Question

in a certain study, womens heights are found to be approximately normally distributed with a mean of 62 inches and a standard deviation of 2.5 inches.
a. what would be the z - score for a woman who is 5 feet 8 inches tall?
b. what percentage of women is she taller than?
click the icon to view a table of z - scores and percentiles.
a. the z - score for the womans height is
(round to one decimal place as needed.)

Explanation:

Step1: Convert height to inches

Since 1 foot = 12 inches, 5 feet 8 inches = 5*12 + 8 = 68 inches.

Step2: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean and $\sigma$ is the standard deviation. Here, $\mu = 62$, $\sigma=2.5$ and $x = 68$.

Step3: Calculate z - score

$z=\frac{68 - 62}{2.5}=\frac{6}{2.5}=2.4$

Step4: Find the percentile

Looking up the z - score of 2.4 in the standard normal distribution table, the area to the left of $z = 2.4$ is approximately 0.9918. So the percentage of women she is taller than is $(1 - 0.9918)\times100\%=0.82\%$

Answer:

a. 2.4
b. 0.82%