Question

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The U.S. Army requires that the heights of women be between 58 and 80 inches. If 200 women want to enlist in the U.S. Army, how many would you expect to meet the height requirements? About 197 women

Answer 1

Solution :

Given that ,

mean = $\mu$ = 63.6

standard deviation = $\sigma$ = 2.5

a) n = 200

P( 58 < x < 80 ) = P[(58 - 63.6 )/ 2.5 ) < (x - $\mu$ ) /$\sigma$  < ( 80 - 63.6 ) / 2.5 ) ]

= P( -2.24 < z < 6.56 )

= P(z < 6.56) - P(z < -2.24 )

Using z table,

= 1 - 0.0125

= 0.9875

= 0.9875 * 200 = 197.5

= 197 women