Question

1. Given that the random variable X is normally distributed with a mean of 30 and a standard deviation of 5, determine the following: P(32.2 <X).

2. The pdf(probability distribution function) of a random variable X is given by f(x) = 3x^2 with support 0<x<1. Determine P(0.4 < X < 0.6).

Answer 1

Solution:

Question 1)

Given : X ~ Normal( )

Find:

P( 32.2 < X ) = ........?

That is:
P( X > 32.2 ) = ...........?

Find z score for x = 32.2

Thus we get:

P( X > 32.2 ) = P( Z > 0.44 )

P( X > 32.2 ) = 1 - P( Z < 0.44 )

Look in z table for z = 0.4 and 0.04 and find corresponding area.

P( Z< 0.44 )= 0.6700

P( X > 32.2 ) = 1 - P( Z < 0.44 )

P( X > 32.2 ) = 1 - 0.6700

P( X > 32.2 ) = 0.33

Question 2)

f(x) = 3x^2 , 0 < x < 1

Find:
P(0.4 < X < 0.6).