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Math Help - Need some help with probability and density funtions.

  1. #1
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    Need some help with probability and density funtions.

    I have two functions:

    f(x)=kx^2 for x = 2, 3, 4, 6
    f(x)=kx^2 for 2<=x<=6

    I need to find the popper value for k, the mean, and the variance.

    For the first function I found k = 1/65 using the fact that the probabilities must sum to 1. Is this correct?

    For the mean of the first function, do I plug in each value into the function, then take this result and multiply it by the value, then sum? Like this:

    (4/65 *2)+(9/65 *3)+(16/65 *4)+(36/65 *6) = 315/65 = 4.85?

    And for the variance my notes say: Sum of (value * mean)^2 * prob. of the value

    3219.68 is what I get, is this correct?

    As far as the second density function, I am not sure where to start? How do you find the proper value for k when the function is continuous?
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  2. #2
    MHF Contributor
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    For the first function, your method for finding k is correct. Check that you have copied the function down correctly (looks like x=5 is missing, but this may be intentional).

    Your method for the expected value of the first function is correct.

    For the variance, i think your notes said "[Sum of (value - mean)^2] * Prob of value"




    In the continuous case, you need to use integration. The pdf must integrate to 1 over its range
    \int_2^6 kx^2 = 1

    k \int_2^6 x^2 = 1

    k \left[ \frac{x^3}{3} \right]^6_2 = 1

    k \left[ 72 - \frac{8}{3} \right] = 1

    k \left[ \frac{208}{3} \right] = 1

    k  = \frac{3}{208}
    Last edited by SpringFan25; July 26th 2010 at 10:51 AM.
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  3. #3
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    Sweet. My notes do say - not *. Guess I was just writing too fast..

    Is the mean of the second function 60/13?
    Last edited by Mattpd; July 26th 2010 at 11:24 AM.
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  4. #4
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    You need to use integration to find the mean of a continuous pdf

    E(x) = \int_2^6 x f(x) dx
    E(x) = \frac{3}{208} \int_2^6 x \times x^2 dx

    E(x) = \frac{3}{208} \int_2^6 x^3 dx
    E(x) = \frac{3}{208} \left[ \fra{x^4}{4} \right]^6_2 dx

    E(x) = \frac{3}{208} \left[ \fra{x^4}{4} \right]^6_2

    E(x) = \frac{3}{208} * 320

    = 60/13
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