Suppose the heights (in inches) of men ages 20-29) ) in the United States are normally distributed with a mean of 693 inches and a standard deviation of 2.92 inches

(a) 0.13567

(b) 0.68023

Complete Problem Statement:

Suppose the heights (in inches) of men ages (20-29) in the United States are normally distributed with a mean of 69.3 inches and a standard deviation of 2.92 inches. A study participant is randomly selected

Find the probability that his height is :

(a) Less than 66 inches

(b) Between 66 inches and 72 inches

Step-by-step explanation:

To find the probability, we will need to calculate the Z-Score.

The formula for which is :

z = (X-μ) / σ

X= raw score

μ= mean

σ = Standard Deviation

(a) Probability that the height of participant is less than 66 inches:

First we find value of z :

z =

Use this to look up at probability table, we get:

Probability (at z equals -1.1 ) = 0.13567

(b) Probability that the height of participant is between 66 inches and 72 inches

First we find value of z for X=72 inches :

z =

Probability ( at z equals 0.9) = 0.8159

We already have value of z for X=66 inches , which is z = -1.1 with Probability (at z equals -1.1 ) = 0.13567

Since we are asked for probability between 66 inches and 72 inches, we subtract the two probabilities , such that:

Probability (at h=72) - Probability (at h=66) = 0.8159-0.13567=0.68023

Probability that height of participant is between 66 inches and 72 inches =0.68023

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