Question

Comparing observations from different populations: The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.66 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.54 inches.

Required:
a. If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?
b. What percentage of men are SHORTER than 6 feet 3 inches?
c. If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?
d. What percentage of women are TALLER than 5 feet 11 inches?

Answer 1

a. Z = 1.99

b. 97.67%

c. Z = 2.56

d. 0.52%

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.66 inches, and thus, we have
`\mu = 69.7, \sigma = 2.66`

6 feet 3 inches = 6*12 + 3 = 75 inches, which means that we have to find z when X = 75. So

`Z = (X - \mu)/(\sigma)`

`Z = (75 - 69.7)/(2.66)`

`Z = 1.99`

b. What percentage of men are SHORTER than 6 feet 3 inches?

The proportion is the p-value of Z = 1.99.

Looking at the z-table, Z = 1.99 has a p-value of 0.9767.

0.9767*100% = 97.67%, which is the answer.

c. If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.54 inches, and thus, we have
`\mu = 64.5, \sigma = 2.54`. We have to find Z when X = 5*12 + 11 = 71. So

`Z = (X - \mu)/(\sigma)`

`Z = (71 - 64.5)/(2.54)`

`Z = 2.56`

d. What percentage of women are TALLER than 5 feet 11 inches?

The proportion is 1 subtracted by the p-value of Z = 2.56.

Looking at the z-table, Z = 2.56 has a p-value of 0.9948.

1 - 0.9948 = 0.0052

0.0052*100% = 0.52%, which is the answer.