Measures of Dispersion
Pulse Rate
(x)
Deviation
(x - µ)
Absolute Deviation
|(x - µ)|
Square of Deviation
(x - µ)2
1
75
-5.00
5.00
25.00
2
84
4.00
4.00
16.00
3
76
-4.00
4.00
16.00
4
84
4.00
4.00
16.00
5
84
4.00
4.00
16.00
6
78
-2.00
2.00
4.00
7
69
-11.00
11.00
121.00
8
90
10.00
10.00
100.00
Sum (∑)
640
0.00
44.00
314.00
Mean (µ)
80.00
0.00
5.50
39.25
6.26
Mean Absolute
Deviation
Variance
Standard Deviation
1. Using the data in the table above, calculate:
 Mean: the average  Median: the middle
 Mode: the most frequent
80
81
84
Symbol for mean is  .
2. Add two more values to the data set that will not affect the mean, median, or mode. How did you determine
these values? Possible answers are 60 and 100; 68 and 92, 75 and 85 – The mean is a balance point. The
distance above the mean must equal the distance below the mean. For example: 60 is 20 units below the mean
and 100 is 20 units above the mean.
3. The deviations tell us the distance a value falls away from the mean. Calculate the deviation as x   , where
x is the data value and  is the mean.
 What do you notice about the mean of the deviations? The mean of the deviations is 0. This is always true.
 Why is this true? The mean is a balance point. The sum of the distances above the mean must equal the sum
of the distances below the mean.
4. The mean absolute deviation tells us the average distance a set of values falls away from the mean. Calculate
mean absolute deviation by finding the average of the absolute value deviations. This example’s mean deviation
is 5.5, which means, on average, the data values are 5.5 units away from 80.
5. The variance tells us the average distance a set of values falls away from the mean. Calculate the variance by
finding the average of the squared deviations. Variance is represented by  (read as sigma squared). This
example’s variance is 39.25, which means, on average, the data values are 39.25 units away from 80. (note this
is not reasonable, because the unit has changed)
 What are the units of the mean? beats per minute What are the units of the variance? (beats per minute)2
 What can be done to the variance to have the same units as the mean? Square root the variance to get
beats per minute. This is needed because we can’t compare beats/min to (beats/min)2.
2
Measures of Dispersion
6. The standard deviation tells us the average distance a set of values falls away from the mean. Calculate by
taking the square root of the variance. Standard deviation is represented by  (read as sigma).This example’s
standard deviation is 6.26, which means, on average, the data values are 6.26 units away from 80.
7. Outliers are values that are significantly larger or smaller than the rest of the data.
 Outliers can affect the:
mean? Yes median? Yes average deviation? NO – always 0
mean absolute deviation? Yes standard deviation? Yes
Which is affected more by outliers – mean absolute deviation or standard deviation? Explain. Standard
deviation is affected more by an outlier because the distance between a value and the mean is squared.
Squaring this difference will magnify the distance between the two values.
8. Mean absolute deviation, variance, and standard deviation all tell us about the spread of the data within a set.
 If the mean absolute deviation, variance, and standard deviation is high, then the data is spread out
 If the mean absolute deviation, variance, and standard deviation is low, then the data is close together
Example 1: Descriptive statistics are shown for two data sets below. Match the description to the data set.
Data Set A
Mean = 45
Absolute Mean Deviation = 9.1
Variance = 106.1
Standard Deviation = 10.3
Data Set B
Mean = 45
Absolute Mean Deviation = 16
Variance = 420.3
Standard Deviation = 20.5
Data Set A
Data Set B
Data Set B has a higher deviation value which tells me that the data values of Data Set B are more
spread out. The 2nd graph shows a wider range (values are more spread).
Example 2: At The Grill, steaks are cut into 12-oz. portions when ordered by a customer. For the
past month, the manager has been trying to find who can most accurately portion out the meat. The
manager collected the standard deviation of sample meat portions each week for each of his cooks.
Joe
Sarah
Holli
Jason
Ben
Week 1
0.2
0.15
0.2
0.15
0.3
Week 2
0.2
0.2
0.18
0.1
0.1
Week 3
0.2
0.22
0.15
0.13
0.2
Week 4
0.15
0.25
0.1
0.11
0.1
Which three should he choose?
Why? Holli, Jason, & Ben should
be chosen because they frequently
have the lowest standard
deviations. This means their cuts
of meat are typically close to the
standard 12 oz. portion. Holli,
Jason, & Ben have the lowest
average standard deviations also.