Question

Suppose that X is normally distributed with mean 105 and standard deviation 16. A. What is the probability that X is greater than 131.4? B. What value of X does only the top 11% exceed?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 105

standard deviation = $\sigma$ = 16

A)

P(x > 131.4) = 1 - P(x < 131.4)

= 1 - P((x - $\mu$ ) / $\sigma$ < (131.4 - 105) / 16)

= 1 - P(z < 1.65)

= 1 - 0.9505 Using standard normal table.

= 0.0495

Probability = 0.0495

B)

The z - distribution of the 11 % is,

P( Z > z ) = 11%

1 - P( Z < z ) = 0.11

P( Z < z) = 1 - 0.11

P( Z < z ) = 0.89

P( Z < 1.23 ) = 0.89

z = 1.23

Using z - score formula,

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

= 1.23 * 16 + 105  

= 124.68

The value of X is 124.68