The pdf of a curve X is
`f(x)={{:(k/sqrtx","0ltxlt4),(0","xle0" or "xge4):}`
Then , `P(Xge1)` is equal to

Options

A. 0.2
B. 0.3
C. 0.4
D. 0.5

Answer

D

Explanation

`therefore" "oversetoounderset(-oo)intf(x)dx=1`
`therefore" "overset4underset0intk/sqrtxdx=1`
`rArr" "k[2sqrtx]_0^4=1`
`rArr" "2k(sqrt4-sqrt0)=1`
`rArr" "4k=1`
`therefore" "k=1/4`
Now, `P(Xge1)=P(1leXlt4)=overset4underset1intf(x)dx`
`=overset4underset1intk/sqrtxdx=2k[sqrtx]_1^4`
`=2(1/4)(2-1)=1/2=0.5`