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Section 2.4
Measures of Variation
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Section 2.4 Objectives
• Determine the range of a data set
• Determine the variance and standard deviation of a population and of
a sample
• Use the Empirical Rule and Chebychev’s Theorem to interpret
standard deviation
• Approximate the sample standard deviation for grouped data
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Range
Range
• The difference between the maximum and minimum
data entries in the set.
• The data must be quantitative.
• Range = (Max. data entry) – (Min. data entry)
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Example: Finding the Range
A corporation hired 10 graduates. The starting salaries for each
graduate are shown. Find the range of the starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
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Solution: Finding the Range
• Ordering the data helps to find the least and greatest salaries.
37 38 39 41 41 41 42 44 45 47
minimum
• Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
maximum
The range of starting salaries is 10 or $10,000.
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Deviation, Variance, and Standard Deviation
Deviation
• The difference between the data entry, x, and the mean of the data
set.
• Population data set:
• Deviation of x = x – μ
• Sample data set:
• Deviation of x  x  x
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Example: Finding the Deviation
A corporation hired 10 graduates. The starting salaries
for each graduate are shown. Find the deviation of the
starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
Solution:
• First determine the mean starting salary.
x 415


 41.5
N
10
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Solution: Finding the Deviation
• Determine the
deviation for each
data entry.
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Deviation ($1000s)
Salary ($1000s), x
x–μ
41
41 – 41.5 = –0.5
38
38 – 41.5 = –3.5
39
39 – 41.5 = –2.5
45
45 – 41.5 = 3.5
47
47 – 41.5 = 5.5
41
41 – 41.5 = –0.5
44
44 – 41.5 = 2.5
41
41 – 41.5 = –0.5
37
37 – 41.5 = –4.5
42
Σx = 415
42 – 41.5 = 0.5
Σ(x – μ) = 0
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Deviation, Variance, and Standard Deviation
Population Variance
•
( x   )
 
N
2
2
Sum of squares, SSx
Population Standard Deviation
•
2

(
x


)
  2 
N
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Finding the Population Variance & Standard
Deviation
In Words
1. Find the mean of the
population data set.
2. Find the deviation of each
entry.
In Symbols
x

N
x–μ
3. Square each deviation.
(x – μ)2
4. Add to get the sum of
squares.
SSx = Σ(x – μ)2
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Finding the Population Variance & Standard
Deviation
In Words
5. Divide by N to get the
population variance.
6. Find the square root of the
variance to get the
population standard
deviation.
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In Symbols
2

(
x


)
2 
N
( x   ) 2

N
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Example: Finding the Population Standard
Deviation
A corporation hired 10 graduates. The starting salaries for each
graduate are shown. Find the population variance and standard
deviation of the starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
Recall μ = 41.5.
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Solution: Finding the Population Standard
Deviation
• Determine SSx
• N = 10
Deviation: x – μ
Squares: (x – μ)2
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
38
38 – 41.5 = –3.5
(–3.5)2 = 12.25
39
39 – 41.5 = –2.5
(–2.5)2 = 6.25
45
45 – 41.5 = 3.5
(3.5)2 = 12.25
47
47 – 41.5 = 5.5
(5.5)2 = 30.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
44
44 – 41.5 = 2.5
(2.5)2 = 6.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
37
37 – 41.5 = –4.5
(–4.5)2 = 20.25
42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5
Salary, x
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Solution: Finding the Population Standard
Deviation
Population Variance
( x   )
88.5

 8.9
•  
N
10
2
2
Population Standard Deviation
•    2  8.85  3.0
The population standard deviation is about 3.0, or $3000.
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Deviation, Variance, and Standard Deviation
Sample Variance
•
( x  x )
s 
n 1
2
2
Sample Standard Deviation
•
2

(
x

x
)
s  s2 
n 1
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Finding the Sample Variance & Standard
Deviation
In Words
In Symbols
x
n
1. Find the mean of the
sample data set.
x
2. Find the deviation of each
entry.
xx
3. Square each deviation.
( x  x )2
4. Add to get the sum of
squares.
SS x  ( x  x ) 2
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Finding the Sample Variance & Standard
Deviation
In Words
5. Divide by n – 1 to get the
sample variance.
6. Find the square root of the
variance to get the sample
standard deviation.
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In Symbols
2

(
x

x
)
s2 
n 1
( x  x ) 2
s
n 1
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Example: Finding the Sample Standard
Deviation
The starting salaries are for the Chicago branches of a corporation. The
corporation has several other branches, and you plan to use the
starting salaries of the Chicago branches to estimate the starting
salaries for the larger population. Find the sample standard deviation of
the starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
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Solution: Finding the Sample Standard
Deviation
• Determine SSx
• n = 10
Deviation: x – μ
Squares: (x – μ)2
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
38
38 – 41.5 = –3.5
(–3.5)2 = 12.25
39
39 – 41.5 = –2.5
(–2.5)2 = 6.25
45
45 – 41.5 = 3.5
(3.5)2 = 12.25
47
47 – 41.5 = 5.5
(5.5)2 = 30.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
44
44 – 41.5 = 2.5
(2.5)2 = 6.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
37
37 – 41.5 = –4.5
(–4.5)2 = 20.25
42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5
Salary, x
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Solution: Finding the Sample Standard
Deviation
Sample Variance
( x  x )
88.5

 9.8
• s 
n 1
10  1
2
2
Sample Standard Deviation
88.5
 3.1
• s s 
9
2
The sample standard deviation is about 3.1, or $3100.
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Example: Using Technology to Find the
Standard Deviation
Sample office rental rates (in dollars
per square foot per year) for Miami’s
central business district are shown
in the table. Use a calculator or a
computer to find the mean rental
rate and the sample standard
deviation. (Adapted from: Cushman &
Wakefield Inc.)
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Office Rental Rates
35.00
33.50
37.00
23.75
26.50
31.25
36.50
40.00
32.00
39.25
37.50
34.75
37.75
37.25
36.75
27.00
35.75
26.00
37.00
29.00
40.50
24.50
33.00
38.00
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Solution: Using Technology to Find the
Standard Deviation
Sample Mean
Sample Standard
Deviation
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Interpreting Standard Deviation
• Standard deviation is a measure of the typical amount
an entry deviates from the mean.
• The more the entries are spread out, the greater the
standard deviation.
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Interpreting Standard Deviation: Empirical
Rule (68 – 95 – 99.7 Rule)
For data with a (symmetric) bell-shaped distribution, the
standard deviation has the following characteristics:
• About 68% of the data lie within one standard
deviation of the mean.
• About 95% of the data lie within two standard
deviations of the mean.
• About 99.7% of the data lie within three standard
deviations of the mean.
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Interpreting Standard Deviation: Empirical
Rule (68 – 95 – 99.7 Rule)
99.7% within 3 standard deviations
95% within 2 standard deviations
68% within 1
standard deviation
34%
2.35%
x  3s
34%
13.5%
x  2s
13.5%
x s
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x
xs
2.35%
x  2s
x  3s
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Example: Using the Empirical Rule
In a survey conducted by the National Center for Health
Statistics, the sample mean height of women in the
United States (ages 20-29) was 64.3 inches, with a
sample standard deviation of 2.62 inches. Estimate the
percent of the women whose heights are between
59.06 inches and 64.3 inches.
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Solution: Using the Empirical Rule
• Because the distribution is bell-shaped, you can use the Empirical Rule.
34% + 13.5% = 47.5% of women are between 59.06
and 64.3 inches tall.
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Chebychev’s Theorem
• The portion of any data set lying within k standard
deviations (k > 1) of the mean is at least:
1
1 2
k
• k = 2: In any data set, at least
1 3
1  2  or 75%
2
4
of the data lie within 2 standard deviations of the mean.
• k = 3: In any data set, at least
1 8
1  2  or 88.9%
3
9
of the data lie within 3 standard deviations of the mean.
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Example: Using Chebychev’s Theorem
The age distribution for Florida is shown in the
histogram. Apply Chebychev’s Theorem to the data
using k = 2. What can you conclude?
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Solution: Using Chebychev’s Theorem
k = 2: μ – 2σ = 39.2 – 2(24.8) = – 10.4 (use 0 since age
can’t be negative)
μ + 2σ = 39.2 + 2(24.8) = 88.8
At least 75% of the population of Florida is between 0 and 88.8 years old.
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Standard Deviation for Grouped Data
Sample standard deviation for a frequency distribution
•
•
( x  x ) 2 f
s
n 1
where n = Σf (the number of
entries in the data set)
When a frequency distribution has classes, estimate the sample mean and the sample standard deviation by using the midpoint of each class.
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Example: Finding the Standard Deviation for
Grouped Data
You collect a random sample of the
number of children per household
in a region. Find the sample mean
and the sample standard deviation
of the data set.
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Number of Children in
50 Households
1
3
1
1
1
1
2
2
1
0
1
1
0
0
0
1
5
0
3
6
3
0
3
1
1
1
1
6
0
1
3
6
6
1
2
2
3
0
1
1
4
1
1
2
2
0
3
0
2
4
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Solution: Finding the Standard Deviation for
Grouped Data
• First construct a frequency distribution.
• Find the mean of the frequency
distribution.
xf 91
x

 1.8
n
50
The sample mean is about 1.8 children.
x
f
xf
0
10
0(10) = 0
1
19
1(19) = 19
2
7
2(7) = 14
3
7
3(7) =21
4
2
4(2) = 8
5
1
5(1) = 5
6
4
6(4) = 24
Σf = 50 Σ(xf )= 91
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Solution: Finding the Standard Deviation for
Grouped Data
• Determine the sum of squares.
x
f
xx
( x  x )2
0
10
0 – 1.8 = –1.8
(–1.8)2 = 3.24
3.24(10) = 32.40
1
19
1 – 1.8 = –0.8
(–0.8)2 = 0.64
0.64(19) = 12.16
2
7
2 – 1.8 = 0.2
(0.2)2 = 0.04
0.04(7) = 0.28
3
7
3 – 1.8 = 1.2
(1.2)2 = 1.44
1.44(7) = 10.08
4
2
4 – 1.8 = 2.2
(2.2)2 = 4.84
4.84(2) = 9.68
5
1
5 – 1.8 = 3.2
(3.2)2 = 10.24
10.24(1) = 10.24
6
4
6 – 1.8 = 4.2
(4.2)2 = 17.64
17.64(4) = 70.56
( x  x )2 f
( x  x )2 f  145.40
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Solution: Finding the Standard Deviation for
Grouped Data
• Find the sample standard deviation.
x 2 x
( x  x )2
( x  x ) f
145.40
s

 1.7
n 1
50  1
( x  x )2 f
The standard deviation is about 1.7 children.
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Section 2.4 Summary
• Determined the range of a data set
• Determined the variance and standard deviation of a population and
of a sample
• Used the Empirical Rule and Chebychev’s Theorem to interpret
standard deviation
• Approximated the sample standard deviation for grouped data
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