Question

The weight of a 5th grader is normally distributed with a mean of 82 pounds and a variance of 80 pounds^2. Let weight, in pounds, be represented by random variable X.

P(76<X<x2)= 0.1330. Find x2

Answer 1

Given,

= 82, = sqrt( 80)

We convert this to standard normal as

P( X < x) = P( Z < x - / )

So,

P( 76 < X < x2) = 0.1330

P( X < x2) - P( X < 76) = 0.1330

P( Z < x2 - 82 / sqrt(80) ) - P( Z < 76 - 82 / sqrt( 80) ) = 0.1330

P( Z < x2 - 82 / sqrt(80) ) - P( Z < -0.6708) = 0.1330

P( Z < x2 - 82 / sqrt(80) ) - 0.2512 = 0.1330

P( Z < x2 - 82 / sqrt(80) ) = 0.3842

From the Z table, z-score for the probability of 0.3842 is -0.2945

x2 - 82 / sqrt(80) = -0.2945

Solve for x2

x2 = 79.366