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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.

Respuesta :

Answer: 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%

Cricetus

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

According to the question,

Mean,

  • [tex]\mu = 69[/tex]

Standard deviation,

  • [tex]\sigma = 2.8[/tex]

As we know,

→ [tex]z = \frac{x- \mu}{\sigma}[/tex]

Now,

The percentage will be:

→ [tex]P(64< x< 78) = P(\frac{64-69}{2.8} < z < \frac{78-69}{2.8} )[/tex]

                            [tex]= P(-1.79< z< 3.21)[/tex]

                            [tex]= P(z< 3.21) - P(z< -1.79)[/tex]

                            [tex]= 0.9993-0.0367[/tex]

                            [tex]= 0.9626[/tex]

or,

                            [tex]= 96.26[/tex] (%)        

Thus the above response is right.

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https://brainly.com/question/15549232

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