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Given the p.d.f. of a continuous r.v. X , f (x) = x23 , for –1 < x < 2 and = 0 otherwise Determine c.d.f. of X hence find P(1 < x < 2)

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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)

Sum
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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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Chapter 7: Probability Distributions - Exercise 7.2 [Page 239]

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