Question

assume the random variable X is normally distributed with a mean 80 in standard deviation 9.4

find the P (x>72)

find the P (95<x<105)

find x so that the area above x is .80

Answer 1

The distribution given here is:

a) The probability here is computed as:

P(X > 72)

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.8026 is the required probability here.

b) P(95 < X < 105)

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.0514 is the required probability here.

c) From standard normal tables, we have:

P(Z > -0.8416) = 0.8

Therefore X = Mean - 0.8416*Std Dev = 80 - 0.8416*9.4 = 72.09

Therefore 72.09 is the required value of x here.