Chapter 20: Measures of Center and Spread 285Calculators and computers often do the work of finding a standard deviation, especially for large
sets of numbers, but let’s go through the steps once, just so you know what’s going on. Let’s use
the test scores of 43, 61, 73, 85, and 98 from the earlier example about spread.
The first step in finding a standard deviation is to find the mean. You know from the earlier
example that the mean of this data is 72. The next step is to subtract that mean from each
number.
Number Number — Mean
43 43 – 72 = -29
61 61 – 72 = -11
73 73 – 72 = 1
85 85 – 72 = 13
98 98 – 72 = 26The basic idea is to average these deviations, but if you add them up right now, the negative num-
bers and the positive numbers will cancel each other out. You could take the absolute value of
each deviation, add the absolute values, and divide by the number of them you have. That would
give you the mean absolute deviation. But the standard deviation calculation squares the deviations
first. Like the absolute value, that makes everything positive.
Number Number — Mean Squared
43 43 – 72 = -29 (-29)^2 = 841
61 61 – 72 = -11 (-11)^2 = 121
73 73 – 72 = 1 12 = 1
85 85 – 72 = 13 132 = 169
98 98 – 72 = 26 262 = 676Now you add up all those squares and get 1,808. Next, divide. If your data set comes from the
whole population of interest, divide by the number of values. If you’re working with just a sample
from a larger group, divide by one less than that. In this data set you have 5 values, so if this set
is everyone’s test, divide by 5. If it’s only a sample of the test takers, divide by 4. 1,808 z 5 =
361.6 or 1,808 z 4 = 452. The last step is to undo the squaring by taking the square root of 361.6
or 452. 361 6..~19 02 or 452 21.26. This set of test scores has a mean of 72 and a standard
deviation of either 19.02 or 21.26. That’s a pretty big standard deviation, either way, telling you
those test scores are very spread out.