square the numbers first. (Note: Taking the square root of the variance gives the
standard deviation.)
For example, take the numbers 3, 5, 8, and 9, with a mean of 6.25 (the sum of the
numbers divided by the total number of numbers). To calculate the variance, deter-
mine the deviation of each number from 6.25 (3.25, 1.25, 1.75, 2.75), square each
deviation (10.5625, 1.5625, 3.0625, 7.5625), then take the average 22.75/4 5.6875,
which is the variance. An easier way to calculate the variance is to square all the num-
bers first (9, 25, 64, 81) and determine the mean (9 25 64 81 divided by 4
44.75). Then subtract the square of the first mean (6.25^2 39.0625)—or 44.75
39.0625 5.6875.
What is the standard deviation?
The standard deviation is considered by some to be the second most important statis-
tic (or statistical measure) in the field; it is the measure of how much the individual
observations are scattered about the mean. In general, the more widely values are
spread out, the larger the standard deviation. For example, if the test results for two
different exams taken by 50 people in a geology class range from 30 to 98 percent for
the first exam and 78 to 95 percent for the second exam, the standard deviation is larg-
er for the first list of exams. This spread (dispersion) of a data set is calculated by tak-
ing the square root of the variance; the notation for standard deviation is most com-
monly seen as follows: V(x) s.d.(or simply s), in which V(x) is the variance.
What is a normal distribution?
A normal distribution is an idealized view of the world, producing the familiar, sym-
metrically shaped “bell-shaped curve.” It is usually based on a large set of measure-
ments of one quantity—such as weights, test scores, or height—which are arranged
by size. In a normal distribution, more than two-thirds of the measurements fall in
the central region of the graph; about one-sixth of them are found on either side. 261
APPLIED MATHEMATICS
What is the chi-square test?
T
he chi-square test is a way to determine the odds for or against a given devia-
tion from the expected statistical distribution. This somewhat complex sta-
tistical test computes the probability that there is no major difference between
the expected frequency of an event with the observed frequency of that event—
and especially to determine if the set of responses is significantly different from
an expected set of responses only because of chance. There are even various ways
to perform this type of test, such as the Pearson’s chi-square test.