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S2 - Continous Random Variables need help watch

    • Thread Starter

    Find the C.D.F for the random variable X with the following probability density function :

    f(x) = {x/9, 0<x<3
    {(6-x)/9 3</=x</=6
    {0, otherwise

    S2 page 47 excerise 2b question 9b

    Please can you show me the correct method i can get the cdf for the first part of x/9 between 0 and 3 as (x^2)/18

    For the second one, I can't get it correctly. I integrated (6-x)/9 between 3 and a variable t. Pls show working using the method that they teach you in the book. I can get the answer using another method, but that is not the method that is meant to be used.

    I don't know what method the book uses. But here's my attempt.

    For 0 <= x <= 3,

    = \int_0^x f(y) dy
    = \int_0^x y/9 dy
    = [(1/18)y^2] (from 0 to x)
    = (1/18)x^2

    For 3 <= x <= 6,

    = \int_0^x f(y) dy
    = \int_0^3 f(y) dy + \int_3^x f(y) dy
    = F(3) + \int_3^x (6 - y)/9 dy
    = 1/2 + [(2/3)y - (1/18)y^2] (from 3 to x)
    = 1/2 + (2/3)x - (1/18)x^2 - 2 + 1/2
    = -1 + (2/3)x - (1/18)x^2

    [Check: When x = 6, -1 + (2/3)x - (1/18)x^2 = -1 + 4 - 2 = 1.]


    You can also do this question without integration. Draw the graph of f: a straight line from (0, 0) to (3, 1/3) and another from (3, 1/3) to (6, 0). (The area under the graph is 1.) F(x) is the area under the graph to the left of x.
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