A mean related to a continuous random variable

Question

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?

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 $\int_{5.5}^{5.7}\left[0.1(x-5.5)\right]f(x)dx$, where $f(x)$ is the pdf?

Answer 1

Hint: Your "new mean" is better described as a conditional mean and can be written $E(X | X > 5.5)$. To proceed in this way you then need to consider the probability that the condition is satisfied, that is, $P(X > 5.5)$.

Answer 2

When a container is filled in such a way that the volume poured does not exceed $5.5$ deciliters, there is no extra cost incurred. When the volume poured does exceed $5.5$ deciliters, there is an extra cost, which is proportional to the amount of volume above $5.5$.

Your misunderstanding is that somehow, containers whose volume poured is below $5.5$ can compensate for the extra cost incurred from those containers where the volume poured is above $5.5$. For instance, if only two containers are filled and one is filled to $5.1$ deciliters and the second is filled to $5.7$, the savings of the first container does not cancel out the extra cost of the second, which is still incurred even though the overall cost may still be acceptable.

Another way to think about it is that the producer would have filled the first container to $5.1$ anyway. The extra cost is conceptualized as answering the question of, "if every container could have been filled so as to guarantee none has more than $5.5$ deciliters, how much money could have been saved?"

With this in mind, we see that the unconditional mean fill volume is not an indicator of the extra cost. If $X$ is the random fill volume, then we are given $$X \sim \operatorname{Uniform}(5.1, 5.7).$$ The extra cost random variable is $$Y = 0.1 \min(X - 5.5, 0) = \begin{cases} 0, & 5.1 \le X \le 5.5 \\ (0.1)(X - 5.5), & 5.5 < X \le 5.7. \end{cases}$$ You want to compute $$\operatorname{E}[Y \mid X > 5.5] = 0.1 \operatorname{E}[X - 5.5 \mid X > 5.5] = 0.1 ( \operatorname{E}[X \mid X > 5.5] - 5.5 ).$$ Note that the extra cost of $0.01$ dollar per centiliter equals $0.1$ dollar per deciliter, which explains the factor of $0.1$.