﻿ Z-Score Transformation Graziano & Raulin
Research Methods (8th edition)

## Z-Score Transformation

### The Normal Curve

The normal curve is based on a complex mathematical equation. Its usefulness stems from the fact that many variables in nature are distributed normally or near normally. This allows us to use the equation for a normal curve to do some useful computations. The bell-shaped curve above represents the shape of a normal curve. In a normal curve, most of the subjects score near the middle of the distribution. More importantly, you can compute exactly what proportion of people score between any two points. For example, 34% of people fall between the mean (the middle of the distribution) and one standard deviation above the mean. Another 34% fall between the mean and one standard deviation below the mean. Therefore, 68% score within one standard deviation of the mean, and roughly 95% fall within two standard deviations from the mean. These values can be computed from the equation for the normal curve, but an easier way is to access the standard normal table, which is available on this CD supplement. To use the standard normal table, you have to convert a score to a Z-score. You compute a Z-score by subtracting the mean of the distribution from the score and dividing the difference by the standard deviation of the distribution as shown in this equation.

Z = (score - mean)/standard deviation

For example, if someone gets a score of 40 on a test with a mean of 50 and a standard deviation of 20, their Z-score would be -.50. In other words, they scored half a standard deviation below the mean for the class.

Z = (40 - 50)/20 = -10/20 = -.50

### Computing the Percentile Rank

What can we do with such a score? Well, if we know that the scores on the test are distributed normally, we can use the standard normal table to compute the percentile rank. To use the standard normal table, we compute the Z-score and then look up the proportion of people using the table. Notice that the standard normal table is divided into sets of three columns. The first column (labeled a) is the Z-score; the second column (labeled b) is the proportion of people between the mean and that Z-score; the third column (labeled c) is the proportion of people beyond that Z-score. If you look closely, you will see that columns (b) and (c) always add up to .5000 because the two together represent the entire area from the mean to the end of one tail. Of course, .5000 of the distribution will fall on the other side of the mean because the normal curve is symmetric.

If someone scored right at the mean, their Z-score would be 0, and their percentile rank would be 50--that is, 50% of the class would score below them. If Z-scores are positive, they are above the mean and therefore the percentile rank would be greater than 50. Conversely, if subjects scored below the mean, their Z-score is negative, as in our example, and their percentile rank is less than 50. We can use the standard normal table to compute the percentile rank.

It is helpful to draw a quick picture when doing the computation for percentile rank. In our example, the Z-score was a -.50. That means that the person scored below the mean. We should draw a line in the middle of the distribution and then another line a bit to the left of the middle to represent a Z-score of -.50. To get a rough idea of proportion, the distance from the mean to near the end of either tail of the distribution is about 2 standard deviations. The percentile rank is the proportion of scores below that line representing -.50. We can see precisely what proportion of the distribution that represents by using the standard normal table. Notice from your drawing that we want to know how many people are below a -.50. In the table, that would be the proportion between a Z of .50 and the tail, which is what is shown in column (c). Notice that that figure is .3085. Multiplying by 100 to convert this proportion to a percentage gives us 30.85%, which we can round to 31%. A score of 40, which represented a Z-score of -.50, is at the 31st percentile. In other words, 31% of the class scores below that score.

What if our student had a score of 60 instead of 40. The student's Z-score would be a +.50 (a half standard deviation above the mean). It helps to draw this by putting a line representing a +.50 on a normal curve. Clearly the percentile rank will be greater than 50 because the area below the line at +.50 includes the 50% below the mean and some percentage between the mean and the +.50 Z-score. Using the standard normal table, we would look up the proportion between the mean and the Z-score and add it to .5000 (the proportion below the mean). Notice that the proportion between the mean and a Z-score of .50 is .1915. Therefore the percentile rank is (.5000 + .1915) * 100 or 69.15%, which we round to 69th percentile. Notice that we could have obtained the same result by looking up in column (c) the proportion above our Z-score and subtracting it from 1.000 and multiplying by 100 to convert it to percentages. That is, the percentage scoring below is equal to 100% minus the percentage scoring above the score. It can be a bit confusing, but with a little practice it will become second nature.

### Exercises

We have included a few exercises to make sure you understand the principles of using a standard normal table to convert a score to a percentile rank. Remember, to do this you must know the mean and that standard deviation of the distribution and the shape of the distribution should be the bell shape of a normal curve. If you have that information, you compute a percentile rank by:

• Computing a Z-score by subtracting the distribution mean from the score and dividing by the standard deviation.
• You draw the situation so that you can see visually what the Z-score represents. Minus Z-scores are below the mean and plus Z-scores are above the mean.
• You then determine the total proportion of the curve below the Z-score. Remember that you will use a different procedure depending on whether you are above or below the mean.
• Finally, you convert that proportion (from the standard normal table) to a percentage by multiplying by 100, which just moves the decimal over two places.

For all of the exercises below, you can assume that the distribution of scores is normal. You can access the standard normal table by clicking on this hyperlink during your exercises. Please note that the standard normal table is formatted as two pages, with the larger Z-scores on the second page.

1. Score = 50; Mean = 35; Standard deviation = 10; Click here to see the correct answer.
2. Score = 105; Mean = 80; Standard deviation = 8; Click here to see the correct answer.
3. Score = 31; Mean = 40; Standard deviation = 6; Click here to see the correct answer.
4. Score = 33; Mean = 36; Standard deviation = 4.5; Click here to see the correct answer.
5. Score = 230; Mean = 310; Standard deviation = 150; Click here to see the correct answer.