Exponential dsn - expected cost of a project

Robb · #1 $Old$ October 29th, 2009, 01:17 AM

Hi all,

The length of time $Y$ necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $C=100+40Y+3Y^2$ relates the cost $C$ of completeting this operation to the square of the time to completion. Find the mean and variance of $C$

So mean of C;
$E(C)=E(100+40Y+3Y^2)=100+40\cdot E(Y)+3\cdot [E(Y)]^2=100+40\cdot 10+3\cdot 100=800$

Just not sure I have calculated variance correctly; $E(C^2)=E[(100+40Y+3Y^2)^2]=$ $E(100^2+1600Y^2+9Y^4+8000Y+600Y^2+240Y^3)=640000$

So $var(C)=640000-800^2=0$

mr fantastic · #2 $Old$ October 29th, 2009, 10:30 PM

Quote:

Originally Posted by Robb $View Post$

Hi all,

The length of time $Y$ necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula $C=100+40Y+3Y^2$ relates the cost $C$ of completeting this operation to the square of the time to completion. Find the mean and variance of $C$

So mean of C;
$E(C)=E(100+40Y+3Y^2)=100+40\cdot E(Y)+3\cdot [E(Y)]^2=100+40\cdot 10+3\cdot 100=800$

Just not sure I have calculated variance correctly; $E(C^2)=E[(100+40Y+3Y^2)^2]=$ $E(100^2+1600Y^2+9Y^4+8000Y+600Y^2+240Y^3)=640000$

So $var(C)=640000-800^2=0$

You have made the very bad mistake of thinking that $E(Y^n) = [E(Y)]^n$ .

You're given $E(Y) = 10$ . To get $E(Y^n)$ you need to use the pdf: $E(Y^n) = \int_0^{+\infty} y^n f(y) \, dy$ .

$E(C) = E(100 + 40Y + 3Y^2) = 100 + 40 E(Y) + 3 E(Y^2)$ .

$Var(C) = E(C^2) - [E(C)]^2$

and note that $C^2 = (100 + 40Y + 3Y^2)^2$ .