Basic Statistics

68 THE NORMAL DISTRIBUTION

men’s heights and is exceeded by just 1%. Such a number is called the 99th

percentile, or Pgg. To find it, we look in the second column (i.e., the X c,olumn)

of Table A.2 and find .9901; in the first column we see 2.33, the approximate

99th percentile. Setting p = 68.0 and g = 2.3 in z = (X - p)/o, we can then

solve for X, the height below which 99% of men’s heights fall.

X - 68.0

2.33 =

2.3

Multiplying both sides of this equation by 2.3 yields

5.359 = X - 68.0

Adding 68.0 to both sides of the equation gives us

X = 73.4 in.

Thus 99% of men’s heights are < 73.4 in.; that is, Pgg, the 99th percentile,

equals 73.4 in.

6.2.2 Linear Interpolation

Usually, when z is computed with the formula z = (X - p)/o, it is found to be a
number that is not given in Table A.2 but that lies between two values in the z[X]
column of the table. We might reasonably proceed in one of four ways: First, use the
closest z available if we are content with an answer that is not very accurate. Second,
we could guess at a number between the two tabled values. Third, we could enter
the value into a statistical program that reports z values. Note that many programs
can give either an accurate z value for a given area, or vice versa. Fourth, we can
interpolate linearly to obtain a reasonably accurate result. Linear interpolation is
more work, but it can be done by hand and it can be done for a variety of reasons.
We now illustrate the various methods of linear interpolation by finding the pro-
portion of men’s heights lying below 7 1 in. Here,

z = (71 - 68)/2.3 = 3/2.3 = 1.304

Note that 1.304 lies between the tabled values of z[X] of 1.30 and 1.3 1. From Table
A.2, we write the information given in Table 6.1.
If not much accuracy is necessary, we might give as an answer .90 or 90%. Or
for a little more accuracy, one could take a value of X halfway between .9032 and
.9049 since 1.304 is about one-half of the way between 1.30 and 1.3 1 ; we then choose
.9040.
When more accuracy is desired, and one is working by hand, linear interpolation
can be done as follows. We note that the distance between 1.310 and 1.300 is .010.
Also, the distance between 1.304 and 1.300 is ,004, or .4 of the distance between the
tabled values (.004/.010 = .4).
The distance between .9049 and .9032 is .0017. Multiplying this distance by the
same .4 results in .4( .0017) = .0007. Thus .0007 is .4 of the distance between .9032
and ,9049. Finally, we add .0007 to .9032 to obtain .9039, an accurate estimate.