Question (FPP pp197, #1)

Consider the following statistics computed from a sample of 1000 families: 
         height of husband: mean=68inches; SD=2.7inches
            height of wife: mean=63inches; SD=2.5inches; r=0.25
Assuming normality, answer the following:

a) What percent of wives are over 68inches.
b) Of women married to men 72inches tall, what percent are over 68inches.


Solution:

a) Since we are interested in all women then regression is not required.
                   z=(68 - 63)/2.5=2
   Hence the desired percentage is (100-95.45)/2=2.275%

b) Since we are interested in wives of men of a particular height then
   height of husband is on the x-axis. We may use our quick method to
   determine the mean height of these wives:
                   rise=run[r(Sy/Sx)]
                       =4[0.25(2.5/2.7)]
                       =0.93
   The mean height is therefore 63.93inches.

   We also need the standard deviation about the regression line (Sy|x):
                   Sy|x=2.5(sqrt(1-0.25^2))
                       =2.5(sqrt(0.9375)
                       =2.42

   The z score may now be computed:
                   z=(68 - 63.93)/2.42=1.68
   Hence the desired percentage is (100-91.09)/2=4.455%