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Chapter
2
Descriptive Statistics
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Chapter Outline
• 2.1 Frequency Distributions and Their Graphs
• 2.2 More Graphs and Displays
• 2.3 Measures of Central Tendency
• 2.4 Measures of Variation
• 2.5 Measures of Position
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Section 2.4
Measures of Variation
.
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Section 2.4 Objectives
• How to find the range of a data set
• How to find the variance and standard deviation of a
population and of a sample
• How to use the Empirical Rule and Chebychev’s
Theorem to interpret standard deviation
• How to approximate the sample standard deviation
for grouped data
• How to use the coefficient of variation to compare
variation in different data sets
.
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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
maximum
• Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
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.
.
Salary ($1000s), x Deviation: 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
42 – 41.5 = 0.5
Σx = 415
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
In Symbols
1. Find the mean of the
population data set.
.
x

N
2. Find deviation of each
entry.
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
.
In Symbols
5. Divide by N to get the
population variance.
2

(
x


)
2 
N
6. Find the square root to get
the population standard
deviation.
( 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 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
In Symbols
5. Divide by n – 1 to get the
sample variance.
6. Find the square root to get
the sample standard
deviation.
.
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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
23
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%
34%
2.35%
2.35%
13.5%
x  3s
.
x  2s
13.5%
x s
x
xs
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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
1 3
• k = 2: In any data set, at least 1  2  or 75%
2
4
of the data lie within 2 standard deviations of the
mean.
1 8
• k = 3: In any data set, at least 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 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.
.
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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
35
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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Coefficient of Variation
Coefficient of Variation (CV)
• Describes the standard deviation of a data set as a
percent of the mean.
• Population data set:
 CV = s i100%
m
• Sample data set:
s

CV  100%
x
.
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Example: Comparing Variation in
Different Data Sets
The table shows the population
heights (in inches) and weights (in
pounds) of the members of a
basketball team. Find the
coefficient of variation for the
heights and the weighs. Then
compare the results.
.
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Solution: Comparing Variation in
Different Data Sets
The mean height is   72.8 inches with a standard
deviation of   3.3 inches. The coefficient of variation
for the heights is
CVheight
s
= i100%
m
3.3
=
i100%
72.8
» 4.5%
.
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Solution: Comparing Variation in
Different Data Sets
The mean weight is   187.8 pounds with a standard
deviation of   17.7 pounds. The coefficient of
variation for the weights is
CVweight
s
= i100%
m
17.7
=
i100%
187.8
» 9.4%
The weights (9.4%) are more variable than the heights
(4.5%).
.
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Section 2.4 Summary
• Found the range of a data set
• Found 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
• Used the coefficient of variation to compare variation
in different data sets
.
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42