Mathematics: 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. A) 31.12% B) 99.93% C) 3.67% D) 96.26%

From -5 to 24 is an additional 29 F. I’m not sure what the options for the expressions are....

Assume that the heights of American men are normally

esdancer3494

22.01.2022

The percent of men heights between 64 and 78 inches = 96.22 $\%$

Step-by-step explanation:

Given -

Mean height $\boldsymbol{(\nu)}$ = 69.0

Standard deviation $\boldsymbol{(\sigma )}$ = 2.8

Let X be the height of american man

The percent of men heights between 64 and 78 inches =

$\mathbf{P(64\leq X \leq 78)}$ = $P(\frac{ 64 - 69}{2.8}\leq \frac{X - \nu }{\sigma}\leq \frac{ 78 - 69}{2.8})$

= $P(\frac{ - 5}{2.8}\leq Z \leq \frac{ 9}{2.8})$ Put [ $\mathbf{Z = \frac{X - \nu }{\sigma}}$ ]

= $P(- 1.785 \leq Z \leq 3.214)$ Using z table

= 0.9993 - .0370

= .9622

= 96.22 $\%$

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