Question

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%

Answer 1

Answer:

Step-by-step explanation:

The northern circumference of the chemical surrounds the formulae x=√yz-6 meaning that the equator of the hairline leading to a clip wire inserting the biggest ever change of reality leading to space pockets that enter another dimension..... Idk I'm just waffling

Answer 2

Answer:

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$\%$