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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(1 < x < 2)
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Solution
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 (1 < x < 2) = F (2) - F (1)
= `((2^3+1)/9)- ( (1^3+1)/9)= 1-2/9=7/9`
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