Question

Assuming that the heights of college women are normally distributed with mean 64 inches and standard deviation 1.5 inches, what percentage of women are between 65.5 inches and 68.5 inches?

Answer 1

Answer:

15.74% of women are between 65.5 inches and 68.5 inches.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 64, \sigma = 1.5$

What percentage of women are between 65.5 inches and 68.5 inches?

This percentage is the pvalue of Z when X = 68.5 subtracted by the pvalue of Z when X = 65.5.

X = 68.5

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

$Z = \frac{68.5 - 64}{1.5}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

X = 65.5

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

$Z = \frac{65.5 - 64}{1.5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

So 0.9987 - 0.8413 = 0.1574 = 15.74% of women are between 65.5 inches and 68.5 inches.