1. ## normal distribution question

Hi,
I'm trying to find an answer to the following problem. Any help will be greatly appreciated.

I have a normally-distributed variable X with a mean of 1000 and a some variance (let's assume 1).
I choose a range of 970-990 (for the example's sake).
For any X above 990, I receive 20 dollars.
For any X below 970, I receive 0 dollars.
For X between 970 and 990, I receive: X - 970

My question is: what is the function for calculating the mean profit of this example?

Thanks!
Eyal

2. Originally Posted by eyal
Hi,
I'm trying to find an answer to the following problem. Any help will be greatly appreciated.

I have a normally-distributed variable X with a mean of 1000 and a some variance (let's assume 1).
I choose a range of 970-990 (for the example's sake).
For any X above 990, I receive 20 dollars.
For any X below 970, I receive 0 dollars.
For X between 970 and 990, I receive: X - 970

My question is: what is the function for calculating the mean profit of this example?

Thanks!
Eyal
Expected profit = 20 Pr(X > 990) - 970 Pr(970 < X < 990).

3. Originally Posted by mr fantastic
Expected profit = 20 Pr(X > 990) - 970 Pr(970 < X < 990).
Could you please elaborate on this function? I understand its first part, but not the second.

Thanks!

4. Originally Posted by eyal
Could you please elaborate on this function? I understand its first part, but not the second.

Thanks!
1. Calculate Pr(X > 990). Multiply by 20.

2. Calculate Pr(970 < X < 990). Multiply by 970.

Subtract the number found in 2 from the number found in 1. The result is the expected profit.

5. Originally Posted by mr fantastic
1. Calculate Pr(X > 990). Multiply by 20.

2. Calculate Pr(970 < X < 990). Multiply by 970.

Subtract the number found in 2 from the number found in 1. The result is the expected profit.
Got it. You use the standard normal distribution, and then transform it to the given distribution.
What would you do if the SD isn't 1?

6. Originally Posted by eyal
Got it. You use the standard normal distribution, and then transform it to the given distribution.
What would you do if the SD isn't 1?
$\displaystyle Z = \frac{X - \mu}{\sigma}$.

7. There's one thing I still have a problem with.
The Pr(970 < x < 990) is the chance of X falling anywhere within that range.
But the amount of money received is depending on where withing that range X falls: If X = 985 I receive 15 dollars, while if X = 972 I receive 2 dollars.

I don't see how the function you gave relates to that fact.

Thanks again,
Eyal

8. Originally Posted by eyal
There's one thing I still have a problem with.
The Pr(970 < x < 990) is the chance of X falling anywhere within that range.
But the amount of money received is depending on where withing that range X falls: If X = 985 I receive 15 dollars, while if X = 972 I receive 2 dollars.

I don't see how the function you gave relates to that fact.

Thanks again,
Eyal
Sorry, I misread your question. I think the term should be $\displaystyle \int_{x=970}^{990}(x - 970) \Pr(970 \leq X \leq x) \, dx$

where the appropriate expression for $\displaystyle \Pr(970 \leq X \leq x)$ needs to be substituted. The integral will need to be calculated numerically.