Statistics with Probability

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  • #13264

    Ezra Halleck
    Participant

    A random variable has the following probability distribution:
    X -2 -1 1 2
    P(X) 0.3 0.2 0.3 ?
    Find and interpret a) probability when X=2 b) mean c) standard deviation.

    #16799

    Maria
    Member

    a) P[X=2] = 1 – (0.3+0.2+0.3) = 0.2
    The probability that outcome would be equal 2 is 20 %

    #16808

    Ezra Halleck
    Participant

    here is Melecia Lee’s answer to b)
    Let E[X] represent the Mean.
    E[X] = -2(0.3) + (-1)(0.2) + 1(0.3) + 2(0.2)
    = -0.6 -0.2 +0.3+0.4
    = -0.1
    This means that the weighted average of the probabily distribution is -0.1

    What you say is fine. I would say “The average for this random variable is -0.1.”

    #16809

    Ciabo
    Member

    Jonathan Ciabotaru

    My resolution for exercise 1

    1)
    -2 = 0.3 -1 = 0.2 1 = 0.3 2 = x
    a) 0.2
    b)E[X]= -0.6 – 0.2 + 0.3 + 0.4 = -0.1

    E[X^2]= 1.2 + 0.2 + 0.3 + 0.8 = 2.5
    V[X] = E[X^2] – E[X]^2 = 2.5 – 0.01 = 2.49

    c)SD[X] = Square root of V[X] = (2.49)^0.5
    SD[X] = 1.57797

    #16830

    Ying
    Member

    when it asks for interpret the standard deviation and mean, we are supposed to write something in words like this “the average outcome is -0.1 and on average, an outcome will be about 1.58 of a unit from the mean”, right?

    #16835

    plara125
    Participant

    Jonathan’s resolution is correct and yes Ying, how you wrote your interpretation is correct

    #16889

    fladmer911
    Member

    V[x]= sum [ ( x-u)^2 – P(x)
    V[x}= [-2(-0.1)]^2 . 0.3 + [-1(-0.1)]^2 . 0.2 + [1(-0.1)]^2 . 0.3 + [2(-0.1)]^2 . 0.2
    = 0.025
    SD[x}= 0.158

    #16890

    #1 fully done, all steps shown.

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