Question

Suppose that X is normally distributed with mean 90 and standard deviation 27.

A. What is the probability that X is greater than 133.2? Probability =

B. What value of X does only the top 12% exceed? X =

Answer 1

Solution :

Given ,

mean = = 90

standard deviation = = 27

(A)P(x > 133.2) = 1 - P(x< 133.2)

= 1 - P[ X - / / (133.2-90) /27 ]

= 1 - P(z <1.6 )

Using z table

= 1 - 0.9452

= 0.0548

probability= 0.0548

(B)

Using standard normal table,

P(Z > z) = 12%

= 1 - P(Z < z) = 0.12

= P(Z < z ) = 1 - 0.12

= P(Z < z ) = 0.88

= P(Z < 1.18) = 0.88

z = 1.18 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x=1.18 *27+90

x= 121.86