# Suppose that in fact $9\%$ of the oranges on the truck do not meet the desired standard. What's the probability that the shipment will be rejected?

I am tutoring a student in AP stats, and came across this question. And I have not been able to solve this problem and get a result that is one of the multiple choices. The closest I got to was B (which I think is the intended answer), but I feel like the answer choices are incorrect.

Problem: When a truckload of oranges arrives at a packing plant, a random sample of 125 is selected and examined, The whole truckload will be rejected if more than $8\%$ of the sample is unsatisfactory. Suppose that in fact $9\%$ of the oranges on the truck do not meet the desired standard. What's the probability that the shipment will be rejected?

A) $0.6966$
B) $0.3483$
C) $0.6517$
D) $0.7803$
E) $0.2197$

• The choice B) is not reasonable. The truckload has $9\%$ bad and we will reject if more than $8\%$ of the sample are bad, so it seems clear that we will have a greater than $50\%$ chance of rejecting. – André Nicolas Mar 24 '16 at 5:51

The exact distribution of the random number $X$ of unsatisfactory oranges drawn from the sample is binomial, under the assumption that there are many, many more oranges than the sample size $n = 125$. The lot will be rejected if $X > 10$. Assuming that the true proportion of unsatisfactory oranges is $p = 0.09$, the resulting probability of rejection is $$\Pr[X > 10] = 1 - \sum_{x=0}^{10} \binom{125}{x} (0.09)^x (0.91)^{125-x} \approx 0.576514.$$ Using the normal approximation, $$X \dot\sim \operatorname{Normal}(\mu = np, \sigma^2 = np(1-p)),$$ and we would calculate using continuity correction $$\Pr[X > 10] \approx \Pr\left[\frac{X - \mu}{\sigma} > \frac{10 - 125(0.09) + 0.5}{\sqrt{125(0.09)(0.91)}} \right] = \Pr[Z > -0.234404].$$ Using a normal table, this is approximately $0.592664$. The quality of the approximation is poor because (a) $p$ is small, and (b) the $z$-score boundary is close to $0$.

But here's what's interesting: if you don't use continuity correction, the $z$-score is $-0.390673$, and the probability is...(drumroll)....

$$0.65198.$$ That is very close to answer choice (C), acceptably so (my calculation was on a computer rather than using a normal table, so it is more precise). Clearly, the writer of the question intended that the solution assumes two things that are not, in my opinion, entirely legitimate assumptions:

1. The solution should use the normal approximation to the binomial.
2. The normal approximation should be applied without continuity correction.

The issue with the first is minor. It's not totally unreasonable to use a normal approximation, even if a reasonably proficient student can use a calculator to compute a binomial sum of relatively few terms. But the second issue, as we can very clearly see from the exact calculation, is quite major. Failure to use continuity correction magnifies the error of the approximation to an extent that is unacceptable, because the real distribution here is binomial, not normal, and that in a finite, integral sample size of $n = 125$ oranges, $8\%$ is exactly 10 oranges.

I should also add that, in a testing situation with multiple choice options, it is often useful to reason in other ways so as to eliminate incorrect answer choices. Right away, I can tell you that the probability of rejection should be greater than $0.5$. This is because if the true proportion of unsatisfactory oranges is greater than the criterion of $8\%$, the test will be more likely to reject than to not reject. It may not be much greater than $0.5$, but intuitively, it shouldn't be less, and certainly not significantly less.

$\newcommand{\Var}{\operatorname{Var}}$Here is my attempt.

We are told that the true percent of bad oranges is $.09$. Throughout, I assume that $p = .09$. Then the question asks $P(\hat p >.08|p=.09)$, where $\hat p$ is the usual proportion $$\hat p = \frac{X_1+\dotsb+X_{125}}{125}$$ and each $X_i$ is an indicator (Bernoulli trial) with chance $p = .09$.

Hence, we see that $\mu = E[\hat p] = .09$, and $\sigma^2 = \Var(\hat p) = \frac{.09(1-.09)}{125}$, and $$P(\hat p >.08|p = .09) = 1-P(\hat p <.08|p = .09) \approx 1-P(Z< (.08-\mu)/\sigma) =0.6519804$$ where I exclude the continuity correction factor since the the question was posed in terms of proportion and $\hat p$ is a continuous random variable.

So I think it is C.

• $\hat p$ is not really continuous, but is in fact a rescaled binomial. The inescapable fact is that saying that a fraction of an orange is unsatisfactory makes no sense at all. It's not possible to observe $0.08 < \hat p < 0.088$ when $n = 125$, and treating it as continuous implies that this event has positive probability. That is the nature of the error in using the normal approximation without continuity correction. – heropup Mar 24 '16 at 6:07