# Homework Help: Expected Monthly Profit

Tags:
1. Jun 16, 2017

### Of Mike and Men

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

2. Jun 16, 2017

### Orodruin

Staff Emeritus
You are missing one possible outcome ...

Edit: Not that that actually fixes the problem ...

3. Jun 16, 2017

### StoneTemplePython

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.

4. Jun 16, 2017

### StoneTemplePython

I'm getting an answer a touch higher than the 3.8 (thousand) mentioned as the official one, though.