Question

CDF of random variable X is given by:

F_X(x) =

0 x < -3

(x/2 + 3/2)   -3 < x < -2

(x/8 + ¾)      -2 < x < 2

1                      x > 2

Find the possible range of values that the random variable can take.

Find E(X) = µ_X, the expected value

Find P(X ≥ 1)

Answer 1

a) Here we are given that: P( X < -3) = 0 and P( X > 2) = 1

Therefore the possible range of values that the random variable X can take here is from -3 to 2

b) The expected value of X here is computed as:

The PDF for X here is computed by differentiating the above CDF with respect to X. Therefore we get the PDF here as:

Therefore -5/4 is the expected value of X here.

c) The required probability here is computed as:

Therefore 1/8 = 0.125 is the required probability here.

answered by: anonymous