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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%

User Trylks
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1 Answer

3 votes

Answer:

A. 16%

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 = (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 = (X - \mu)/(\sigma)


Z = (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%

User Dnalow
by
8.6k points

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