MHB Find μ and σ^2 of Y when X~N(2,4)

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If X follows a normal distribution X ~ N(2,4), then the transformed variable Y = -X - 1 is also normally distributed as Y ~ N(μ,σ^2). The expected value E(Y) is calculated as E(Y) = -E(X) - 1, resulting in μ = -3. The variance of Y remains the same as that of X, which is σ^2 = 4, since the variance is unaffected by adding or subtracting a constant. Therefore, the parameters for Y are μ = -3 and σ^2 = 4.
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If X ~ N(2,4) then Y = -X-1 is also a normal random variable such that Y ~ N(μ,σ^2).
Find μ and σ^2.

I know that E(X) = 2 and Var(X) = 4.

E(Y) = -E(X) - 1
E(Y) = -2 - 1 = -3
So I found μ, but I'm not sure how to find the variance. Help?
 
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dlee said:
If X ~ N(2,4) then Y = -X-1 is also a normal random variable such that Y ~ N(μ,σ^2).
Find μ and σ^2.

I know that E(X) = 2 and Var(X) = 4.

E(Y) = -E(X) - 1
E(Y) = -2 - 1 = -3
So I found μ, but I'm not sure how to find the variance. Help?

Welcome to MHB, dlee! :)

There are a couple of basic properties for expectations and variances as you can see here.

In particular $\sigma^2(X + a)=\sigma^2(X)$ and $\sigma^2(aX)=a^2\sigma^2(X)$, where $a$ is some arbitrary constant.
 
AH the variance is 4! Thank you!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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