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I understand that for X>3, Y=6-X and for X<3, Y=X.

For X = 3, Y=3

For part b, I got P(Y>y)= (3-y)/3, for 0≤y<3

Now for part c, I know P(Y>y) relates to the cdf.

But the definition of cdf relates to P(Y<y), so I'm guessing I have to

do 1-P(Y>y) to get the cdf which is y/3, 0≤y<3.

I'm thinking the pdf would be 1/3 for 0≤y<3?

I know for sure Y is a uniform distribution.

I'm not too sure on the interval (x-a)/(b-a)

Is it Y~Uniform distribution(0,3)?