Question

In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67 inches and a standard deviation of 3.0 inches. A study participant is randomly selected. Complete parts​ a and b below.a.) Find the probability that the participant is less than 64.5 inches?b.) Find the probability that the participant is more than 68.25 inches?

Answer 1

Answer:

a) 20.33% probability that the participant is less than 64.5 inches.

b) 33.72% probability that the participant is more than 68.25 inches

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:

$\mu = 67, \sigma = 3$

a.) Find the probability that the participant is less than 64.5 inches?

This is the pvalue of Z when X = 64.5.

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

$Z = \frac{64.5 - 67}{3}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033

20.33% probability that the participant is less than 64.5 inches.

b.) Find the probability that the participant is more than 68.25 inches?

This is 1 subtracted by the pvalue of Z when X = 68.25.

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

$Z = \frac{68.25 - 67}{3}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628

1 - 0.6628 = 0.3372

33.72% probability that the participant is more than 68.25 inches