# pmf help

• Sep 15th 2009, 01:32 PM
kanikedude03
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.
• Sep 15th 2009, 08:55 PM
mr fantastic
Quote:

Originally Posted by kanikedude03
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, $\sum_{k = 1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$)
• Sep 16th 2009, 08:30 AM
kanikedude03
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.