The pdf of a discrete random variable is defined as `f(x)={{:(kx^2","0lexle6),(0", ""elsewhere"):}`
Then the value of F(4) is

Options

A. `(30)/(91)`
B. `(30)/(97)`
C. `(15)/(47)`
D. None of these

Answer

A

Explanation

We know that the total probability is always one
`thereforek(0)^2+k(1)+k(2)^2+k(3)^2+k(4)^2+k(5)^2+k(6)^2=1`
`rArr0+k+4k+9k+16k+25k+36k=1`
`rArr" "k=1/(91)`
`therefore` For decreate random variable, cumulative distribution function is
`F(x)=P(Xlex)`
`therefore" "F(4)=P(Xle4)`
`=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)`
`=k(0)^2+k(1)^2+k(2)^2+k(3)^2+k(4)^2`
`=0+k+4k+9k+16`
`=30k=30xx1/(91)=(30)/(91)`