Let X be continuous with pdf fx 3x2 if 0 x 1 and zero oth

Let X be continuous with pdf f(x) = 3x² if 0 < x < 1, and zero otherwise. Find:

(a) E(X).

(b) Var(X).

(c) E(X’).

(d) E(3X – 5X² + 1).

Solution

a)

E(X) = integral (0->1) xf(x) dx

= integral (0->1) 3x^3 dx

= (0->1) 3x^4/4 dx

= 3/4

= 0.75 Answer

b)

E(X^2) = integral (0->1) x^2f(x) dx

= integral (0->1) 3x^4 dx

= (0->1) 3x^5/5 dx

= 3/5

=0.6

Therefore ,

V(X) = E(X^2) - (E(X))^2

= 0.6 - (0.75)^2

= 0.0375 Answer

c)

E(X \') = integral (0->1) x\' f(x) dx

= integral (0->1) 3x^2 dx

= (0->1) x^3 dx

= 1 Answer

d)

E(3X-5X^2+1) = integral (0->1) (3x-5x^2+1)f(x) dx

= integral (0->1) (9x^3-15x^4+3x^2) dx

= (0->1) (9x^4/4 - 3x^5 + x^3) dx

= 9/4 - 3 +1

= 0.25 Answer