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  • Few probability questions

    I'm having trouble with a couple of problems.

    Let a chip be taken at random from a bowl that contains 6 white chips, 3 red chips, and 1 blue chip. Let the random variable X = 1 if the outcome is a white chip; let X = 5 if the outcome is a red chip; and let X = 10 if the outcome is a blue chip.
    a. Find the p.m.f. of X
    b. Find the mean and standard deviation of X.

    For a I need a function where you can plug in X=1 and get 6/10, X=5 and get 3/10, etc..

    Once I have that I can find the mean and standard deviation. I've been trying to figure out that equation.
    I have (x^2+5)/10x
    It gives the correct x and y values for the 1st 2 x's but not the 3rd.

    Also


    Suppose X is a random variable such that E(X+4)=10 and E((X+4)^2)=116

    How do I find the mean and variance?
    Any help would greatly be appreciated.

    Thanks
    Last edited by sebastianzx6r; October 31 2005, 07:10 PM. Reason: typos

  • #2
    Originally posted by sebastianzx6r
    I'm having trouble with a couple of problems.

    Let a chip be taken at random from a bowl that contains 6 white chips, 3 red chips, and 1 blue chip. Let the random variable X = 1 if the outcome is a white chip; let X = 5 if the outcome is a red chip; and let X = 10 if the outcome is a blue chip.
    a. Find the p.m.f. of X
    b. Find the mean and standard deviation of X.

    For a I need a function where you can plug in X=1 and get 6/10, X=5 and get 3/10, etc..

    Once I have that I can find the mean and standard deviation. I've been trying to figure out that equation.
    I have (x^2+5)/10x
    It gives the correct x and y values for the 1st 2 x's but not the 3rd.

    Also


    Suppose X is a random variable such that E(X+4)=10 and E((X+4)^2)=116

    How do I find the mean and variance?
    Any help would greatly be appreciated.

    Thanks
    Why don't we try:
    P(X=x) = 6/10, x = 1
    3/10, x = 5
    1/10, x = 10

    Also, the fact that it is a probability mass function of a discrete (as opposed to a continuous) distribution implies that over all x, the sum of the probabilities must equal 1. This certainly is true for our P(X).

    For the second question, since the expected value is a linear operator and the expected value of a constant is the constant itself, E(X+4)=E(X)+4=10-->E(X)=6.

    Also, E((X+4)^2)=E(X^2+8X+16)=E(X^2)+E(8X)+E(16)
    =E(X^2)+8*E(X)+16=E(X^2)+8*6+16=116-->E(X^2)=52
    And, Var(X)=E(X^2)-E(X)^2=52-6^2=16.

    Help?
    act justly. walk humbly. .

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    • #3
      I'd really appreciate if you don't make duplicate threads next time. You can post anywhere and I will find it. I see I just wasted time answering your question in your duplicate thread.
      Whether you are the lion or the gazelle, when the sun comes up, you better be running.

      Comment


      • #4
        Originally posted by Ken
        I'd really appreciate if you don't make duplicate threads next time. You can post anywhere and I will find it. I see I just wasted time answering your question in your duplicate thread.

        And I answered it in the triplicate thread.

        Comment

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