# The Sum and Difference of Two Normal Distributions

Questions in the attached file.

Â© BrainMass Inc. brainmass.com October 7, 2022, 7:20 pm ad1c9bdddfhttps://brainmass.com/statistics/normal-distribution/sum-difference-normal-distributions-177502

## SOLUTION This solution is **FREE** courtesy of BrainMass!

The solution file is attached.

Please explain your work and reasoning:

Let X be a random variable, normally distributed with mean 100 and variance 64.

Mean = Î¼ = 100,

Variance = Ïƒ2 = 64,

Standard Deviation = Ïƒ = 8.

1. If there are two random variables, X1 and X2, independent of each other, and each distributed as X as given above, then find the probability that X1+X2 is less than or equal to 220.

X = X1 + X2 is normally distributed with Î¼ = Î¼1 + Î¼2 and variance Ïƒ^2 = Ïƒ1^2 + Ïƒ2^2

Î¼ = 100 + 100 = 200, Ïƒ^2 = 64 + 64 = 128

Ïƒ = âˆš128 = 11.314

z = (X - Î¼)/Ïƒ = (220 - 200)/11.314 = 1.77

P(X â‰¤ 200) = Area under the Standard Normal Curve to the left of z = 1.77, which is 0.961

2. For the above problem, find L such that X1+X2 is greater than or equal to L is 0.10.

P(X) â‰¥ 0.10 corresponds to a z score of 1.645

SE = 1.645 * Ïƒ = 1.645 * 11.314 = 18.61

L = [Î¼ - SE, Î¼ + SE] = [200 - 18.61, 200 + 18.61] = [181.39, 218.61]

3. If another random variable X3 is also distributed as X as given above and is independent of X1 and X2, what is the expected value of X1+X2+X3? What is the variance of (X1+X2+X3)?

Expected value of X1 + X2 + X3 = Î¼1 + Î¼2 + Î¼3 = 300, Variance of X1 + X2 + X3 = Ïƒ1^2 + Ïƒ2^2 + Ïƒ3^2

= 64 + 64 + 64 = 192.

https://brainmass.com/statistics/normal-distribution/sum-difference-normal-distributions-177502