# [SOLVED] Mathematical Expectation (Continuous Prob Dist.)

• Dec 6th 2009, 01:12 AM
hazel
[SOLVED] Mathematical Expectation (Continuous Prob Dist.)
4.56 walpole

The total time, measure in units of 100 hours, that a teenager runs her stero set over a period of one year is a continuous random variable X that has the density function:

f(x) = x, 0<x<1
2-x, 1<=x<2
0, elsewhere

Evalue the mean of the random variable Y=60X^2 + 39X , where Y is equal to the number of kilowatt hours expended annually.

what i did was :
E(X) = int 0-1 x dx + int 1-2 (2-x) dx
E(X^2) = int 0-1 X^2 dx + int 1-2 x(2-x) dx

and then E(Y) = 60((EX^2)) + 39 (E(X))

what did i do wrongly ?
• Dec 6th 2009, 03:04 AM
mr fantastic
Quote:

Originally Posted by hazel
4.56 walpole

The total time, measure in units of 100 hours, that a teenager runs her stero set over a period of one year is a continuous random variable X that has the density function:

f(x) = x, 0<x<1
2-x, 1<=x<2
0, elsewhere

Evalue the mean of the random variable Y=60X^2 + 39X , where Y is equal to the number of kilowatt hours expended annually.

what i did was :
E(X) = int 0-1 x dx + int 1-2 (2-x) dx Mr F says: You got 1, right?

E(X^2) = int 0-1 X^3 dx + int 1-2 x^2(2-x) dx Mr F says: Note the corrections in red. So E(X^2) = 7/6.

and then E(Y) = 60((EX^2)) + 39 (E(X))

what did i do wrongly ?

Fix the mistake and you will get 109.
• Dec 8th 2009, 07:18 AM
hazel
but why is there an additional x ?
• Dec 8th 2009, 12:39 PM
mr fantastic
Quote:

Originally Posted by hazel
but why is there an additional x ?

Surely you know that $E(X^n) = \int x^n f(x) \, dx$.
• Dec 10th 2009, 06:57 AM
hazel
urgh. oh yes. now i got it.
thanks.