt{\frac{\sum f m^{2}}{n} - {\bar{x}}^{2}} = \sqrt{\frac{440,051.5}{86} - {70.66}^{2}} = 11.14$

Homework

Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005.

μ = 1,000 FTES
median = 1,014 FTES
σ = 474 FTES
first quartile = 528.5 FTES
third quartile = 1,447.5 FTES
n = 29 years

A sample of 11 years is taken. About how many are expected to have an FTES of 1,014 or above? Explain how you determined your answer.

The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the sixth number in order. Six years will have totals at or below the median.

Seventy-five percent of all years have an FTES

at or below ______.
at or above ______.

The population standard deviation = ______.

474 FTES

What percentage of the FTES were from 528.5 to 1,447.5? How do you know?

What is the IQR? What does the IQR represent?

919

How many standard deviations away from the mean is the median?

Additional Information: The population FTES for 2005–2006 through 2010–2011 was given in an updated report. The data are reported here.

Year	2005–2006	2006–2007	2007–2008	2008–2009	2009–2010	2010–2011
Total FTES	1,585	1,690	1,735	1,935	2,021	1,890

Calculate the mean, median, standard deviation, the first quartile, the third quartile, and the IQR. Round to one decimal place.

mean = 1,809.3
median = 1,812.5
standard deviation = 151.2
first quartile = 1,690
third quartile = 1,935
IQR = 245

Construct a box plot for the FTES for 2005–2006 through 2010–2011 and a box plot for the FTES for 1976–1977 through 2004–2005.

Compare the IQR for the FTES for 1976–1977 through 2004–2005 with the IQR for the FTES for 2005-2006 through 2010–2011. Why do you suppose the IQRs are so different?

Hint: think about the number of years covered by each time period and what happened to higher education during those periods.

Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer.

Student	GPA	School Average GPA	School Standard Deviation
Thuy	2.7	3.2	.8
Vichet	87	75	20
Kamala	8.6	8	.4

A music school has budgeted to purchase three musical instruments. The school plans to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean 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.

For pianos, the cost of the piano is .4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.

An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran one mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.

Why is Kenji considered a better runner than Nedda even though Nedda ran faster than he?
Who is the fastest runner with respect to his or her class? Explain why.

Scientists are studying a particular disease. They found that countries that have the highest rates of people who have ever been diagnosed with this disease range from 11.4 percent to 74.6 percent.

Percentage of Population with Disease	Number of Countries
11.4–20.45	29
20.45–29.45	13
29.45–38.45	4
38.45–47.45	0
47.45–56.45	2
56.45–65.45	1
65.45–74.45	0
74.45–83.45	1

What is the best estimate of the average percentage of people with the disease for these countries? What is the standard deviation for the listed rates? The United States has an average disease rate of 33.9 percent. Is this rate above average or below? How unusual is the U.S. obesity rate compared to the average rate? Explain.

$\bar{x} = 23.32$
Using the TI 83/84, we obtain a standard deviation of: $s_{x} = 12.95.$
The obesity rate of the United States is 10.58 percent higher than the average obesity rate.
Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the disease percentage that is one standard deviation from the mean. The U.S. disease rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, although 34 percent have the disease, does not have an unusually high percentage of people with the disease.

fs-idm116546656 gives the percentage of children under age five diagnosed with a specific medical condition.

Percentage of Children with the Condition	Number of Countries
16–21.45	23
21.45–26.9	4
26.9–32.35	9
32.35–37.8	7
37.8–43.25	6
43.25–48.7	1

What is the best estimate for the mean percentage of children with the condition? What is the standard deviation? Which interval(s) could be considered unusual? Explain.

Bringing It Together

Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:

Number of Movies	Frequency
0	5
1	9
2	6
3	4
4	1

Find the sample mean $x$ .
Find the approximate sample standard deviation, s.

1.48
1.12

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, $x$
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 percentage 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.

Following are the published weights (in pounds) of all of the football 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 percent 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 a California-based football team, would the above data be a sample of weights or the population of weights? Why?
Assume the population was a California-based football team. Find
1. the population mean, μ,
2. the population standard deviation, σ, and
3. the weight that is two standard deviations below the mean.
4. In addition, when the team's most famous quarterback, played football, he weighed 205 pounds. Also find how many standard deviations above or below the mean was he?
That same year, the mean weight for a player from a Texas football team was 240.08 pounds with a standard deviation of 44.38 pounds. One player weighed in at 209 pounds. With respect to his team, who was lighter, the California quarterback or the Texas player? How did you determine your 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
View original figure (external; image not reproduced in this text edition)

Original figure description: A box plot with a whisker between 174 and 205.5, a solid line at 205.5, a dashed line at 241, a solid line at 272.5, and a whisker between 272.5 and 302.
205.5, 272.5
sample
population
1. 236.34
2. 37.50
3. 161.34
4. .84 standard deviations below the mean
young

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.

Refer to fs-idm70725344 to determine which of the following are true and which are false. Explain your solution to each part in complete sentences.

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows three graphs. The first is a histogram with a mode of 3 and fairly symmetrical distribution between 1 (minimum value) and 5 (maximum value). The second graph is a histogram with peaks at 1 (minimum value) and 5 (maximum value) with 3 having the lowest frequency. The third graph is a box plot. The first whisker extends from 0 to 1. The box begins at the firs quartile, 1, and ends at the third quartile,6. A vertical, dashed line marks the median at 3. The second whisker extends from 6 on.

The medians for all three graphs are the same.
We cannot determine if any of the means for the three graphs are 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 are different.

true
true
true
false

In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six 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 percent 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.

A survey of enrollment at 35 community colleges across the United States yielded the following figures:

6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622

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 8,000 would be how many standard deviations away from the mean?

Enrollment Frequency

1,000–5,000 10

5,000–10,000 16

10,000–15,000 3

15,000–20,000 3

20,000–25,000 1

25,000–30,000 2
Check student’s solution.
mode
8,628.74
6,943.88
–0.09

Enrollment	Frequency
1,000–5,000	10
5,000–10,000	16
10,000–15,000	3
15,000–20,000	3
20,000–25,000	1
25,000–30,000	2

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

The 80^th percentile is ________.

The number that is 1.5 standard deviations below the mean is approximately ________.

0.7
4.8
–2.8
cannot be determined

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 table23.

Number of Books	Frequency	Relative Frequency
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 farther than two standard deviations away from the mean? In some situations, statisticians may use this criterion to identify data values that are unusual, compared to the other data values. Note that this criterion is most appropriate to use for data that is mound shaped 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
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 1

Attribution. Statistics, OpenStax, Rice University, and the contributing authors credited in the book. Licensed under Creative Commons Attribution 4.0 International. Source revision 7dea80ae28 of openstax/osbooks-statistics (2026-04-23). CC BY 4.0. Original source ↗. Text edition of a pinned CC BY 4.0 source revision of the OpenStax book, converted from CNXML by the EvidenceHelix importer. Original prose, learning objectives, tables, exercises, glossary and captions are preserved; images are not copied (figures link to the original source and keep their credits). No new content was added. Later revisions of this book can carry different licence terms and are not used.

Sample		$z = \frac{x - \bar{x}}{s}$
Population		$z = \frac{x - μ}{σ}$