A continuous random variable X has a pdf of the form fx 2x9

A continuous random variable X has a pdf of the form f(x) = 2x/9 for 0 < x < 3, and zero otherwise.

a) Find the CDF of X.

b) find [x<= 2]

c) Find P[-1<X<1.5].

d) find a number m such that P[X<= m] = P[X>=m] (median)

e) find E(X)

Solution

a)

F(x) = Integral [2x/9 dx] = x^2 / 9 + C

As F(3) = 1, by properties of cdf\'s,

F(3) = 3^2/9 + C = 1 --> C = 0.

Thus,

F(x) = x^2/9, 0<x<3
0, otherwise [answer]

************

b)

P(x<=2) = x^2/9 | (0,2) = 2^2 / 9 - 0^2/9 = 4/9 [answer]

**************

c)

P(-1<x<1.5) = P(0<x<1.5) = x^2/9|(0,1.5) = (1.5^2)/9 - 0^2/9 = 0.25 [answer]

***************

d)

If m is the median, then its left tailed area is 0.5. Thus,

F(m) = m^2/9 = 0.5

--> m = 2.121320344 [answer]

*****************

e)

E(x) = Integral [x f(x) dx]|(0,3) = Integral [2x^2 / 9 dx]|(0,3)

= 2x^3/27|(0,3)

= 2(3^3)/27 - 2(0^3)/27

E(x) = 2 [answer]