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Math Help - CDF and Quartiles

  1. #1
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    CDF and Quartiles

    I am given a probability density function -

    f(x) = (1/2)(1-x), for -1 <= x <= 1

    For finding the cdf, I distribute the (1/2) and integrate to

    F(x) = (1/2)x - (1/4)x^2

    My question is how am I able to use this to find the upper and lower quartiles? Every other problem I've done finding the quartiles has just been a linear F(x). Does having a quadratic mean I will have...two upper and low quartiles? That doesn't seem right to me. Any help would be appreciated. Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by SterlingM View Post
    I am given a probability density function -

    f(x) = (1/2)(1-x), for -1 <= x <= 1

    For finding the cdf, I distribute the (1/2) and integrate to

    F(x) = (1/2)x - (1/4)x^2

    My question is how am I able to use this to find the upper and lower quartiles? Every other problem I've done finding the quartiles has just been a linear F(x). Does having a quadratic mean I will have...two upper and low quartiles? That doesn't seem right to me. Any help would be appreciated. Thanks.
    You have the cumulative distribution function wrong.

    What you shuold have is:

    F(x)=\frac{2x-x^2}{4}+c

    for x \in [-1,1] and that F(-1)=0.

    Which makes c=0.75

    Now to find the first and third quartiles you solve the equations (which are quadratics and solvable using the quadratic formula):

    F(x_{25})=0.25

    and

    F(x_{75})=0.75

    for roots in [-1,1]

    CB
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  3. #3
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    .

    The integration is definitely right, but how did you get it? Is there a quotient rule for integration?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by SterlingM View Post
    The integration is definitely right, but how did you get it? Is there a quotient rule for integration?
    I got it in exactly the same way you did, I integrated the powers and then tidied up. You left off the constant of integration the value of which is determined by the end point conditions.

    CB
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