Question

Suppose that X is normally distributed with mean 115 and standard deviation 20. A. What is the probability that X is greater than 152? Probability = B. What value of X does only the top 11% exceed? X =

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 115

Standard deviation = 20

A. P(X is greater than 152) = 1 - P(X < 152)

= 1 - P(Z < (152 - 115)/20)

= 1 - P(Z < 1.85)

= 1 - 0.9678

= 0.0322

B. Let E be the value exceeded by only the top 11%

P(X > E) = 0.11

P(X < E) = 1 - 0.11 = 0.89

P(Z < (E - 115)/20) = 0.89

Take the value of Z corresponding to probability of 0.89 from standard normal distribution table.

(E - 115)/20 = 1.23

E = 139.6