The GM Recall Example, Part I

Example 1: The First Distribution

The General Motors building in downtown Detroit, MI. Photo courtesy the Library of Congress.

You have read recently about several automotive recalls. In each case, the media reported that there was a defective part (the ignition switch in the GM case). This is only partly true, however (pun intended). It is not that all ignition switches were defective, it is that too many were. Parts need to meet certain specifications, such as length (as in the case of the ignition switch). Due to normal manufacturing processes, these lengths are random variables with a mean (μ) and a standard deviation (σ).

To illustrate this with made-up numbers, let us say that the ignition switch needs to be longer than 10.4 cm in order to fully engage the mechanism, but less than 10.5 cm to keep from wearing the steering column too much. It turns out that the average length of the ignition switches is actually μ = 10.45 cm, which is great! The problem is that the standard deviation of the length is too large, say σ = 0.02 cm. This is about the length of a mite that lives in human hair follicles. However, this tiny variation causes 1.242% of all ignition switches to fail these requirements.

Here is the calculation: If we define L as the length of an ignition switch in centimeters, then we know

$$ L \sim N(\mu=10.45;\ \sigma=0.02) $$

Next, we also know P[ 10.40 < L < 10.50 ] is the proportion of ignition switches that work properly. This is a Type 3 probability statement, which can be rewritten as

P[ 10.40 < L < 10.50 ] = P[ L < 10.50 ] − P[ L < 10.40 ]

Using our program of choice (StatCrunch or Excel), we have this probability is 0.9938 − 0.0062 = 0.9876.

Since this is the proportion of ignition switches that actually work, the proportion that fail is one less that value, or $p=0.0124$. This is also the probability that you get a faulty ignition switch. Pretty small, huh?

If that probability is so small, why the extensive brouhaha? The answer is this: There were a whole lot of cars made. In fact, the number of cars exposed to that ignition switch was $n = 1,\!620,\!665$. This means approximately $np = 20,\!129$ cars have a faulty ignition switch. This has reportedly led to 13 deaths.

Example 2: The Second Distribution

A 1941 Buick convertible on Route 40 outside of Denver, CO (1942). Photo courtesy the Library of Congress.

In my town, the GM dealer is trying to figure out how many replacement ignition switches he will need to stock. We can use probability to get this answer, too.

Before we get started here, we need to determine our random variable and its distribution. The dealer wants to estimate the number of ignition switches he will need. That is our random variable, X.

Now, to determine the distribution of X. What do we know about it? Well, we know that it depends on the number of cars in my area. Each car in that area can be classes as either needing a new ignition switch or not. Thus, X has a Binomial distribution. Now, we need to determine the two parameters, n and p.

From data, the dealer estimates that there are approximately 524 cars in the recall years in the area. Thus, $n = 524$. From the previous part of this example, we know $p = 0.0124$. And so, we have

$$ X \sim Bin(n=524;\ p=0.0124) $$

Now that we know this, we can calculate the expected number of cars in my area needing a replacement ignition. As X has a Binomial distribution, $E[X] = np$. So, in this case, $\mu = np = 6.4976$. The expected number of replacement ignition switches needed is about 6.5.

Remember, this is an expected value, a long-run average. We may need more replacements. We may need fewer. Because the actual number we need, X, is a random variable, we cannot know with absolute certainty the number we will need. We can calculate probabilities of each possible outcome, however. And that knowledge is what we will have to base our decisions on.

Example 3: Excess Inventory

The Crystal Motors GM dealership in Brooklin, NY (1950). Photo courtesy the Library of Congress.

In the previous example, we discovered that the expected number of ignition parts needed is about 6.5. However, that tells us very little about the actual number we will need. As good money-handlers, we want to ensure that we have no excess inventory, but we also want to ensure that we have no upset customers.

For instance, if the dealer buys eight switches, what is the probability he will have too many parts? Note that excessive inventory means lost money, something businesses want to avoid. These ignition switches are not free.

Let us define X as the number of ignition switches needing replacement, which we did above. With this, the distribution of X is

$$ X \sim Bin(n=524,;\ p=0.0124) $$

Now, all we have to do is calculate P[X < 8] = P[X ≤ 7]. Our favorite program calculates this probability to be 0.6720. Thus, buying 8 switches means there is more than a 67% chance of having leftover switches. If the dealer buys only 5 switches, the probability drops to 0.3665. This is a more manageable value. In fact, if the dealer buys only 1 switch, the probability of having excess inventory is just 0.0110. This is awesome!

… or is it?

Example 4: Mad Customers

Young North Carolinians in an old Ford (1939). Are they angry because their ignition switch works? Photo courtesy the Library of Congress.

However, the other side of the coin is that if the dealer stocks too few ignition switches, there will be upset customers and bad press. That seems a bit more costly than having too many switches laying about. While working through these, remember that

$$ X \sim Bin(n=524;\ p=0.0124) $$

So, if the dealer does buy 1 switch, as suggested in the previous example, what is the probability he will have to turn away customers? P[X > 1] = 1 − P[X ≤ 1] = 0.9890. Wow! That is a problem.

What if the dealer buys the originally-planned 8 switches? What is the probability of turning away customers? P[X > 8] = 1 − P[X ≤ 8] = 0.2083, which is just shy of 21%. That is still rather high.

What if the dealer buys 100 ignition switches? What is the probability of having too few? That is, what is the probability that he has too many customers; P[X > 100] = 1 − P[X ≤ 100]. This probability is 0.0000. There we go! The probability of disappointing no customers has been achieved. But, at the cost of 100 ignition switches and an almost guarantee of having to throw out inventory.

Example 5: Playing it Safe

Considering the cost of the ignition switch and the cost of the bad press, it is much more likely that he would like to ensure he has a very low probability of having to turn away customers. Perhaps, he would like to be 99% sure that no one is turned away. With this goal, how many ignition switches should be bought?

Here, we have the probability, but we need to calculate the value of X. That is, we need to find the value x_99% (read as “x sub 99%”) such that P[X ≤ x_99%] = 0.99. We call this value of X the “ninety-ninth percentile”. In StatCrunch, this requires trial and error (but not too much). Increase the value of x until the calculated probability is above 0.99 the first time. In Excel, the formula is =BINOM.INV(524,0.0124199268,0.99). Both give an answer of 13.

Thus, if the dealer buys 13 ignition switches, the probability that he will be able to service all of his customers is no less than 99%. If I were the car dealer, I would buy at least 13 ignition switches. The cost of the part is much less than the cost of bad press. Of course, if I were a car dealer, I’d have a newer car!

And that is it. In this series of examples, I showed several ways that these last two Modules could be helpful in making business decisions. We used both the Normal distribution and the Binomial distribution. These two distributions worked together here to help us answer some complicated questions. The first step is always to figure out what you want to measure. This is the random variable. The second step is to determine the distribution of that random variable. Sometimes, it is easy. Sometimes, it is hard. Sometimes, it is so hard that we start to look at expected values (means) instead of observations. In such cases, we rely on the Central Limit Theorem.