Question

What proportion of US women have a height greater than 69.5 inches?

Answer 1

Using the Normal distribution, it is found that 0.0359 = 3.59% of US women have a height greater than 69.5 inches.

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

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

• It 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, which is the percentile of X.

US women’s heights are normally distributed with mean 65 inches and standard deviation 2.5  inches, hence $\mu = 65, \sigma = 2.5$.

The proportion of US women that have a height greater than 69.5 inches is 1 subtracted by the p-value of Z when X = 69.5, hence:

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

$Z = \frac{69.5 - 65}{2.5}$

$Z = 1.8$

$Z = 1.8$ has a p-value of 0.9641.

1 - 0.9641 = 0.0359

0.0359 = 3.59% of US women have a height greater than 69.5 inches.

You can learn more about the Normal distribution at brainly.com/question/24663213