Calculating a mean related to a continuous random variable

[archaic]

Homework Statement

We are filling some liquid in multiple containers. The volume poured is uniformly distributed between 5.1 and 5.7 deciliters. Every centiliter costs 0.01 dollar.
Knowing that any centiliter added beyond 5.5 deciliters represents an extra cost for the producer, what is the mean of this extra cost?

Relevant Equations

N/A

I am not sure about how to approach this.
Since the volume is uniformly distributed, the mean volume is $(5.7+5.1)/2=5.4$, which is less than $5.5$. From this, I could say that, on average, the producer won't spend any extra dollars.

But then I thought that maybe I should interpret this as the average if we only consider the cases where the volume is more than $5.5$, so that my random variable is now $5.5\leq X\leq 5.7$.
If so, my new mean is $5.6$, which is just one centiliter above the lower-bound volume for there being an extra cost, resulting in a mean of $0.01$$.
Or maybe $\int_{5.5}^{5.7}\left[0.01(x-5.5)\right]f(x)dx$, where $f(x)$ is the pdf?

 

[pasmith]

Your relevant equations box is missing the definiiton of the expectation of a function of a continuous random variable X taking values in an event space \Omega: <br /> E(f(X)) = \int_{\Omega} f(x) p(x)\,dx where p is the PDF of X.

In this case, the extra cost is given by <br /> f(V) = \begin{cases} 0 &amp; V \leq 55 \\<br /> 0.01(V - 55) &amp; V &gt; 55\end{cases} where V \sim U(51,57) is in centiliters.

 

[etotheipi]

$$f(V) = \begin{cases} 0 & V \leq 55 \\
0.01(V - 55) & V > 55\end{cases}$$

Why don't we just have $f(V) = 0.01(V-55)$ across the whole $51 \leq V \leq 57$? Wouldn't a negative $f(V)$ would just be money saved by not filling the container up enough, and integrating up with the p.d.f. would result in you expecting to spend actually less than if all containers were $V=55$ exactly?

 

[pasmith]

Why don't we just have $f(V) = 0.01(V-55)$ across the whole $51 \leq V \leq 57$? Wouldn't a negative $f(V)$ would just be money saved by not filling it up enough?

I think because the question asks for expectation of "extra cost" (due to a container having more than 55 cl), not "cost".

It is of course true that this extra cost is completely offset by the "extra saving" resulting from short-filling a container with less than 53 cl.

 

[Office_Shredder]

I agree the way I read the question you don't get to win the cost back when V is smaller than 55. If you're wondering how this makes sense in the real world, I would guess that the company charges the customer less if the fill it up to less than 55, but eats the extra cost if they overfill it. Or perhaps 55 is full, and they have a second machine that operates very slowly to get it to exactly 55 - if you undershoot it fills it up, and if you overshoot it has to throw the liquid out.

 

[archaic]

Your relevant equations box is missing the definiiton of the expectation of a function of a continuous random variable X taking values in an event space \Omega: <br /> E(f(X)) = \int_{\Omega} f(x) p(x)\,dx where p is the PDF of X.

In this case, the extra cost is given by <br /> f(V) = \begin{cases} 0 &amp; V \leq 55 \\<br /> 0.01(V - 55) &amp; V &gt; 55\end{cases} where V \sim U(51,57) is in centiliters.

Aha, so my last guess was the correct one.
What does $V\sim U(51,57)$ mean, though?

 

[Office_Shredder]

U(51,57) is a uniform random variable distributed between 51 and 57, and V~ means V has the same distribution. The reason for using ~ instead of = is so you can write things like $V\sim U(51,57)$ and $X\sim U(51,57)$ and you haven't claimed anything about whether V and X are literally equal, or just have the same distribution (but e.g are independent, or have some other weird correlation).

 

[archaic]

U(51,57) is a uniform random variable distributed between 51 and 57, and V~ means V has the same distribution. The reason for using ~ instead of = is so you can write things like $V\sim U(51,57)$ and $X\sim U(51,57)$ and you haven't claimed anything about whether V and X are literally equal, or just have the same distribution (but e.g are independent, or have some other weird correlation).

Thank you!