1 8 points Suppose a probability density function pdf of a r

1. (8 points) Suppose a probability density function (pdf) of a random variable X given by Find; (a) The mean E(X) and the variance Var(X). (b) The variance Var(Y) with Y = 3X - 1.

Solution

Mean of the function would be the integral of the [x.f(x)] from a to b

Thus,

Mean :

[ x ( 1/3 (4 - 3x²) ]

= 1/3 [ 4x - 3x³ ] Integrated from 0 to 1

Then, we get:

1/3 [2x² - 3x⁴/4 ] from 0 to 1

= (1/3) [ 2 - 3/4]

= 5/12

Variance = Integral of (x - 5/12)² * (1/3) * ( 4 - 3x²) from 0 to 1

On simplifying the integral, we get:

Integral of : (4/3)x² - 10x/9 + 25 / 108 - x⁴ + 5x³ /6 - 25x² /144

= 4x³ / 9 - 5x² / 9 + 25x/108 - x⁵ / 5 + 5x⁴/24 - 25x³ /432

= 17 /420

Hope this helps.

For formulae and other solved examples, this link might help

http://ltcconline.net/greenl/courses/117/DoubIntProb/ExpValVariance.htm