MHB Calc Expected Value & Variance of Multivar. Func.

AI Thread Summary
The discussion focuses on calculating the expected value and variance of a multivariable function involving two normally distributed variables, X and Y. The user struggles to derive the probability density functions f(x) and g(y) necessary for applying the properties of expected values. They correctly identify the normal distribution parameters for X, including its mean and variance. The response provides the correct form of the probability density function for X and suggests a similar approach for Y. Understanding these functions is essential for solving the problem effectively.
TheFallen018
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Hey, I've got this problem that I've been trying to crack for a while. I can't find any info for multi-variable expected values in my textbook, and I couldn't find a lot of stuff that made sense to me online. Here's the problem.

View attachment 8906
Find $E(C)$
Find $Var(C)$

I tried to get the limits from the normal distributions that were given. If I was doing it right, I had
$90\leq X \leq 130 $
$1.9\leq Y \leq 2.9$
for X and Y.

I think my main problem is that I'm not sure how to get $f(x)$ and $f(y)$ so that I can use the property
$E(aX+bY) = aE(X)+bE(Y)$
and
$E(x) = \int_{n}^{m}x*f(x)dx$
where $n \leq f(x) \leq m$.

I think I might be able to figure it out once I can work out what the functions should be, but I'm a little stuck here. Any help would be awesome. Thanks.
 

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You are told that X is normally distributed with mean 110 and standard deviation 20 (I am assuming that the second number, "$20^2$" is the variance that is the square of the standard deviation.) Whoever gave you this problem clearly expects you to know what the "normal distribution" is! The "f(x)" you want is $\frac{1}{\sqrt{2(\pi)(20^2)}}e^{-\frac{(x- 110)^2}{2(20^2)}} = \frac{1}{20\sqrt{2\pi}}e^{\frac{(x- 110)^2}{800}}$ and similarly for g(y). (You have "f(x)" and "f(y)" but they are not the same function!)

Look at https://en.wikipedia.org/wiki/Normal_distribution
 
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