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Math Help - Distribution help

  1. #1
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    Distribution help

    The question:

    p(X = k) = \frac{c}{k!}, k = 0, 1, 2, ...

    a) Determine the value of c.
    b) Calculate P(X = 2)
    c) Calculate P(X < 2)
    d) Calculate P(X \ge 4)

    I think I know how to do b), c) and d), but a) is giving me trouble.

    I tried this:

    \sum\limits_{i = 0}^{k} \frac{c}{i!} = 1 (since it's a distribution)

    \sum\limits_{i = 0}^{k} \frac{1}{i!} = \frac{1}{c}


    I'm confident I've done this completely wrong. Any advice?
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:

    p(X = k) = \frac{c}{k!}, k = 0, 1, 2, ...

    a) Determine the value of c.
    b) Calculate P(X = 2)
    c) Calculate P(X < 2)
    d) Calculate P(X \ge 4)

    I think I know how to do b), c) and d), but a) is giving me trouble.

    I tried this:

    \sum\limits_{i = 0}^{k} \frac{c}{i!} = 1 (since it's a distribution)

    \sum\limits_{i = 0}^{k} \frac{1}{i!} = \frac{1}{c}


    I'm confident I've done this completely wrong. Any advice?
    You need to use the standard power series \displaystyle e^x = \sum_{i=0}^{+\infty}\frac{x^i}{i!}.
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  3. #3
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    Ahh, so 'x' in the power series is just '1', which makes c = 1/e?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:

    p(X = k) = \frac{c}{k!}, k = 0, 1, 2, ...

    a) Determine the value of c.
    b) Calculate P(X = 2)
    c) Calculate P(X < 2)
    d) Calculate P(X \ge 4)

    I think I know how to do b), c) and d), but a) is giving me trouble.

    I tried this:

    \sum\limits_{i = 0}^{k} \frac{c}{i!} = 1 (since it's a distribution)

    \sum\limits_{i = 0}^{k} \frac{1}{i!} = \frac{1}{c}


    I'm confident I've done this completely wrong. Any advice?
    ... that isn't completely wrong... only the condition for c is...

    \displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{c} (1)

    ... i.e. the sum has infinite terms... Once You have c the other questions are relatively easy, for example...

    \displaystyle  P (x<2) = c\ (1+1)= 2\ c

    Kind regards

    \chi \sigma
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