The pdf of a random variable X is
`f(x)=3(1-2x^(2)),0ltxlt1`
`=0" otherwise"`
The `P((1)/(4)ltXlt(1)/(3))=` . . .

Options

A. `(179)/(864)`
B. `(159)/(864)`
C. `(169)/(864)`
D. `(189)/(864)`

Answer

A

Explanation

We have, p.d.f of a random variable X is `f(x)=3(1-2x^(2)),0 lt x lt 1`
`=0`, otherwise
`thereforeP((1)/(4)ltXlt(1)/(3))=int_(1//4)^(1//3)f(x)dx`
`=int_(1//4)^(1//3)3(1-2x^(2))dx`
`=3[x-(2)/(3)x^(3)]_(1//4)^(1//3)`
`=3[((1)/(3)-(2)/(3)((1)/(3))^(3))-((1)/(4)-(2)/(3)((1)/(4))^(3))]`
`=3[((1)/(3)-(1)/(4))-(2)/(3)((1)/(3^(3))-(1)/(4^(3)))]`
`=3[(1)/(12)-(2)/(3)xx(37)/(1728)]`
`=3xx(179)/(2592)`
`=(179)/(864)`