Question

Suppose that X is normally distributed with mean 100 and standard deviation 25.

A. What proportion of X is greater than 142?
(Provide answer rounded to four decimal places.)
Proportion =  

B. What value of X does only the top 15% exceed?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 100

standard deviation = $\sigma$ = 25

A.

P(x > 142) = 1 - P(x < 142)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (142 - 100) / 25)

= 1 - P(z < 1.68)

= 1 - 0.9535

= 0.0465

proportion = 0.0465

B.

Using standard normal table ,

P(Z > z) = 15%

1 - P(Z < z) = 0.15

P(Z < z) = 1 - 0.15

P(Z < 1.04) = 0.85

z = 1.04

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.04 * 25 + 100 = 126