Basic Statistics

AREAS UNDER THE NORMAL CURVE 67

Figure 6.4 Proportions of men’s heights over 70.3 in.

Figure 6.5 Proportions of men’s heights < 65.7 in.

we have t = (65.7 - 68.0)/2.3 = -1. TableA.2 does not give negative values

of 2, but since the curve is symmetric, the area below z = -1 is equal to the

area above t = 1, or 15.87% (see Figure 6.5).

4. Subtracting both the percentage of heights that are < 65.7 in. and the percentage
   of heights that are > 70.3 in. from loo%, we obtain 68.26% as the percentage of
   heights lying between 65.7 and 70.3 in. Here, we see that approximately two-
   thirds of the observations lie between one standard deviation above the mean
   and one standard deviation below the mean if the data are normally distributed
   (see Figure 6.6).

Figure 6.6 Proportions of men’s heights between 65.7 and 70.3 in.

5. It is sometimes of interest to know some percentiles for the distribution of
   men’s heights. Suppose we wish to find the height that exceeds 99% of the