Here is a full example of how we use the Central Limit Theorem. We’ll first look at a probability calculation for a single event, then we’ll look at a number of samples. Here’s what you’ll see:
Here is the scenario that we will work with:
A company makes generic, low-cost ink cartridges for printers. They claim that the average number of copies a cartridge will produce is 1000 copies with a standard deviation of 120 copies. The number of copies a cartridge makes follows a normal distribution.
I am going to ask you to calculate the probability that a single cartridge prints fewer than 800 copies. Then I’ll ask you to calculatoe the probability that the average of 5 randomly selected cartridges is fewer than 800.
Let’s think about this logically before we do it mathematically. If I get one cartridge that prints fewer than 800 cartridges, is that unusual? Perhaps it is unusual, but not really cause for alarm. I would expect that every once in a while there is a printer cartridge that doesn’t perform to expectation. It happens.
But if I randomly select 5, what are the chances that the average printing performance is less than 800 copies? If the company claims that their average is 1000, then a random selection’s average should definitely not be less than 800. If that happens, that would definitely be unusual and cause for alarm.
So, I would expect from my calculations that a single cartridge would have a small but possible probability, while my sample will show an extremely small probability.
Now, let’s find out.
Let’s start with a single event. Let’s answer this question:
Find the probability that a single cartridge chosen at random will print no more than 800 copies
So, we want to know the probability that we have a cartridge that prints fewer than 800 copies. That is, $P(x < 800)$. We’ll start by finding the z-score.
\[z = \frac{x-\mu}{\sigma} = \frac{800-1000}{120} = \frac{-200}{120} = -1.67\]Now, we look at a z-table and find -1.67 on the z-table. We find that,
\[P(x < 800) = P(z < -1.67) = 0.0475 = 4.75\%\]There is a 4.75% chance that we randomly select a single cartridge that will print no more than 800 copies.
Now, let’s try work with a random sample of 5 cartridges. In order to do this, we must make sure that the Central Limit Theorem applies. Let’s restate the problem, then check the two conditions for the central limit theorem.
A company makes generic, low-cost ink cartridges for printers. They claim that the average number of copies a cartridge will produce is 1000 copies with a standard deviation of 120 copies. The number of copies a cartridge makes follows a normal distribution.
Do you remember the two conditions?
Since we satisfied both conditions, the central limit theorem applies and we can continue.
We are now working with a sample. So, let’s find our sampling mean and sampling standard deviation.
\[\mu_\bar{x} = \mu = 1000 \qquad \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} = \frac{120}{\sqrt{5}} = \frac{120}{2.2360} = 53.67\]Now, we can find our z-score with our new mean and standard deviation for our sample.
\[z = \frac{\bar{x} - \mu_\bar{x}}{\sigma_\bar{x}} = \frac{800 - 1000}{53.67} = \frac{-120}{53.67} = -3.73\]We can now look up this probability. (This z-score actually won’t be on a z-table. That means it’s going to be really small! We’ll have to use a calculator for this one.)
\[P(\bar{x} < 800) = P(z < -3.73) = 0.000097 = 0.0097\%\]So, the probability that a single individual cartridge printing fewer than 800 copies is 4.75%. On the other hand, the probability that the average of 5 random cartridges is 0.0097%. This is exactly what we expected before we started our calculations.