Question

Tall Club International has a requirement that women must be at least 70 inches tall. Given that women have normally distributed heights with a mean of 63.7 inches and a standard deviation of 2.9 inches, find the percentage of women who satisfy that height requirement. (Round to the nearest ten-thousandth.)

Answer 1

Answer:

0.0150 = 1.50% of women satisfy that height requirement.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$\mu = 63.7, \sigma = 2.9$

Find the percentage of women who satisfy that height requirement.

This is 1 subtracted by the pvalue of Z when X = 70. So

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

$Z = \frac{70 - 63.7}{2.9}$

$Z = 2.17$

$Z = 2.17$ has a pvalue of 0.9850

1 - 0.9850 = 0.0150

0.0150 = 1.50% of women who satisfy that height requirement.