Question

2) (10 pts) A height study(X) is conducted for a group of normally distributed employees at a small company. The average height of the employees is 66 inches and the standard deviation is 4 inches.

1. (4 pts) Find P(X >= 67.5 inches).

2. (6 pts) Find P(61 inches <= X <= 64.5 inches).

Answer 1

Solution :

Given that,

mean = $\mu$ = 66

standard deviation = $\sigma$ = 4

a ) P ( x > 67.5 )

= 1 - P (x < 67.5 )

= 1 - P ( x -  $\mu$/ $\sigma$) < ( 67.5 - 66 / 4)

= 1 - P ( z < 1.5 / 4 )

= 1 - P ( z < 0.375)

Using z table

= 1 - 0.6462

= 0.3538

Probability = 0.3538

b ) P ( 61 < x < 64.5 )

P ( 61 - 66 / 4) < ( x -  $\mu$/ $\sigma$ ) < ( 64.5 - 66 / 4)

P ( - 5 / 4 < z < -1.5 / 4 )

P (- 1.25 < z < - 0.375 )

P (z < - 0.375 ) - p ( z < - 1.25 )

Using z table

= 0.3538 - 0.1056

= 0.2482

Probability = 0.2482