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# S2 - Continous Random Variables need help watch

1. 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.
2. I don't know what method the book uses. But here's my attempt.

For 0 <= x <= 3,

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

For 3 <= x <= 6,

F(x)
= f(y) dy
= f(y) dy + f(y) dy
= F(3) + (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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