Question

In 2014, the CDC estimated that the mean height for adult women in the U.S. was 64 inches with a standard deviation of 4 inches. Suppose X, height in inches of adult women, follows a normal distribution. Which of the following gives the probability that a randomly selected woman has a height of greater than 68 inches?
A. 16%B. 34%C. 68%D. 84%E. 95%F. 99.7%

Answer 1

Answer:

A. 16%

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:

$\mu = 64, \sigma = 4$

Which of the following gives the probability that a randomly selected woman has a height of greater than 68 inches?

This is 1 subtracted by the pvalue of Z when X = 68. So

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

$Z = \frac{68 - 64}{4}$

$Z = 1$

$Z = 1$ has a pvalue of 0.84.

1 - 0.84 = 0.16

So the correct answer is:

A. 16%