Assume that the heights of American men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. The U. S. Marine Corps requires that men have heights between 64 and 78 inches. Find the percent of men meeting these height requirements.

"96.26%" would be the percent of men meeting these height requirements.

According to the question,

Mean,

Standard deviation,

As we know,

→

Now,

The percentage will be:

→

or,

                            (%)        

Thus the above response is right.

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96.2%

Step-by-step explanation:

Assume that the heights of American men are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = heights of American men.

µ = mean height

σ = standard deviation

From the information given,

µ = 69.0 inches

σ = 2.8 inches

the probability of men that have heights between 64 and 78 inches is expressed as

P(64 ≤ x ≤ 78)

For x = 64,

z = (64 - 69)/2.8 = - 1.79

Looking at the normal distribution table, the probability corresponding to the z score is 0.037

For x = 78,

z = (78 - 69)/2.8 = 3.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.999

Therefore,

P(64 ≤ x ≤ 78) = 0.999 - 0.037 = 0.962

Therefore, the percent of men meeting these height requirements is

0.962 × 100 = 96.2%

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