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Math Help - random variable probability question

  1. #1
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    random variable probability question

    Given a random variable x whose probability distribution is:

    x = 0, P(x) = .03125
    x = 1, P(x) = .15625
    x = 2, P(x) = .3125
    x = 3, P(x) = .3125
    x = 4, P(x) = .15625
    x = 5, P(X) = .03125

    determine the probability that (\mu - 2\sigma \leq x \leq \mu + 2\sigma).

    I get \mu = 2.5, \sigma \approx 1.118, so the equation should be P(.264 \leq x \leq 4.736) but I can't remember how to figure the probability from that! I know that the Empirical rule tells us that for a mound shaped distribution, the probability should be approx. .95, but how to figure it exactly?
    Last edited by earachefl; August 8th 2007 at 02:06 PM. Reason: meh
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  2. #2
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    For starters, you had better read up on WHEN the Emperical Rule might apply. You may wish to restrict its use to Normal or Near Normal distributions. This little guy with discrete data and only a few possible outcomes may not fit very well. Tchebychev may have similar things to say, but that still may not conclude your examination.

    For enders, you're done! You have a good interval. What data points fall within it? Add up their probabilities and call it a day.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by earachefl View Post
    Given a random variable x whose probability distribution is:

    x = 0, P(x) = .03125
    x = 1, P(x) = .15625
    x = 2, P(x) = .3125
    x = 3, P(x) = .3125
    x = 4, P(x) = .15625
    x = 5, P(X) = .03125

    determine the probability that (\mu - 2\sigma \leq x \leq \mu + 2\sigma).

    I get \mu = 2.5, \sigma \approx 1.118, so the equation should be P(.264 \leq x \leq 4.736) but I can't remember how to figure the probability from that! I know that the Empirical rule tells us that for a mound shaped distribution, the probability should be approx. .95, but how to figure it exactly?
    P(.264 \leq x \leq 4.736)=P(x=1) + P(x=2)+P(x=3)+P(x=4)

    or:

    P(.264 \leq x \leq 4.736)=1 - P(x=0) + P(x=5)

    RonL
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