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

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

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

Explanation:

Given -

Mean height
`\boldsymbol{(\\u)}` = 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(( 64 - 69)/(2.8)\leq (X - \\u )/(\sigma)\leq ( 78 - 69)/(2.8))`

=
`P(( - 5)/(2.8)\leq Z \leq ( 9)/(2.8))` Put [
`\mathbf{Z = (X - \\u )/(\sigma)}`]

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

= 0.9993 - .0370

= .9622

= 96.22
`\%`