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Mathematics 17 Online
OpenStudy (anonymous):

The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula C = 100 + 40Y + 3Y^2 relates to the cost C completing this operation to the square of the time to completion. Find the mean and variance of C.

OpenStudy (anonymous):

I have a mean of 1100. I need help with the Variance. I get so far and then I need and mgf, which i'm not sure how to calculate

OpenStudy (anonymous):

@ganeshie8 - I was told maybe you can help me?

OpenStudy (anonymous):

How did you get the mean?

OpenStudy (anonymous):

E(100+40Y+3Y^2) = 100 + 40(E(Y)) +3(E(Y^2))

OpenStudy (anonymous):

E(Y^2) = sd^2+mean^2, being exponential, mean is 10 and sd is 10^2 so E(Y^2) = 100+100 = 200. Insert into above equation and it comes to 1100

OpenStudy (anonymous):

There are similar properties with variance as well.

OpenStudy (anonymous):

The properties were unhelpful as far as I found anyways. :(

OpenStudy (anonymous):

\[ \text{Var}(100 + 40Y + 3Y^2)=\text{Var}(100)+\text{Var}(40Y )+\text{Var}(3Y^2) \]

OpenStudy (anonymous):

This works, right?

OpenStudy (anonymous):

no.

OpenStudy (anonymous):

V(40Y+3Y^2) is what it would turn into. the Variance properties are different.

OpenStudy (anonymous):

Okay, let \(X=Y^2\)\[ \text{Var}(40Y+3X) =40^2 \text{Var}(Y) +3^2\text{Var}(X) +2(40)(3)\text{Cov}(Y,X) \]

OpenStudy (anonymous):

I don't think that's very helpful either :/ I got it down to V(C) = E(C^2)-E(C)^2 by a property.

OpenStudy (anonymous):

from my understanding I'd just need mgf's from there to calculate. My prof recommended this method.

OpenStudy (anonymous):

You can do that, by there isn't anything tricky about my method.

OpenStudy (anonymous):

I just don't know how to work your method either lol

OpenStudy (anonymous):

The exponential distribution is: \[ \lambda e^{-\lambda x} \]And the mean is \(1/\lambda\).

OpenStudy (anonymous):

I'm not exactly certain how that helps me?

OpenStudy (anonymous):

Because if you square it you can notice interesting things. \[ \lambda^2e^{-2\lambda x} = \frac \lambda 2(2\lambda e^{-2\lambda x)} =\frac \lambda 2(\lambda' e^{-\lambda' x}) \]

OpenStudy (anonymous):

I'm sorry (my brain has died) please explain this to me slowly and step by step. How does that help me? What do I do with it?

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