Results 1 to 4 of 4

Math Help - N: P(N=n)=(3/5)[(2/5)^n], find E(N)

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    38

    N: P(N=n)=(3/5)[(2/5)^n], find E(N)

    Suppose the number of children N of a random family satifies P(N=n)=(3/5)[(2/5)^n] where n=0,1,2,... Compute E(N).

    I went through the math and I come up with 1, which is the probability space. So, I know I'm not going about this the right way.

    E(N=n)=\frac{3}{5}(\frac{2}{5})^0 + \frac{3}{5}(\frac{2}{5})^1 + \frac{3}{5}(\frac{2}{5})^2 + ...

    =\sum_{n=0}^{\infty}\frac{3}{5}(\frac{2}{5})^n

    =\frac{3}{5}\sum_{n=0}^{\infty}(\frac{2}{5})^n

    =\frac{3}{5}\frac{5}{3}

    =1

    Could somone point out the basic understanding of expected value that I seem to be missing in this problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32

    Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)

    I'm going to assume you mean expected value. Here is the definition

    \operatorname{E}[N] = n_1p_1 + n_2p_2 + \dotsb + n_kp_k \;.

    What you have is the sum of all possible probabilities, which of course must be equal to 1!
    However, each of your probabilities P[N=i] has a weight ni
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2008
    Posts
    38

    Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)

    E(N=n)=0*\frac{3}{5}(\frac{2}{5})^0 + 1*\frac{3}{5}(\frac{2}{5})^1 + 2*\frac{3}{5}(\frac{2}{5})^2 + ...

    =\sum_{n=0}^{\infty}n*\frac{3}{5}(\frac{2}{5})^n

    =\frac{2}{3}

    At least I was right about missing a basic understanding. The unfortunate part is that I need to learn how to go from the summation to the answer using paper and pencil, I won't always have access to a calculator that can do the sum for me...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32

    Re: N: P(N=n)=(3/5)[(2/5)^n], find E(N)

    I'm trying to remember back in my statistics class. If you were to take the natural log of both sides, would that help? when I see the product of n and (2/5)^n, that's what I want to do. EDIT:// but it is a sum still so it might not.
    Last edited by MacstersUndead; November 15th 2012 at 05:17 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 05:57 PM
  2. Replies: 2
    Last Post: July 5th 2010, 09:48 PM
  3. Replies: 1
    Last Post: February 17th 2010, 04:58 PM
  4. Replies: 0
    Last Post: June 16th 2009, 01:43 PM
  5. Replies: 2
    Last Post: April 6th 2009, 09:57 PM

Search Tags


/mathhelpforum @mathhelpforum