Question

The distribution of heights of adult men in the U.S. is approximately normal with mean 69 inches and standard deviation 2.5 inches. Use what you know about the EMPIRICAL RULE to answer the following.

a)Approximately what percent of men are taller than 69 inches?

b)Approximately what percent of men are between 64 and 66.5 inches?

Answer 1

Solution:

We are given that heights are normally distributed.

Mean = 69

SD = 2.5

Part a

We have to find P(X>69)

P(X>69) = 1 – P(X<69)

Z = (X – mean) / SD

Z = (69 – 69)/2.5

Z = 0

P(Z<0) = P(X<69) = 0.5000

(by using z-table)

P(X>69) = 1 – P(X<69)

P(X>69) = 1 – 0.5000

P(X>69) = 0.5000

Required probability = 0.5000

According to empirical rule, approximately 99.7% of the data lies within three standard deviations from mean. Total area is 1. Area above the mean is 0.5 and area below the mean is 0.5.

Part b

We have to find P(64<X<66.5)

P(64<X<66.5) = P(X<66.5) – P(X<64)

Find P(X<66.5)

Z = (X – mean) / SD

Z = (66.5 - 69)/2.5 = -1

P(Z<-1) = P(X<66.5) = 0.158655

(by using z-table)

Find P(X<64)

Z = (64 - 69)/2.5 =-2

P(Z<-2) = P(X<64) = 0.02275

(by using z-table)

P(64<X<66.5) = P(X<66.5) – P(X<64)

P(64<X<66.5) = 0.158655 - 0.02275

P(64<X<66.5) = 0.135905

Required probability = 0.135905

According to empirical rule diagram, area between z = -2 and -1 is approximately 13.6%.