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Math Help - stat Q

  1. #1
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    stat Q

    so for binomial distribution if I have my n=15 and p=.2

    and I have to calculate the probability of exactly 8 fails,

    I know I can do something like this,
    P(X=8) = P(X<=8) - P(X<=7) = B(8;15,.2) - B(7;15,.2)

    in which case,I'm subtracting the two cdf's to obtain the value at X=8,

    but can I do something like this

    B(8;15,.2) = (15 (.2)^8 (1-.2)^(15-8)
    8)


    do I have to have to use the cdf's of the binomial random variable or I can also use the simple formula of the binomial theorem?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by NidhiS View Post
    so for binomial distribution if I have my n=15 and p=.2

    and I have to calculate the probability of exactly 8 fails,

    I know I can do something like this,
    P(X=8) = P(X<=8) - P(X<=7) = B(8;15,.2) - B(7;15,.2)

    in which case,I'm subtracting the two cdf's to obtain the value at X=8,

    but can I do something like this

    B(8;15,.2) = (15 (.2)^8 (1-.2)^(15-8)
    8)


    do I have to have to use the cdf's of the binomial random variable or I can also use the simple formula of the binomial theorem?
    In cases like this (X=x), you can just say that P\!\left(X=8\right)={15\choose 8}\left(.2\right)^8\left(.8\right)^{15-8}

    Otherwise, if you have P\!\left(x_1\leq X\leq x_2\right), you can evaluate \sum_{x=x_1}^{x_2}{15\choose x}\left(.2\right)^x\left(.8\right)^{15-x} (which in a sense is taking the difference between two cdfs...

    Does this make sense?
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