Let Y have the pdf fy y218 3 y 3 zero elsewhere Find PY

Let Y have the pdf f(y) = y2/18, 3 < y < 3, zero elsewhere. Find P(|Y | < 1)
and P(Y2 <9)

Solution

a)

P(|y| < 1) = P(-1<y<1)

= Integral(y^2/18 dy)|(-1,1)

= y^3/54 |(-1,1)

= (1^3/54) - ((-1)^3/54)

= 1/27 [ANSWER]

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b)

P(y^2 < 9) = P(-3<y<3)

which is the domain itself. Hence, by the total integral must be 1,

P(y^2 < 9) = 1 [ANSWER]