Assume that the weight of American males is normally distributed with a mean of
ID: 3358640 • Letter: A
Question
Assume that the weight of American males is normally distributed with a mean of 160 pounds, with a standard deviation of 15 pounds. As a store manager, you wish to order 20000 (male) bathrobes. These bathrobes come in three sizes, Small, Medium, and Large. The weight/size table for the male bathrobe is:
Male weight Bathrobe size
Under 145 pounds -> Small
145-178 pounds -> Medium
178 pounds or more -> Large
If you wish to order the same proportion of sizes as in the male population, how many medium size bathrobes should you order?
Explanation / Answer
Proportion of males weighing between 145 and 178 = P(145 < X < 178)
Mean = 160 pounds
Standard deviation = 15 pounds
P(X < A) = P(Z < (A - mean)/standard deviation)
Proportion of males weighing between 145 and 178 = P(145 < X < 178)
= P(X < 178) - P(X < 145)
= P(Z < (178-160)/15) - P(Z < (145-160)/15)
= P(Z < 1.2) - P(Z < -1)
= 0.8849 - 0.1587
= 0.7262
Number of medium size bathrobes to be ordered = 2000x0.7262 = 1452