Question

The height of an adult male in the United States is approximately normally distributed with a mean of 69.3 inches and a standard deviation of 2.8 inches. Assume that such an individual is selected at random. What is the probability that his height will be less than 66.7 inches?

Answer 1

Answer:

17.62% probability that his height will be less than 66.7 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:

$\mu = 69.3, \sigma = 2.8$

What is the probability that his height will be less than 66.7 inches?

This is the pvalue of Z when X = 66.7. So

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

$Z = \frac{66.7 - 69.3}{2.8}$

$Z = -0.93$

$Z = -0.93$ has a pvalue of 0.1762

17.62% probability that his height will be less than 66.7 inches