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Question
Given the p.d.f. of a continuous r.v. X , f (x) = `x^2/3` ,for –1 < x < 2 and = 0 otherwise
Determine c.d.f. of X hence find
P( x < 1)
Solution
Let F(x) be the c.d.f. of X
Then F(x) = ` int_(-∞)^x f (x) dx`
=` int_(-∞)^-1 f (x) dx + int_(-1)^x f (x) dx`
= 0 + `int_(-1)^x x^2/3 dx = 1/3int_(-1)^x x^2 dx`
= `1/3[x^3/3]_-1^x`
= `1/3[x^3/3-(-1/3)]`
∴ f(x) = `(x^3+1)/9`
P( x < 1) = `F (1) - F (-1) - [(1^3 + 1)/9]- [((-1)^3 + 1)/9]- (2-0)/9-2/9`
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