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Math Help - pmf help

  1. #1
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    pmf help

    my problem is Let X be a random variable with pmf pk = c/k^2 for k = 1, 2, ....

    a.) estimate the value of c numerically.
    b) find P[X > 4]
    c) find P[6 <= X <= 8]

    i'm thinking that i need to do the summation from k=1 to infinite of c/k^2. And i think i can pull out the c and just make it c * the summation of k=1 to infinite of 1/k^2. I then get 1.64493c as the answer for a. I'm not sure what to do for part b or c. any help would be much appreciated.
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  2. #2
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    Quote Originally Posted by kanikedude03 View Post
    my problem is Let X be a random variable with pmf pk = c/k^2 for k = 1, 2, ....

    a.) estimate the value of c numerically.
    b) find P[X > 4]
    c) find P[6 <= X <= 8]

    i'm thinking that i need to do the summation from k=1 to infinite of c/k^2. And i think i can pull out the c and just make it c * the summation of k=1 to infinite of 1/k^2. I then get 1.64493c as the answer for a. I'm not sure what to do for part b or c. any help would be much appreciated.
    b) \Pr(X > 4) = 1 - \Pr (X \leq 3) = 1 - \left( \frac{c}{3^2} + \frac{c}{2^2} + \frac{c}{1^2} \right) = 1 - c \left( \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{1^2} \right).


    c) \Pr(6 \leq X \leq 8) = c \left( \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} \right).


    (By the way, \sum_{k = 1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6})
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  3. #3
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    thanks man. can you briefly explain why you solve it like that? i'm having trouble understanding some basic things in this class. I haven't had statistics in about 4 years so i forget a lot of things.
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