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Math Help - 2 probability theory questions!

  1. #1
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    2 probability theory questions!

    1)A small manufacturer sells 1 car each month with a probability of 0.3; it sells 2 cars each month with a probability of 0.1 and never sells more than 2 cars per month. Let X be the number of cars sold each month

    -Find the mean and the variance of X
    -If the monthly profit is 2(X^2)+3(X)+1, what is the expected montly profit?

    2) In a game, Eric gives Heather three well-balanced quarters to flip. Heather will keep all the quarters that land in heads, and will give back the ones that land in teails. Yet, if all of them land in tails, Heather will pay two bucks to Eric. WHat is the expected value and the variance of Heather's net gain?
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    Quote Originally Posted by Andreamet View Post
    1)A small manufacturer sells 1 car each month with a probability of 0.3; it sells 2 cars each month with a probability of 0.1 and never sells more than 2 cars per month. Let X be the number of cars sold each month

    -Find the mean and the variance of X
    -If the monthly profit is 2(X^2)+3(X)+1, what is the expected montly profit?

    [snip]
    Note that Pr(X = 0) = 0.6.

    E(X) = (0)(0.6) + (1)(0.3) + (2)(0.1) = 0.5

    E(X^2) = (0^2)(0.6) + (1^2)(0.3) + (2^2)(0.1) = 0.7

    Variance = E(X^2) - [E(X)]^2 = 0.45


    Let Y = 2X^2 + 3X + 1.

    Then E(Y) = 2 E(X^2) + 3 E(X) + 1 = ......
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  3. #3
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    Quote Originally Posted by Andreamet View Post
    [snip]
    2) In a game, Eric gives Heather three well-balanced quarters to flip. Heather will keep all the quarters that land in heads, and will give back the ones that land in teails. Yet, if all of them land in tails, Heather will pay two bucks to Eric. WHat is the expected value and the variance of Heather's net gain?
    Let X = number of heads tossed.

    Pr(X = 0) = 1/8. Gets -$2.00, that is loses $2.
    Pr(X = 1) = 3/8. Gets $0.25.
    Pr(X = 2) = 3/8. Gets $0.50.
    Pr(X = 3) = 1/8. Gets $0.75.

    E(X) = (-2.00)(1/8) + (0.25)(3/8) + (0.50)(3/8) + (0.75)(1/8) = ....

    E(X^2) = ([-2.00]^2)(1/8) + (0.25^2)(3/8) + (0.50^2)(3/8) + (0.75^2)(1/8) = ....

    Variance = E(X^2) - [E(X)]^2 = ......
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    thanks!l
    Last edited by Andreamet; March 8th 2008 at 02:54 PM.
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    Quote Originally Posted by Andreamet View Post
    wouldn't the variance be 0.7 instead of 0.45?



    woul
    No. The variance is NOT the value of E(X^2).

    Variance = E(X^2) - [E(X)]^2 = 0.7 - (0.5)^2 = 0.7 - 0.25 = 0.45.
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