I'm stuck on the following problem:
A batch of components has arrived at a distributor.
The batch can be characterized only if the proportion of defective components is at most 0.10.The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2.
a. What is the probability that the batch will be accepted when the actual proportion of defectives is 0.01? 0.05? 0.10? 0.20?
b. Let $p$ denote the actual proportion of the defectives in the batch.
A graph of P(batch is accepted) as a function of $p$, with $p$ on the horizontal axis and $P($batch is accepted$)$ on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan.Use the results of part (a) to sketch this curve for $0 \le p \le 1$.
c. Repeat parts (a) and (b) with "1" replacing "2" in the acceptance sampling plan.
d. Which of the two sampling plans, that of part (a), (c), appears most satisfactory and why?
I've got the following at the moment for part a but I get the feeling that I'm doing this slightly wrong and don't want to do all of it incorrectly.
a.) $n = 10$
$k($number of success$) = 2$
$p($population proportion$) = .01$
$P(k \le 2) = \binom{n}{k} \times p^k \times (1-p)^(n-k)$
$P(k \le 2) = .00441$
Also I'm unsure of what is being asked for part b.
Best Answer
If $X$ is Binomial($n,p)$, then $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Thus, $P(X \leq k) = \sum_{l \leq k} \binom{n}{l} p^l (1-p)^{n-l}$. So, your formula for (a) is wrong.