Question

A continuous random variable X has the p.d.f. F(x)={Ax(6-x)² if 0<x<6 ,{0 if otherwise Calculate the mean and standard deviation of X.Construct the distribution function F(x) and hence evaluate P(X>1/4).?

Answer 1

`P(x>1/4)~=0.99`

Explanation:

As probability distribution function is

`F(x)=Ax(6-x)^2` `{0," if "0 < x < 6}`

As `int_0^6Ax(6-x)^2dx=1`, we have

`Aint_0^6(36x-12x^2+x^3)dx=1`

or `A[18x^2-4x^3+x^4/4]_0^6=1`

or `A[18*36-4*6^3+6^4/4]=1`

or `A(648-864+324)=1` i.e. `A=1/108`

and distribution function is `F(x)=x/108(6-x)^2`

and `{0," if "0 < x < 6}`

and `P(x>1/4)=1/108int_(1/4)^6(36x-12x^2+x^3)dx`

= `1/108[18x^2-4x^3+x^4/4]_(1/4)^6`

= `1/108[18*36-4*6^3+6^4/4-(18/16-4/64+1/4^5)]`

= `1/108(108-1.0634765625)`

= `106.9365234375/108=0.990153~=0.99`

graph{x/108(6-x)^2 [-0.1, 6.05, -1.03, 2]}