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Section 3.2
Measures of Dispersion
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Objectives
o Compute the range, variance, and standard
deviation.
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Range
Range
Range = Maximum Data Value − Minimum Data Value
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Example 3.10: Calculating the Range
The following data were collected from samples of call
lengths (in minutes) observed for two different mobile
phone users. Calculate the range of each data set.
a. 2, 25, 31, 44, 29, 14, 22, 11, 40
b. 2, 2, 44, 2, 2, 2, 2, 2
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Example 3.10: Calculating the Range (cont.)
Solution
a. The maximum value is 44 minutes and the minimum
value is 2 minutes, so the range is as follows.
Range  Maximum Data Value  Minimum Data Value
 44  2
 42 minutes
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Example 3.10: Calculating the Range (cont.)
b. The maximum value for the second data set is also
44 minutes and the minimum value is also 2
minutes, so the range is calculated in the same way.
Range  Maximum Data Value  Minimum Data Value
 44  2
 42 minutes
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Standard Deviation
Standard Deviation
The standard deviation is a measure of how much we
might expect a typical member of the data set to differ
from the mean.
The population standard deviation is given by
s
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  x  m
2
i
N
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Standard Deviation
Standard Deviation (cont.)
where xi is the ith value in the population,
μ is the population mean, and
N is the number of values in the population.
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Standard Deviation
Standard Deviation (cont.)
The sample standard deviation is given by
s
 x  x 
2
i
n 1
where xi is the ith data value,
x ̄ is the sample mean, and
n is the number of data values in the sample.
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Example 3.11: Calculating Standard Deviation
Calculate the sample standard deviation of the
following data collected regarding the numbers of
hours students studied for a physics exam.
5, 8, 7, 6, 9
Solution
Let’s calculate the sample standard deviation by hand
using the following formula.
s
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 x  x 
2
i
n 1
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Example 3.11: Calculating Standard Deviation
(cont.)
To start, we need the mean and sample size of the data
set. We calculate that x ̄ = 7 and n = 5. Next, we need to
subtract the mean from each number in the sample,
and then square each of these differences. Let’s use a
chart to keep everything organized.
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Example 3.11: Calculating Standard Deviation
(cont.)
Deviations and Squared Deviations of the Data
xi
(xi – x̄)
(xi – x̄)2
5
8
7
6
9
5  7 = 2
87=1
77=0
6  7 = 1
97=2
4
1
0
1
4
Next, find the sum of the squared deviations by adding
up the values in the last column.
 x  x 
i
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2
 4  1  0  1  4  10
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Example 3.11: Calculating Standard Deviation
(cont.)
Finally, substitute the appropriate values into the
sample standard deviation formula as follows.
s
 x  x 
2
i
n 1
10

51
 2.5
 1.581139  1.6
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Example 3.11: Calculating Standard Deviation
(cont.)
Alternate Calculator Method
To find the sample standard deviation on a TI-83/84
Plus calculator, follow the steps below.
• Press
.
• Choose option 1:Edit and press
.
• Enter the data in L1.
• Press
again.
• Choose CALC.
• Choose option 1:1-Var Stats.
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Example 3.11: Calculating Standard Deviation
(cont.)
• Press
twice. (Note: If your data are not in L1,
before pressing
the second time, enter the list
where your data are located, such as L3 or L5.)
The fourth value in the output, seen in the screenshot
below, gives the sample standard deviation, which is
Sx = 1.58113883 ≈ 1.6.
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Standard Deviation
Where Does the Formula Come From?
You can think of the standard deviation as a sort-of
average distance that values in a set lie from the mean.
While not the actual average, the standard deviation is
usually very close to the average and, conceptually, that
is a good way to think about standard deviation as we
discuss how the formula is derived. To derive the
formula for standard deviation, we will use a method
similar to finding average distance, or deviation, from
the mean.
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Standard Deviation
Where Does the Formula Come From? (cont.)
First, we must know the actual deviation from the
mean for every data value in the set. The deviation is
simply the difference between each value and the
mean.
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Standard Deviation
Where Does the Formula Come From? (cont.)
Next, we need to find the sum of all the deviations.
Here we run into a problem because the sum is zero. In
fact, the sum of the deviations from the mean for any
data set is always equal to zero because the positive
deviations cancel out the negative deviations. To help
eliminate the problem with the negative values
canceling the positive values, we can make all the
deviations positive by squaring each one.
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Standard Deviation
Where Does the Formula Come From? (cont.)
Now, find the average by summing up all the squared
deviations and dividing by the population size just as
you would find a traditional mean. To find the standard
deviation, we must take one additional step and
“undo” the previous squaring by taking the square
root. The completed population standard deviation
formula is as follows.
s
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  x  m
2
i
N
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Example 3.12: Using a TI-83/84 Plus Calculator
to Find Standard Deviation
Use a TI-83/84 Plus calculator to find the standard
deviation of the data shown below, given the following
conditions.
11, 18, 25, 51, 44, 29, 30, 17, 29, 47, 52, 60
a. Assume that the values represent the ages (in years)
of patients randomly sampled from an urgent care
clinic.
b. Assume that the values represent the ages (in years)
of all patients seen by Dr. Dabbs one afternoon.
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Example 3.12: Using a TI-83/84 Plus Calculator
to Find Standard Deviation (cont.)
Solution
• Press
.
• Choose 1:Edit.
• Enter the data in L1.
• Press
again.
• Choose CALC.
• Choose option 1:1-Var Stats.
• Press
twice.
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Example 3.12: Using a TI-83/84 Plus Calculator
to Find Standard Deviation (cont.)
A list of numerical summaries will be generated for the
data. The beginning of the list is shown below. Use
these values to find the answers for this example.
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Example 3.12: Using a TI-83/84 Plus Calculator
to Find Standard Deviation (cont.)
a. We are told that the values in this case represent a
random sample of patients. We will then need to
use the sample standard deviation, denoted on the
calculator by Sx. From the list, we see that s ≈ 15.9
years.
b. In this scenario, the values represent all patients
seen in one afternoon. The population standard
deviation is most appropriate here, and it is denoted
on the calculator by x. From the list, we see that
 ≈ 15.2 years.
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Example 3.13: Interpreting Standard Deviations
Mark is looking into investing a portion of his recent bonus
into the stock market. While researching different
companies, he discovers the following standard deviations
of one year of daily stock closing prices.
Profacto Corporation: Standard deviation of stock
prices = $1.02
Yardsmoth Company: Standard deviation of stock
prices = $9.67
What do these two standard deviations tell you about the
stock prices of these companies?
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Example 3.13: Interpreting Standard Deviations
(cont.)
Solution
A smaller standard deviation indicates that the data
values are closer together, while a larger standard
deviation indicates that the data values are more
spread out. In this example, the standard deviation of
stock prices for the Profacto Corporation is
considerably smaller than that of the Yardsmoth
Company. Hence, there is less variability in the daily
closing prices of the Profacto stock than in the
Yardsmoth stock prices.
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Example 3.13: Interpreting Standard Deviations
(cont.)
If Mark wants a stable long-term investment, then
Profacto appears to be the better choice. If, however,
Mark is looking to make a quick profit and is willing to
take the risk, then the Yardsmoth stock would seem to
better suit his purposes. Note that looking at the
standard deviations is just one component of
evaluating market prices.
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Standard Deviation
Coefficient of Variation
The coefficient of variation for a set of data is the ratio
of the standard deviation to the mean as a percentage.
For a population, it is given by
s
CV   100%
m
Where  is the population standard deviation and
m is the population mean.
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Standard Deviation
Coefficient of Variation (cont.)
For a sample, it is given by
s
CV   100%
x
where s is the sample standard deviation and
x ̄ is the sample mean.
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Example 3.14: Calculating and Interpreting
Coefficient of Variation
Suppose that Graph A from Figure 3.1 represents
average amounts of annual rainfall for a sample of
farms in the United States and Graph B represents
prices per 20 acres of farmland for the same farms. The
mean and standard deviation of Data Set A are 26.08
inches and 7.55 inches, respectively, whereas the mean
and standard deviation of Data Set B are $117,000 and
$42,931, respectively.
Which of the two graphs has the larger standard
deviation relative to its mean?
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Example 3.14: Calculating and Interpreting
Coefficient of Variation (cont.)
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Example 3.14: Calculating and Interpreting
Coefficient of Variation (cont.)
Solution
s
Data Set A: CV   100%
x
7.55

 100%  28.9%
26.08
s
Data Set B: CV   100%
x
42,931

 100%  36.7%
117,000
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Example 3.14: Calculating and Interpreting
Coefficient of Variation (cont.)
Now, it is clear to see that Data Set B has the larger
standard deviation relative to its own mean. Thus, our
comparison of CV shows that there is more variability
in Data Set B than in Data Set A.
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Variance
Variance
The variance is the square of the standard deviation.
The population variance is given by
s
2
x  m



2
i
N
where xi is the ith value in the population,
μ is the population mean, and
N is the number of values in the population.
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Variance
Variance (cont.)
The sample variance is given by
s
2
x  x



2
i
n 1
where xi is the ith data value,
x ̄ is the sample mean, and
n is the number of data values in the sample.
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Example 3.15: Calculating Variance
Calculate the variance of the data shown below, given
the following conditions.
3, 2, 5, 6, 4
a. Assume that the data represent the actual weight
changes (in pounds) for a sample of fitness club
members during the month of April.
b. Assume that the data represent the actual weight
changes (in pounds) of every member of a book club
during the month of April. Use a TI-83/84 Plus
calculator to perform the calculation.
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Example 3.15: Calculating Variance (cont.)
Solution
a. As these data represent a sample, we need to
calculate the sample variance. The formula that we
need is given below.
2
xi  x 


2
s 
n 1
To start, we need the mean and sample size of the
data set. We calculate that x ̄ = 4 and n = 5. Next, we
need to subtract the mean from each number in the
sample, and then square each of these differences.
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Example 3.15: Calculating Variance (cont.)
Let’s use a chart to keep everything organized.
Deviations and Squared Deviations of the Data
xi
(xi – x̄)
(xi – x̄)2
3
2
5
6
4
3  4 = 1
2  4 = 2
54=1
64=2
44=0
1
4
1
4
0
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Example 3.15: Calculating Variance (cont.)
Next, find the sum of the squared deviations by adding
up the values in the last column.
 x  x 
i
2
 1 4 1 4  0
 10
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Example 3.15: Calculating Variance (cont.)
Finally, substitute the appropriate values into the
formula for sample variance as follows.
s
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2
x  x



2
i
n 1
10

51
 2.5
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Example 3.15: Calculating Variance (cont.)
b. To calculate a variance on a TI-83/84 Plus calculator,
you must actually calculate the standard deviation
and then square that value to get the variance. The
steps for calculating the standard deviation of a data
set were presented in Example 3.12, and are
repeated below.
• Press
.
• Choose 1:Edit.
• Enter the data in L1.
• Press
again.
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Example 3.15: Calculating Variance (cont.)
• Choose CALC.
• Choose option 1:1-Var Stats.
• Press
twice.
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Example 3.15: Calculating Variance (cont.)
The calculator actually presents the values for both the
population and sample standard deviations. Since the
data set in this scenario represents a population (all
members of the book club), we are looking for the
population variance. Thus, we need to square the
population standard deviation, given by
 = 1.414213562.
  1.414214 
2
2
 2.0
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Empirical Rule
Empirical Rule for Bell-Shaped Distributions
• Approximately 68% of the data values lie within one
standard deviation of the mean.
• Approximately 95% of the data values lie within two
standard deviations of the mean.
• Approximately 99.7% of the data values lie within
three standard deviations of the mean.
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions
The distribution of weights of newborn babies is bellshaped with a mean of 3000 grams and standard
deviation of 500 grams.
a. What percentage of newborn babies weigh between
2000 and 4000 grams?
b. What percentage of newborn babies weigh less than
3500 grams?
c. Calculate the range of birth weights that would
contain the middle 68% of newborn babies’ weights.
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
Solution
a. Since we know that the distribution of the data is
bell-shaped, we can apply the Empirical Rule. We
need to know how many standard deviations 2000
grams and 4000 grams are from the mean. By
subtracting, we can find how far each of these
figures is from the mean. Then, dividing by the
standard deviation, we can convert these
differences into numbers of standard deviations.
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
Here are the calculations.
2000  3000  1000 and 4000  3000  1000
1000
1000
 2
2
500
500
Thus, these weights lie two standard deviations above
and below the mean. According to the Empirical Rule,
approximately 95% of values lie within two standard
deviations of the mean. Therefore, we can say that
approximately 95% of newborn babies weigh between
2000 and 4000 grams.
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
b. To begin, let’s find out how many standard
deviations a weight of 3500 grams is away from the
mean by performing the same calculation as before.
3500  3000  500
500
1
500
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
Thus it is one standard deviation above the mean. The
Empirical Rule says that 68% of data values lie within
one standard deviation of the mean. Because of the
symmetry of the distribution, half of this 68% is above
the mean and half is below. Putting the upper 34%
together with the 50% of data that is below the mean,
we have that approximately
50%  34%  84%
of newborn babies weigh less than 3500 grams.
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
c. From the Empirical Rule, we know that 68% of the
data values lie within one standard deviation of the
mean for bell-shaped distributions. The standard
deviation of this distribution is 500; thus, by adding
500 to and subtracting 500 from the mean of the
distribution, we will get the range of birth weights
that contain the middle 68% of newborn babies’
weights.
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Example 3.16: Applying the Empirical Rule for
Bell-Shaped Distributions (cont.)
Upper end: 3000 + 500 = 3500
Lower end: 3000  500 = 2500
Thus, 68% of newborn babies weigh between 2500 and
3500 grams.
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Chebyshev’s Theorem
Chebyshev’s Theorem
The proportion of data that lie within K standard
1
deviations of the mean is at least 1  2 , for K > 1.
K
When K = 2 and K = 3, Chebyshev’s Theorem says the
following.
1 3
• K = 2: At least 1  2   75% of the data lie
2 4
within two standard deviations of the mean.
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Chebyshev’s Theorem
Chebyshev’s Theorem (cont.)
1 8
• K = 3: At least 1  2   88.9% of the data
3 9
lie within three standard deviations of the mean.
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Example 3.17: Applying Chebyshev’s Theorem
Suppose that in one small town, the average household
income is $34,200, with a standard deviation of $2200.
What percentage of households earn between $27,600
and $40,800?
Solution
Since we are not told in the problem whether the
distribution of the data is bell-shaped, we cannot apply
the Empirical Rule here. However, we can apply
Chebyshev’s Theorem to find a minimum estimate.
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Example 3.17: Applying Chebyshev’s Theorem
(cont.)
In order to do so, we need to know how many standard
deviations $27,600 and $40,800 are from the mean. By
subtracting, we can find how far each of these figures is
from the mean. Then, dividing by the standard
deviation, we can convert these differences into
numbers of standard deviations. Here are the
calculations.
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Example 3.17: Applying Chebyshev’s Theorem
(cont.)
$27,600  $34,200  $6600
$6600
 3
$2200
and
$40,800  $34,200  $6600
$6600
3
$2200
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Example 3.17: Applying Chebyshev’s Theorem
(cont.)
Thus these incomes lie three standard deviations above
and below the mean. Chebyshev’s Theorem can then
be applied for K = 3. Using the calculation previously
shown in the box with the theorem, we can say that at
least 88.9% of the household incomes lie within this
range.
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