Question

The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches. What are the z-scores for a woman 6 feet tall and a man 5'10" tall? (You may round your answers to two decimal places) z-scores for a woman 6 feet tall: z-scores for a man 5'10" tall:

Answer 1

Answer: The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.

Step-by-step explanation:

Let x and y area the random variable that represents the heights of women and men.

Given : The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches.

i.e. $\mu_1 = 64$   $\sigma_1=2.7$

Since , $z=\dfrac{x-\mu}{\sigma}$

Then, z-score corresponds to  a woman 6 feet tall (i.e. x=72 inches).

[∵  1 foot = 12 inches , 6 feet = 6(12)=72 inches]

$z=\dfrac{72-64}{2.7}=2.96296296\approx2.96$

Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.

i.e. $\mu_2 = 69.3$   $\sigma_2=2.8$

Then, z-score corresponds to a man 5'10" tall (i.e. y =70 inches).

[∵  1 foot = 12 inches , 5 feet 10 inches= 5(12)+10=70 inches]

$z=\dfrac{70-69.3}{2.8}=0.25$

∴ The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.