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```Chapter
3
Numerically
Summarizing Data
Section
3.2
Measures of
Dispersion
Objectives
1. Determine the range of a variable from raw data
2. Determine the standard deviation of a variable from
raw data
3. Determine the variance of a variable from raw data
4. Use the Empirical Rule to describe data that are bell
shaped
5. Use Chebyshev’s Inequality to describe any data set
3-3
To order food at a McDonald’s restaurant, one must choose
from multiple lines, while at Wendy’s Restaurant, one
enters a single line. The following data represent the wait
time (in minutes) in line for a simple random sample of 30
customers at each restaurant during the lunch hour. For
(a) What was the mean wait time?
(b) Draw a histogram of each restaurant’s wait time.
(c ) Which restaurant’s wait time appears more dispersed?
Which line would you prefer to wait in? Why?
3-4
Wait Time at Wendy’s
1.50
2.53
1.88
3.99
0.90
0.79
1.20
2.94
1.90
1.23
1.01
1.46
1.40
1.00
0.92
1.66
0.89
1.33
1.54
1.09
0.94
0.95
1.20
0.99
1.72
0.67
0.90
0.84
0.35
2.00
Wait Time at McDonald’s
3.50
0.00
1.97
0.00
3.08
3-5
0.00
0.26
0.71
0.28
2.75
0.38
0.14
2.22
0.44
0.36
0.43
0.60
4.54
1.38
3.10
1.82
2.33
0.80
0.92
2.19
3.04
2.54
0.50
1.17
0.23
(a) The mean wait time in each line is 1.39
minutes.
3-6
(b)
3-7
Objective 1
• Determine the Range of a Variable from Raw
Data
3-8
The range, R, of a variable is the difference
between the largest data value and the smallest data
values. That is,
Range = R = Largest Data Value – Smallest Data Value
3-9
EXAMPLE Finding the Range of a Set of Data
The following data represent the travel times (in
minutes) to work for all seven employees of a
start-up web development company.
23, 36, 23, 18, 5, 26, 43
Find the range.
Range = 43 – 5
= 38 minutes
3-10
Objective 2
• Determine the Standard Deviation of a
Variable from Raw Data
3-11
The population standard deviation of a
variable is the square root of the sum of squared
deviations about the population mean divided by
the number of observations in the population, N.
That is, it is the square root of the mean of the
squared deviations about the population mean.
The population standard deviation is
symbolically represented by σ (lowercase Greek
sigma).
3-12
x1     x2   
2


2
 L  x N   
2
N
 x
i
 
2
N
where x1, x2, . . . , xN are the N observations in
the population and μ is the population mean.
3-13
A formula that is equivalent to the one on the
previous slide, called the computational
formula, for determining the population
standard deviation is
x



2

3-14
x
2
i
i
N
N
EXAMPLE
Computing a Population Standard
Deviation
The following data represent the travel times (in
minutes) to work for all seven employees of a startup web development company.
23, 36, 23, 18, 5, 26, 43
Compute the population standard deviation of this
data.
3-15
xi
μ
xi – μ
(xi – μ)2
23
36
23
18
24.85714
24.85714
24.85714
24.85714
-1.85714
11.14286
-1.85714
-6.85714
3.44898
124.1633
3.44898
47.02041
5
26
43
24.85714
24.85714
24.85714
-19.8571
1.142857
18.14286
394.3061
1.306122
329.1633
 x   
i

3-16
 x
i
 
N
2

2

902.8571
902.8571
 11.36 minutes
7
Using the computational formula, yields the same
result.
xi
(xi )2
23
36
529
1296
23
18
5
26
529
324
25
676
43
1849

Σ xi = 174 Σ (xi)2 = 5228
3-17
x



2
x
2
i
i
N
N
174 

5228 
2

7
7
 11.36 minutes
The sample standard deviation, s, of a variable is
the square root of the sum of squared deviations
about the sample mean divided by n – 1, where n
is the sample size.
 x
s
 x
2
i
n 1
x1  x   x2  x 
2

2
 L  xn  x 
2
n 1
where x1, x2, . . . , xn are the n observations in the
sample and x is the sample mean.
3-18
A formula that is equivalent to the one on the
previous slide, called the computational
formula, for determining the sample standard
deviation is
x



2
s
3-19
x
2
i
i
n 1
n
We call n - 1 the degrees of freedom because
the first n - 1 observations have freedom to be
whatever value they wish, but the nth value has
no freedom. It must be whatever value forces
the sum of the deviations about the mean to
equal zero.
3-20
EXAMPLE Computing a Sample Standard
Deviation
Here are the results of a random sample taken
from the travel times (in minutes) to work for all
seven employees of a start-up web development
company:
5, 26, 36
Find the sample standard deviation.
3-21
x
xi
xi  x
xi  x 
2
5
26
22.33333
22.33333
-17.333
3.667
300.432889
13.446889
36
22.33333
13.667
186.786889
x  x   500.66667
2
i
s
3-22
 x
i
 x
n 1
2

500.66667
 15.82 minutes
2
Using the computational formula, yields the same
result.
xi
(xi )2
5
26
25
676
36
1296
Σ xi = 67
Σ
(xi)2
x



2

x
2
i
i
n 1
n
67 

1997 
2
= 1997

3
2
 15.82 minutes
3-23
3-24
EXAMPLE
Comparing Standard Deviations
Determine the standard deviation waiting time
for Wendy’s and McDonald’s. Which is
larger? Why?
3-25
Wait Time at Wendy’s
1.50
2.53
1.88
3.99
0.90
0.79
1.20
2.94
1.90
1.23
1.01
1.46
1.40
1.00
0.92
1.66
0.89
1.33
1.54
1.09
0.94
0.95
1.20
0.99
1.72
0.67
0.90
0.84
0.35
2.00
Wait Time at McDonald’s
3.50
0.00
1.97
0.00
3.08
3-26
0.00
0.26
0.71
0.28
2.75
0.38
0.14
2.22
0.44
0.36
0.43
0.60
4.54
1.38
3.10
1.82
2.33
0.80
0.92
2.19
3.04
2.54
0.50
1.17
0.23
EXAMPLE
Comparing Standard Deviations
Sample standard deviation for Wendy’s:
0.738 minutes
Sample standard deviation for McDonald’s:
1.265 minutes
Recall from earlier that the data is more dispersed
for McDonald’s resulting in a larger standard
deviation.
3-27
Objective 3
• Determine the Variance of a Variable from
Raw Data
3-28
The variance of a variable is the square of the
standard deviation. The population variance is
σ2 and the sample variance is s2.
3-29
EXAMPLE
Computing a Population Variance
The following data represent the travel times (in
minutes) to work for all seven employees of a start-up
web development company.
23, 36, 23, 18, 5, 26, 43
Compute the population and sample variance of this
data.
3-30
EXAMPLE
Computing a Population Variance
Recall that the population standard deviation (from
slide #49) is σ = 11.36 so the population variance is
σ2 = 129.05 minutes
and that the sample standard deviation (from slide
#55) is s = 15.82, so the sample variance is
s2 = 250.27 minutes
3-31
Objective 4
• Use the Empirical Rule to Describe Data that
are Bell Shaped
3-32
The Empirical Rule
If a distribution is roughly bell shaped, then
• Approximately 68% of the data will lie within
1 standard deviation of the mean. That is,
approximately 68% of the data lie between
μ – 1σ and μ + 1σ.
• Approximately 95% of the data will lie within
2 standard deviations of the mean. That is,
approximately 95% of the data lie between
μ – 2σ and μ + 2σ.
3-33
The Empirical Rule
If a distribution is roughly bell shaped, then
• Approximately 99.7% of the data will lie
within 3 standard deviations of the mean. That
is, approximately 99.7% of the data lie
between μ – 3σ and μ + 3σ.
Note: We can also use the Empirical Rule based
on sample data with x used in place of μ and s
used in place of σ.
3-34
3-35
EXAMPLE Using the Empirical Rule
The following data represent the serum HDL
cholesterol of the 54 female patients of a family
doctor.
41
62
67
60
54
45
3-36
48
75
69
60
54
47
43
77
69
60
55
47
38
58
70
61
56
48
35
82
65
62
56
48
37
39
72
63
56
50
44
85
74
64
57
52
44
55
74
64
58
52
44
54
74
64
59
53
(a) Compute the population mean and standard
deviation.
(b) Draw a histogram to verify the data is bell-shaped.
(c) Determine the percentage of all patients that have
serum HDL within 3 standard deviations of the
mean according to the Empirical Rule.
(d) Determine the percentage of all patients that have
serum HDL between 34 and 69.1 according to the
Empirical Rule.
(e) Determine the actual percentage of patients that
have serum HDL between 34 and 69.1.
3-37
(a) Using a TI-83 plus graphing calculator, we find
  57.4 and   11.7
(b)
3-38
22.3
34.0
45.7
57.4
69.1
80.8
92.5
(c) According to the Empirical Rule, 99.7% of the all patients that
have serum HDL within 3 standard deviations of the mean.
(d) 13.5% + 34% + 34% = 81.5% of all patients will have a serum
HDL between 34.0 and 69.1 according to the Empirical Rule.
(e) 45 out of the 54 or 83.3% of the patients have a serum HDL
between 34.0 and 69.1.
3-39
Objective 5
• Use Chebyshev’s Inequality to Describe Any
Set of Data
3-40
Chebyshev’s Inequality
For any data set or distribution, at least
1

 1  2  100% of the observations lie within k
k
standard deviations of the mean, where k is any
number greater than 1. That is, at least
1

 1  2  100% of the data lie between μ – kσ
k
and μ + kσ for k > 1.
Note: We can also use Chebyshev’s Inequality
based on sample data.
3-41
EXAMPLE Using Chebyshev’s Theorem
Using the data from the previous example, use
Chebyshev’s Theorem to
(a) determine the percentage of patients that have serum
HDL within 3 standard deviations of the mean.
1

 1  2  100%  88.9%
3
(b) determine the actual percentage of patients that have
serum HDL between 34 and 80.8 (within 3 SD of mean).
52/54 ≈ 0.96 ≈ 96%
3-42