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Oct 7 2013 03:02pm
Okay, so the problem gives us mean=58 and SD=1.5, normally distributed.
The first 5 parts of the problem ask for various probabilities, which I'm fine with.

The last part of the problem says to find the expected value, E(W^2 + 2W)
I really have no idea how to do this, when I look through my notes, I can't find any mention of expected values in normal distributions (considering they would just be the mean...)

Any help is appreciated, thanks.
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Oct 8 2013 08:27pm
Anyone?
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Oct 8 2013 08:37pm
What's W? Gaussians are "Linear", which means that any linear combinations of Gaussian RVs will still be Normally distributed.
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Oct 8 2013 08:48pm
Quote (zackill4 @ Oct 8 2013 09:37pm)
What's W? Gaussians are "Linear", which means that any linear combinations of Gaussian RVs will still be Normally distributed.


W is just a variable, in this case it represents WEIGHT, which is what's normally distributed (weight of pizzas in the problem).

So would E(W^2 + 2W) just be the same as E(W), since it's originally Normally Distributed?
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Oct 8 2013 08:50pm
Quote (furbyjs @ Oct 8 2013 09:48pm)
W is just a variable, in this case it represents WEIGHT, which is what's normally distributed (weight of pizzas in the problem).

So would E(W^2 + 2W) just be the same as E(W), since it's originally Normally Distributed?


No actually. I read your post wrongly but that's not correct even if that was a linear combo
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Oct 8 2013 08:52pm
Quote (furbyjs @ Oct 8 2013 09:48pm)
W is just a variable, in this case it represents WEIGHT, which is what's normally distributed (weight of pizzas in the problem).

So would E(W^2 + 2W) just be the same as E(W), since it's originally Normally Distributed?


So we know expectation is a linear operator right?

So first split this into

E(W^2) + E(2W)

also by linearity of expectation

=E(W^2) + 2E(W)

now we know that variance is E(W^2) - E(W)^2 right?

so this is the same as


= variance(W) + E(W)^2 + 2E(W)
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Oct 9 2013 06:16am
Quote (zackill4 @ Oct 8 2013 09:52pm)
So we know expectation is a linear operator right?

So first split this into

E(W^2) + E(2W)

also by linearity of expectation

=E(W^2) + 2E(W)

now we know that variance is  E(W^2) - E(W)^2 right?

so this is the same as


= variance(W) + E(W)^2 + 2E(W)


I'm not familiar with all these "linearity" rules, but it makes sense.
Thanks.
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