Advanced Placement Statistics

Question

Cassandra's Cogs states in its advertisement that 90% of its orders are ready to ship within three working days. A simple random sample of 100 of the 2,500 orders received during the past month are pulled for an audit. The audit shows that 84 of these orders were shipped on time.  If Cassandra's Cogs really ships 90% of its orders within three working days, what is the probability that the proportion in the simple random sample of 100 orders is as small as the proportion in the auditors' sample or smaller? Explain why you were able to use the method you used? Explain your results in the context of the problem.

Answer 1

There are two ways of solving this problem with the tools you have available on the AP exam. I'll explain both of them:

Method 1: the binomial distribution

Recall the assumptions behind the binomial distribution:

1. Each trial can only go one of two ways. This is true. Each shipment is either late or on time.
2. The trials are identically distributed. We assume this is true. There's no reason to expect that the first order that month is more or less likely to be delayed than the 2nd order, or the 89th order, or the last order.
3. The trials are independent. We have to assume this is true in order to use the techniques in this class. However, this assumption might not be accurate in the real world. Realistically, one accident at the warehouse or heavy snowstorm could cause a lot of shipments to be delayed together. Nonetheless, we have to assume that the trials are independent or it would go beyond the scope of this class.

Now that we can assume that the number of late shipments is binomially distributed, we can use our calculator to calculate the probability of 84 or fewer shipments arriving on time.

We turn on our calculator, and hit 2nd and then the VARS key. This should bring up the DISTR menu. We scroll down to the eleventh item on the list, called binomcdf(

we select it, and enter 100 for the number of trials, 0.9 for the success probability, and 84 for the maximum number of successes that we're interested in, in that order. We get binomcdf(100, 0.9, 84) = 0.0399, so there is a 3.99% chance that the number of on-time shipments would be that low.

Method 2: Using the normal approximation

We need several more assumptions to use the normal approximation

1. Each trial can only go one of two ways. This is true. Each shipment is either late or on time.
2. The trials are identically distributed. We assume this is true. There's no reason to expect that the first order that month is more or less likely to be delayed than the 2nd order, or the 89th order, or the last order.
3. The trials are independent. We have to assume this is true in order to use the techniques in this class. However, this assumption might not be accurate in the real world. Realistically, one accident at the warehouse or heavy snowstorm could cause a lot of shipments to be delayed together. Nonetheless, we have to assume that the trials are independent or it would go beyond the scope of this class.
4. There are at least 5 expected successes and at least 5 expected failures. Because there are 90 expected successes and 10 expected failures, this condition is satisfied, meaning that the overall distribution is sufficiently unskewed, so we can use the normal approximation.

Next, we calculate the mean number of successes (90% of 100, or 90), and the standard deviation of the number of successes.

The standard deviation formula is

s.d. = sqrt(N*p*(1-p), where N is the number of trials (100), and p is the success probability (0.9)

s.d. = sqrt(100*0.9*0.1) = sqrt(9) = 3.

So, our number of successes will be approximately normally distributed with mean 90 and standard deviation 3.

What does that mean about the probability of the number of successes being 84 or lower? There's one more subtlety we have to account for. The normal distribution can take any value, including fractions. The true number of successes HAS TO be a whole number. To account for the difference, we use rounding. (This is called the continuity correction). We assume that if the normal distribution tells us the number of successes would be between 83.5 and 84.5, then the actual number of successes would be 84. For this reason, we are interested in the probability that this normal random variable would be 84.5 or less.

Now we have to calculate the z-score corresponding to 84.5 successes.

z = (actual - average)/s.d

z = (84.5 - 90)/3

z = -5.5/3

z = -1.8333

We use a table or a calculator to find the probability that a normal random variable will have a z-score of -1.8333 or less

We get a probability of 0.0333

Why are the numbers obtained with the two methods a little bit different? One uses the exact binomial distribution and the other only uses an approximation.