Compute the standard deviation of the following measures of bearing diameter: 5.80 millimeters, 5.83 millimeters, 5.82 millimeters, 6.11 millimeters, 5.80 millimeters.
We can call the standard deviation of these values SD. And to begin, we can get an idea of what these measurements might mean. We’re told they have to do with the diameter of bearings, perhaps ball bearings manufactured at a facility. We’re given five diameters, presumably the measured diameters of five different bearings. Yet, we know that on a production scale, many more bearings would be made.
This means that we’re working with a sampling of a larger collection of values. We don’t have the measured diameters of every bearing made. But we do have those of five of them. The reason this matters has to do with the mathematical relationship we’ll use to solve for standard deviation. In one mathematical formulation for standard deviation, where we calculate what’s called the population standard deviation, we take all of our values, find their average, then take the difference between each individual value and that average and square it, and then sum together those squares. And then we divide by the total number of elements in our population 𝑛. Finally, taking the square root of that fraction.
However, there is another type of standard deviation called a sample standard deviation. In this framework, we’re working with values which represent just a sample or a subsection of an overall population. When we calculate the sample standard deviation instead of the total number of elements in the population 𝑛, we use 𝑛 minus one. So we see, our result will depend on whether we’re calculating a population or a sample standard deviation.
Going back to our problem statement, assuming again that many more bearings than the five listed here will be manufactured. That means we’re working with a sample of an overall population. So the standard deviation we’ll calculate will use the sample standard deviation formulation. With that settled, let’s begin going about calculating SD.
The first step we’ll want to take is to calculate the average of the five values we’ve been given. That average value, which we’ll call 𝑥 bar, is equal to the sum of our five values divided by the number of values we have. Keeping all of our resulting digits for now, that average is 5.872 millimeters.
The next step we can take is to subtract this value from each of our individual five measured bearing diameters, which we’ll then square and sum. This sum is equal to the numerator under our square root sign. And in the denominator, we have the number of measured diameters minus one.
Taking the square root of this fraction lets us solve for the standard deviation SD. We find it’s equal to 0.13 millimeters. That’s the sample standard deviation of these five values.