Question

Suppose that ? is normally distributed with mean 115 and standard deviation 15.

A. What is the probability that ? is greater than 134.8?
Probability =  

B. What value of ?X does only the top 14% exceed?
? =  

Answer 1

Solution :

Given that ,

mean = = 115

standard deviation = = 15

A.

P(x > 134.8) = 1 - P(x < 134.8)

= 1 - P[(x - ) / < (134.8 - 115) / 15)

= 1 - P(z < 1.32)

= 1 - 0.9066

= 0.0934

Probability = 0.0934

B.

Using standard normal table ,

P(Z > z) = 14%

1 - P(Z < z) = 0.14

P(Z < z) = 1 - 0.14

P(Z < 1.08) = 0.86

z =

Using z-score formula,

x = z * +

x = 1.08 * 15 + 115 = 131.2