The distribution of weights for the population of males in t

The distribution of weights for the population of males in the United States is approximately normal with mean = 172.2 pounds and standard deviation = 29.8 pounds.

d) Suppose five males are selected at random from the population. What is the probability that exactly three will weigh more than 130 pounds?

e) What is the 20th percentile of the distribution of weights for the male population in the U.S., i.e., the weight that divides the bottom 20% from the top 80%?

f) What is the 90th percentile of the distribution?

Solution

N(172.2, 29.8)

d) P(one weighs >130)

= P(Z>-1.42) = 0.9222

OUt of 5, 3 weigh > 130 pounds can be calculated using binomial with p =0.9222, n =5, r =3

Required prob =0.0475

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e) 20th percentile

From z table 20th percentile = -0.84

x score = 172.2-0.84(29.8)

= 147.168

20th percentile = 147.168

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f) 90th percentile z = 1.28

90th percentile x = 172.2+1.28(29.8) =

210.344