For the purposes of this post, we will refer to the data values in a data set as "scores."
In the last post, we used an example N(14, 2) to illustrate the 68-95-99.7 Rule, which stands for the various percents of scores lying within 1, 2, and 3 standard deviations from the mean. We can generalize the diagram we used to represent N(0, 1), where 0 is the mean and 1 is the standard deviation. This makes the model easier to apply, because the units we're most accustomed to seeing -- -1, 0,1,2, and so on -- appear as standard deviation units. Take a look:
We call N(0,1) the Standard Normal model. We now can use the number line to locate points that are any number of standard deviations from the mean...even fractional numbers.
In any Normal model, we're going to want to see what how many standard deviations a particular score in the data set is from its mean. We can do this for any score, and it has to do with converting a "raw" score (a score from our data) to a "standardized" score (a score from the Standard Normal model. How do we do this? There's a formula, and it's really an easy one:
X stands for the score you're trying to convert,
stands for the mean, and
stands for the standard deviation.
Suppose we're back in the Normal model N(14, 2) and we want to see how many standard deviations a score of 15 is from the mean. We would subtract our mean from 15, then divide by the standard deviation, 2. That is: (15 - 14) / 2 = 0.5. This mean that our score of 15 is 0.5 standard deviations from the mean. 0.5 is the score on the Standard Normal model that represents our score from N(14, 2). We call 0.5 our standardized score, also known as a z-score. Z-scores tell us how many standard deviations a given "raw" score is from the mean.
(Self-Test): In the Normal model N(50, 4), standardize a score of 55.
(Answer): Find the z-score using the formula:
z = (55 - 50) / 4 = 5 / 4 = 1.25. The score is 1.25 standard deviations above the mean.
(Self-Test): In the Normal model N(50, 4), find the z-score for 42.
(Answer): z-scores can be negative, too. in this problem, 42 is less than the mean. So it will lie to the left of 50, and its z-score will be negative. z = (42 - 50) / 4 = -8 / 4 = -2. The score is 2 standard deviations below the mean.
Why would we want to standardize our scores? There are actually two reasons.
1. It can help us see how unusual a score might be.
How? Well, the percents we have been talking about can also be thought of as probabilities. For example, the probability that a score is greater than the mean is 50%, the same as the probability that a score is less than the mean. In the last post, we marked off regions and computed percents. In statistics, we consider any z-score of 3 or more, or -3 or less, as unusual, because (as we saw in the last post) only 0.15% of scores are in each of those regions. In other words, the probability of seeing a score in one of these regions is at most 0.15%, less than even a quarter-percent. That's unusual.
2. It can allow us to compare apples to oranges.
How? Well, suppose you have just gotten back 2 tests you took: one in algebra and one in earth science. Suppose further that both score distributions follow a [different] Normal model. The algebra test's scores follow N(80, 5) and the earth science test's scores follow N(85, 8). Now imagine that you got a 90 on the algebra test and a 93 on the earth science test. You can easily see that percentage-wise, your score on the earth science test is higher than your algebra score. But relative to the distributions, on which test did you perform better?
To find out, figure out the z-score for each test score. For your algebra test, your z-score is (90 - 80) / 5 = 10 / 5 = 2 (2 standard deviations above the mean). For your earth science test, your z-score is (93 - 85) / 8 = 8 / 8 = 1 (1 standard deviation above the mean. Relatively speaking, your performance was better on the algebra test; that is, your score was more exceptional. Think of the probabilities. The probability of a score that's 2 standard deviations or more above the mean is 2.5%, whereas a score that's 1 standard deviation or more above the mean is 16%. Get the idea?
(Self-Test) Suppose Tom's algebra test score was 86 and his earth science test score was 86. In which test did Tom perform better, given that the test scores follow the Normal models we used above?
(Answer): Tom's z-score on the algebra test was z = (86 - 80) / 5 = 6/5 = 1.20. His z-score on the earth science test was z = (86 - 85) / 8 = 1/8 = 0.125. Because Tom's z-score on his algebra test (1.20) is higher than his earth science test z-score (0.125), his algebra performance was better than his earth science performance.
One more thing...let's use the z-score formula to go backwards: to convert a standardized score back to a raw score. Suppose, Nancy earned a score on the algebra test that was 1.6 standard deviations below the mean. (In other words, her z-score was 1.6.) What actual percentage score would that represent for her, assuming N(80, 5)? In this case, we would start with the z-score formula and fill in what we know. Then, using algebra (!) we would solve for the "raw" score...
Nancy's algebra test score was 72.
In the next post, we'll expand on the probability side of the Standard Normal model.