X is continuous random variable with probability density function `f(x)=(x^(2))/(8),0 le x le 1`. Then, the value of `P(0.2leXle0.5)` is

A continuous random varibale X has probability density function f(x)=kx^(2) 0 le x lt 1 find k (X gt6)

A continuous random varibale X has probability density function f(x)=kx^(2) 0 le x lt 1 find k P(1 lt x lt 5)

A continuous random varibale X has probability density function f(x)=kx^(2) 0 le x lt 1 find k find P(X lt 5)

Let X is a continuous random variable with probability density function f(x)={{:(x/6+k,0lexle3),(0," otherwise"):} The value of k is equal to

Let X is a continuous random variable with probability density function f(x)={{:(x/6+k,0lexle3),(0," otherwise"):} The value of k is equal to

A continuous random varibale X has probability density function f(x)=kx^(2) 0 le x lt 1 find k find distrubution function

If X is the random variable with probability density function f(x) is given by f(x)={{:(x+1, -1 le x lt 0), (-x+1, 0 le x lt 1), (0,"otherwise"):} then find (i) the distribution function F(x) (ii) P(-0.5 le X le 0.5)

The probability density function of X is given by f(x)={{:(kxe^(-2x), "for " x gt 0), (0, "for " x le 0):} Find the value of k.

The probability density function of X is f(x)={{:(x, 0 lt x lt 1), (2-x, 1 le x lt 2), (0, "otherwise"):} Find P(0.2 le X lt 0.6)