The following is the p.d.f (Probability Density Function ) of a continous random variable X :
`f(x) = (x)/(32), 0 lt x lt 8`
= 0 , otherwise
Find the expression for c.d.f (Cumulative Distribution Function ) of X

The following is the p.d.g. (Probability Density Function) of a continuous random varible X: f(x)=(x)/(32), 0ltxlt8 =0, otherwise (a) Find following the expression for c.d.f. (Cumulative Distribution Function) of X. (b) Also find its value at x=0.5 and 9.

The following is the p.d.f (Probability Density Function ) of a continous random variable X : f(x) = (x)/(32), 0 lt x lt 8 = 0 , otherwise Also, find its value at x = 0.5 and 9.

The following is the p.d.f of continuous random variable X. f(x) = (x)/(8), 0 lt x lt 4 =0 , otherwise. Find the expression for c.d.f of X

The p.d.f of a continous random variable X is f(x)=x/8, 0 lt x lt 4 =0 , otherwise Then the value of P(X gt 3) is

The p.d.f of a continous random variable X is f(x)=x/8, 0 lt x lt 4 =0 , otherwise Then the value of P(X gt 3) is

Thep.d.f of a continous random variable X is f(x) = (x^(2))/(3), - 1 lt x lt 2 0 = otherwise Then the c.d.f of X is

The p.d.f of a continuous random variable X is f(x) = (x^(2))/(3), - 1 lt x lt 2 0 = otherwise Then the c.d.f of X is

If the p.d.f of a continuous random variable X is f(x)=kx^(2)(1-x), 0 lt x lt 1 = 0 otherwise Then the value of k is

If the p.d.f of a continuous random variable X is f(x)=kx^(2)(1-x), 0 lt x lt 1 = 0 otherwise Then the value of k is

The p.d.f. of a continuous random variable X is f(x) = K/(sqrt(x)), 0 lt x lt 4 = 0 , otherwise Then P(X ge1) is equal to