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Putting Statistics to Work
Copyright © 2011 Pearson Education, Inc.
Unit 6B
Measures of Variation
Copyright © 2011 Pearson Education, Inc.
Slide 6-3
6-B
Why Variation Matters
Consider the following waiting times for 11 customers
at 2 banks.
Big Bank (three lines):
4.1 5.2 5.6 6.2 6.7 7.2
7.7 7.7 8.5 9.3 11.0
Best Bank (one line):
6.6 6.7 6.7 6.9 7.1 7.2
7.3 7.4 7.7 7.8 7.8
Which bank is likely to have more unhappy customers?
→ Big Bank, due to more surprise long waits
Copyright © 2011 Pearson Education, Inc.
Slide 6-4
6-B
Range
The range of a data set is the difference between its
highest and lowest data values.
range = highest value (max) – lowest value (min)
Copyright © 2011 Pearson Education, Inc.
Slide 6-5
6-B
Quartiles

The median of the data set divides the data set into the
lower half and the upper half.

The lower quartile (or first quartile) divides the lowest
fourth of a data set from the upper three-fourths. It is the
median of the data values in the lower half of a data set.

The middle quartile (or second quartile) is the overall
median.

The upper quartile (or third quartile) divides the lower
three-fourths of a data set from the upper fourth. It is the
median of the data values in the upper half of a data set.
Copyright © 2011 Pearson Education, Inc.
Slide 6-6
6-B
The Five-Number Summary

The five-number summary for a data set
consists of the following five numbers:
low value

lower quartile
median
upper quartile
high value
A boxplot shows the five-number summary
visually, with a rectangular box enclosing the
lower and upper quartiles, a line marking the
median, and whiskers extending to the low and
high values.
Copyright © 2011 Pearson Education, Inc.
Slide 6-7
6-B
The Five-Number Summary
Five-number summary of the waiting times at each bank:
Big Bank
Best Bank
low value (min) = 4.1
lower quartile = 5.6
median = 7.2
upper quartile = 8.5
high value (max) = 11.0
low value (min) = 6.6
lower quartile = 6.7
median = 7.2
upper quartile = 7.7
high value (max) = 7.8
The corresponding boxplot:
Copyright © 2011 Pearson Education, Inc.
Slide 6-8
6-B
Standard Deviation
The standard deviation is the single number most
commonly used to describe variation.
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-9
6-B
Calculating the Standard Deviation
The standard deviation is calculated by completing
the following steps:
1. Compute the mean of the data set. Then find the
deviation from the mean for every data value.
deviation from the mean = data value – mean
2. Find the squares of all the deviations from the mean.
3. Add all the squares of the deviations from the mean.
4. Divide this sum by the total number of data values
minus 1.
5. The standard deviation is the square root of this
quotient.
Copyright © 2011 Pearson Education, Inc.
Slide 6-10
6-B
Standard Deviation
Let A = {2, 8, 9, 12, 19} with a mean of 10. Find the sample
standard deviation of the data set A.
x (data value)
2
8
9
12
19
x – mean
(deviation)
2 – 10 = –8
8 – 10 = –2
9 – 10 = –1
12 – 10 = 2
19 – 10 = 9
Total
(deviation)2
(-8)2 = 64
(-2)2 = 4
(-1)2 = 1
(2)2 = 4
(9)2 = 81
154
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
154

 6.2
5 1
Copyright © 2011 Pearson Education, Inc.
Slide 6-11
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
x – mean
(deviation)
(deviation)2
1
7
8
11
18
Total
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-12
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
1
7
8
11
18
x – mean
(deviation)
1 – 9 = -8
(deviation)2
64
Total
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-13
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
1
7
8
11
18
x – mean
(deviation)
1 – 9 = -8
7 – 9 = -2
(deviation)2
64
4
Total
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-14
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
1
7
8
11
18
x – mean
(deviation)
1 – 9 = -8
7 – 9 = -2
8 – 9 = -1
(deviation)2
64
4
1
Total
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-15
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
1
7
8
11
18
x – mean
(deviation)
1 – 9 = -8
7 – 9 = -2
8 – 9 = -1
11 – 9 = 2
(deviation)2
64
4
1
4
Total
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-16
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
1
7
8
11
18
x – mean
(deviation)
1 – 9 = -8
7 – 9 = -2
8 – 9 = -1
11 – 9 = 2
18 – 9 = 9
Total
(deviation)2
64
4
1
4
81
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
Copyright © 2011 Pearson Education, Inc.
Slide 6-17
6-B
Standard Deviation
Let B = {1, 7, 8, 11, 18} with a mean of 9. Find the sample
standard deviation of the data set B.
x (data value)
1
7
8
11
18
x – mean
(deviation)
1 – 9 = -8
7 – 9 = -2
8 – 9 = -1
11 – 9 = 2
18 – 9 = 9
Total
(deviation)2
64
4
1
4
81
154
sum of (deviation s from the mean) 2
standard deviation 
total number of data values  1
154

 6.2
5 1
Copyright © 2011 Pearson Education, Inc.
Slide 6-18
6-B
The Range Rule of Thumb

The standard deviation is approximately related to
the range of a data set by the range rule of
thumb:
range
standard deviation 
4

If we know the standard deviation for a data set,
we estimate the low and high values as follows:
low value  mean  2  standard deviation
high value  mean  2  standard deviation
Copyright © 2011 Pearson Education, Inc.
Slide 6-19