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    maithany's Avatar
    maithany Posts: 11, Reputation: 1
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    #1

    Oct 15, 2011, 08:56 AM
    Probability Distribution: P(x)=(x-3)^2/55
    Suppose that the probability distribution of a random variable x can be described by the formula :

    P(x)=(x-3)^2/55

    for each of the values x= -2, -1, 0, 1, and 2.

    a) write the probability distribution of x.

    b) show that the probability distribution of x satisfies the properties of a discrete probability distribution.

    c) calculate the mean of x.

    d) calculate the variance and standard deviation of x.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Oct 16, 2011, 07:07 AM
    For the probability distribution, you can make a table like this one:



    To get each, just put the value of x of the column in the given equation.

    Can you do that first? Post back please :)
    maithany's Avatar
    maithany Posts: 11, Reputation: 1
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    #3

    Oct 16, 2011, 09:01 AM

    this was my answer for part a.

    X P(x)
    -2 25/55
    -1 16/55
    0 9/55
    1 4/55
    2 1/55

    b. 0.45+0.29+0.16+0.07+0.02 = 0.99

    c. -0.91-0.29+0.07+0.04 = -1.091

    d. I didn't know how to answer this part..
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Oct 16, 2011, 09:26 AM
    b. Better use the fractions themselves, then you will get one.

    And that said, you have been asked to explain. This isn't really an explanation, you just showed that the total probability gives 1. What is the meaning of this? It is that each outcome is independent of each other and that there are no values in between each outcome (ie. X = 1.5 does not exist)

    c. Good! :)

    d. Variance is obtained like this:



    The first is the same as squaring each value of x, giving 4, 1, 0, 1, 4 and multiply it by the probability, then add each result.

    The second is the square of the expectation (or the mean), so will be (-1.091)^2 = 1.19

    Then evaluate Var(X).

    Sd(X) is the square root of Var(X).

    Can you post what you get? :)

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