Measures of Dispersion
Section 4.3

Sometimes measures of central tendency alone do not give enough information for inferential statistics.  Consistency or spread of data must be included before conclusions can be drawn.

Example 1 on page 143 is a perfect example of how the mean, median and mode may not be enough to draw conclusions.

George: mean = 185, median = 185, mode = 185

Danny: mean = 185, median = 185, mode = 185

Relying solely on central tendency draws an inaccurate conclusion about both bowlers.

Measures of dispersion determine how data points differ from the average (mean).

The difference between a single data point x, and the mean x bar is called deviation from the mean.  (x – x )

If all the deviations from the mean were added together, the total = 0 (by definition of mean).  To use deviation, first square the difference before summing.

Variance:  s2   =   1/(n-1)  2

s2 reminds us that deviations have been squared.  Working with sample data, dividing by n – 1 instead of n gives a closer estimate to the population data.

standard deviation =

Example 1:

Find the standard deviation of the following:

 Errors in typing  per page mean deviation deviation squared 1 x = 2 1 – 2 = -1 1 0 x = 2 0 – 2 = -2 4 3 x = 2 3 – 2 = 1 1 2 x = 2 2 – 2 = 0 0 1 x = 2 1 – 2 = -1 1 5 x = 2 5 – 2 = 3 9 Total 16

x  =  1 + 0 + 3 + 2 + 1 + 5  =  12  =  2
6                                                   6

First find variance:

s2 = 16    = 16/5 = 3.2
6 – 1

s = standard deviation =  = 1.78 = 1.8

(Keep data on board for later)

Alternate Formula:    s2   =   1     [ Sx 2 - (Sx)2 ]  looks difficult, but is easier to use.
n-1                  n

Finding Variance and Standard Deviation of grouped data.

Recall with grouped data; first find the midpoint xm of each interval, then multiply by the frequency.

s2   =   1    [ Sfx 2 - ( Sfx )2 ]
n -1                   n

Using both measures of central tendency and measures of dispersion gives the best analysis of a collection of data.  Combining both measures provides a look at what percentage lies within a specified number of standard deviations of the mean.

Using EX 1 of typing errors, s = 1.8, x = 2, then

2 – 1.8 = .2                 2 + 1.8 = 3.8

___________|_________________

0  .2      1          2          3      3.84          5

4/6 = 2/3 = .66 = 66% lie within one standard deviation of the mean.

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