66 THE NORMAL DISTRIBUTION
Figure 6.3 Proportions of men’s heights < 70.3 in.
6.2.1 Computing the Area Under a Normal Curve
If the X’s are normally distributed and their population mean is p and their population
standard deviation is 0, then if p is subtracted from every observation, the resulting
new population (the population of X - p) is also normally distributed; its mean is 0
and its standard deviation is cr. If, then, each X - p is divided by cr, the (X - p)/o
will again be normally distributed, with mean 0 and standard deviation 1. So, if we
make the transformation z = (X - p)/cr, the z’s are normally distributed with mean
0 and standard deviation 1. The new variable z is called the standard normal variate,
and the areas under its distribution curve are tabulated in Table A.2. In Table A.2,
the columns are labeled .[A] and A; A percent of the area under the standard normal
curve lies to the left of z[A]. For example, under .[A] we read 2.00 and under X in the
same row we read .9772; thus 97.72% of the area lies below 2.00.
The use of Table A.2 is illustrated below, where we assume that the heights of
a large group of adult men are approximately normally distributed; further, that the
population mean height is 68 in. and that the population standard deviation of the
heights is 2.3 in.- What proportion of the men’s heights is < 70.3 in.? Changing scale with the
formula z = (X - p)/cr, we have
z = (70.3 - 68.0)/2.3 = 2.3/2.3 = 1
In Table A.2 we look for z = 1 under the column heading .[A]; we then read
across to the next column and find 3413. The column heading of this second
column is A. Since the areas under the curve are proportions, the proportion
of men’s heights < 70.3 in. is .8413. Or, equivalently, the percentage of men’s
heights below 70.3 in. is 84.13% (see Figure 6.3).- The proportion of men’s heights that are > 70.3 in. can be found by subtracting
,8413 (the answer in step 1) from 1 .OOOO (the area under the entire curve). Thus
15.87% of the men are taller than 70.3 in. (see Figure 6.4). - What proportion of the heights is < 65.7 in.? From
z = (X - p)/0