Uniform Distribution Homework: Mean and Variance of Profit Y

[Mdhiggenz]

Homework Statement

The proportion of time x that an industrial robot is in operation during a 40 hour work week is a random variable wth probability density function
f(x)= 2x 0≤x≤1

I already found my E(x) to be 2/3 " Mean" and V(x) to be 1/18 " variance"

Here is where things get confusing for me.

For the robot under study, the profit Y for a week is given by

Y=200*x-60

Find E(Y) and V(Y).

So I found E(Y) to be just y=200(E(x))-60

which makes sense.

But for V(Y)

we have to ignore the constant and sqareroot the first term so it would be

V(Y)=v[200x-60]=200²*v(x) "ignoring the constant"=200²*1/18

I don't understand the logic to computing the variance. Why would we ignore the constant and just square the first term?

Thanks

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

I presume that you know that E(x)= \int xf(x)dx and V= \int (x- x_0)^2f(x)dx. Given that y= 2x- 60, yes, you can just put E(x) in for x and get E(y)= 2(2/3)- 60= 4/3- 60= -176/3 or -58 and 2/3.

If you were to replace x by 2x- 60 and E(x) by that mean, you should be able to see that in ((2x- 60)- (4/3- 60))^2 you can cancel the two "-60" terms. to find the variation of Y, integrate (2x- 4/3)^2.

 

[jbunniii]

I don't understand the logic to computing the variance. Why would we ignore the constant and just square the first term?

In general, if $y = ax + b$, then $E[y] = aE[x] + b$, so
$$\begin{align}
y - E[y] &= (ax + b) - (aE[x] + b)\\
&= ax - aE[x]\\
&= a(x - E[x])
\end{align}$$
Therefore,
$$\begin{align}
var(y) &= E\bigl[(y - E[y])^2\bigr]\\
&= E\bigl[\bigl(a(x - E[x])\bigr)^2]\\
&= E\bigl[a^2(x - E[x])^2\bigr]\\
&= a^2E\bigl[(x - E[x])^2\bigr]\\
&= a^2 var(x)
\end{align}$$