Expected Monthly Profit for a Small Manufacturing Firm

Of Mike and Men
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Homework Statement


A small manufacturing firm sells 1 machine per month with 0.3 probability; it sells 2 machines per month with 0.1 probability; it never sells more than 2 machines per month. If X represents the number of machines sold per month and the monthly profit is 2X2 + 3X + 1 (in thousands of dollars), find the expected monthly profit.

Homework Equations

The Attempt at a Solution


E(2X2 + 3X + 1) = ∑(2X2 + 3X + 1)f(x), x = 1, 2
= (2+3+1)(.3) + (8+6+1)(.1)
= 1.8 + 1.5
= 3.3
3.3 * 1000 = $3,300

The answer in the back is $3,800.

If I take my 3.3 and add the mean of E(X) = .3(1) + 2(.1) = 0.5, I get 3.8 * 1000 = $3,800. I'm not sure if this is coincidence or actually how you solve the problem. If it's how you solve the problem, I don't understand why you add the mean of X. If it's not the way you solve it, I'm not sure what to do and would like some hints (but not a solution -- if possible) as to where to go.

Thanks.
 
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You are missing one possible outcome ...

Edit: Not that that actually fixes the problem ...
 
Of Mike and Men said:

Homework Statement


A small manufacturing firm sells 1 machine per month with 0.3 probability; it sells 2 machines per month with 0.1 probability; it never sells more than 2 machines per month. If X represents the number of machines sold per month and the monthly profit is 2X2 + 3X + 1 (in thousands of dollars), find the expected monthly profit.

Homework Equations

The Attempt at a Solution


E(2X2 + 3X + 1) = ∑(2X2 + 3X + 1)f(x), x = 1, 2
...the mean of E(X) = .3(1) + 2(.1) = 0.5

So profit ##= 2X^2 + 3X + 1##

##E[profit] = E[2X^2 + 3X + 1] = E[2X^2] + E[3X] + E[1]##

by linearity of expectations.

This last line should be simplified a bit -- how would you do it? You correctly calculated ##E[X]##. What is ##E[X^2]##? Or if you prefer, what is the Variance -- you can recover ##E[X^2]## from that.
 
I'm getting an answer a touch higher than the 3.8 (thousand) mentioned as the official one, though.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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