Question

The heights of adult women are approximately normally distributed about a mean of 65 inches, with a standard deviation of 2 inches. If Rachael is at the 99th percentile in height for adult women, then her height, in inches, is closest to:a) 60 b) 62 c) 69 d) 70e) 74

Answer 1

Answer:

Her height, in inches, is closest to 70.

Step-by-step explanation:

This can be solved by the the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$, the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

Each z-score value has an equivalent p-value, that represents the percentile that the value X is:

In this problem, we have that:

$\mu = 65, \sigma = 2$

A z-score of 2.33 has a p-value of 0.9901. This means that a height with a z-score of 2.33 is in the 99th percentile.

So, what is the value of X when $Z = 2.33$

$Z = \frac{X - \mu}{\sigma}$

$2.33 = \frac{X - 65}{2}$

$X - 65 = 4.66$

$X = 69.66$

Her height, in inches, is closest to 70.