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Math Help - normal distribution question

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by eyal View Post
    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).
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    Quote Originally Posted by mr fantastic View Post
    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!
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  4. #4
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    Quote Originally Posted by eyal View Post
    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.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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?
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    Quote Originally Posted by eyal View Post
    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?
    Z = \frac{X - \mu}{\sigma}.
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    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
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  8. #8
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    Quote Originally Posted by eyal View Post
    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 \int_{x=970}^{990}(x - 970) \Pr(970 \leq X \leq x) \, dx

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