Studying for a statistics course an stumped on how to go bout solving a problem.
The following is a problem I have solved – the problem I'm having trouble with is related to this one:
Impurities in the batch of final product of a chemical process often reflect a serious problem. From considerable plant data gathered, it is known that the proportion $y$ of impurities in a batch has a density function given by
$$f(y) = 10(1−y)^9,\quad 0 ≤ y ≤ 1.$$
A batch is considered not acceptable if the percentage of impurities exceeds $60\%$. With the current quality of the process, what is the percentage of batches that are not acceptable?
I found this quantity as such:
We can take integral $\int_{0.60}^1 10(1-y)^9dy$ and use transformation to see that it is equivalent to the negative integral $\int_{0.4}^0 10x^9dx$, which, then, is equivalent to $\int_0^{0.4}10x^9\, dx$. I perform the integration and get the value
$$0.00010408576 \quad \text{or} \quad 0.01048576\%~\text{of batches!}$$
That's all fine and good. But what of the follow-up questions?
For the random variable Y in the previous problem
a. Find the expected percentage of impurities.
b. Find the expected value of the proportion of quality material (i.e., $E[1-y]$).
What is meant by this? Did we not just find the $\%$ of batches impure? What exactly is being asked by a? I imagine once we have a, finding b my be simple.
Thoughts? Direction?
Best Answer
You found the percent of batches for which the percent of impurities is more than a certain amount. You did not find the average percent of impurities in a batch. Notice that if that cut-off amount beyond which a batch is unacceptable were different, the answer you got would have been different, but the average percent of impurities in a batch would still be the same! You just have to read the words carefully.
The random variable $Y$ is the proportion of impurities, and its density is $$ f_Y(y) = 10(1-y)^9\quad\text{for }0\le y\le 1. $$ The expected value of $Y$ is what is sought. It is $$ \mathbb E(Y) = \int_0^1 y f_Y(y) \, dy. $$ That is $$ \int_0^1 y10(1-y)^9\,dy = -\int_1^0 (1-x)10x^9\,dx. $$