:10b.  Solution.  Since (56 - 76)/5 = -2, the members who are
56 inches tall are two standard deviations below average.  Since 
0.90 x -2 = 1.8, their average weight is 1.8 SD's above the
average weight of the sample, which is 140 + (1.8 x 40) = 140 +
72 = 212 lbs.