Question

Assume that the heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.2 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If 150 women want to enlist in the U.S. Army, how many would you expect to meet the height requirements?

Answer 1

Answer:

The number of women expected to meet the height requirements is 149 women.

Step-by-step explanation:

To solve the question, we note that

The population mean = 63.5 inches

The population standard deviation = 2.2

The range of required height = 58 to 80

We are required to find the proportion of women between the height range

The z value for 58 is given by

$z = \frac{x- \mu}{\sigma}$

Where:

x = Required statistic

μ = Population mean

σ = Standard deviation

= (58 - 63.5)/2.2 = -2.5 and

From the z-score table, we have P(z= -2.5) = 0.0062

The z value for 80 is given by

= (80 - 63.5)/2.2 = 7.5

From the z-score table, we have P(z= 7.5) ≈ 1

Therefore the proportion of women that satisfy the given range is between the two given probabilities, that is P(z= 7.5) - P(z= -2.5) or

1 - 0.0062 ≈ 0.994

Therefore the the number of women expected to  meet the criteria is

150 × 0.994 = 149.07 ≈ 149 women out of 150 women meet the criteria.