IQ test scores

Z-stats for IQ score
Z stats and IQ test scores

IQ test scores are meaningless without the accompanying standard deviation of the test in question. I will walk through an example to help you understand IQ test scores and percentiles. Most IQ tests are assumed to have an average score of 100 (or put differently, the test publishers will norm the IQ tests to have a mean of 100 within a given country). The standard deviation of the test can vary however. Most of the well-known tests typically have a standard deviation of either 15, 16 or 24 points although tests with a standard deviation of 15 and 16 points are the most commonly seen.

We know from properties of the standard normal distribution that 50% of the population will have a score that is lower than the mean of 100, while the other 50% will have an IQ which is greater than the mean. This observation is clearly seen when you think about a bell curve.

We know from statistical properties of the standard normal distribution that about 68% of all IQ observations will fall within 1 standard deviation of the mean,  95% of all observations will fall within two standard deviations of the mean and 99.7% of all observations would fall within 3 standard deviations of the mean.

IQ test scores and percentiles – an example

So let’s contextualize this with an example. Suppose that I take a culture-fair IQ test with a standard deviation of 16 points. From the above-mentioned statistical properties of the standard normal deviation, we know that about 68% of the population should achieve IQ test scores between 84 and 116 points (i.e. the mean of 100 + or – 16), while just over 95% of the population would have IQ test scores between 68 and 132 (i.e. 100 + or –  2 x 16), and 99.7% of the population would score between 52 and 148. Put differently, most observations of the entire population of IQ test scores will be covered within 3 standard deviations of the mean (in this case 100 + or – 48 points * i.e. 16 x 3).

Although this is interesting to people who like statistics, IQ test scores are most interesting when converted into a percentile. A percentile provides a ranking of the score within the context of the population of test takers.

Suppose that on this culture-fair IQ test, I get a score of 117 points (IQ of 117). To understand the percentile, I will need a standard normal table (also known as a Z table). You can access a Z-table online for free here.

I  then need to convert my IQ score into a Z-score before I can work out the percentile. The Z-score can be calculated as follows: Z =(X – u) / SD. Where X is the test result (117 in this example), u = the average score of the entire population (100 for most IQ tests), and SD is the standard deviation of the test in question (SD=16 in this example).

The Z score for my theoretical test result is calculated as follows Z = (117 – 100) / 16 = 1.060

With this IQ test score of 117 now converted into a Z-score of 1.060, you can turn to the Wikipedia Z table and you start by looking up the first two digits of the Z score (in this case 1.060) in the vertical column of the table, and then following on the relevant row until you land in the cell which corresponds to the next two digits of the Z-score (in this case second and third decimals of  1.060). The Z-table percentage is 0.35543 (or about 35.5%). This percentage is the probability of the IQ score being between 100 and 117. However, to turn it into a percentile, we would need to add the left hand side of the Normal distribution. That is, we know that each tail of the bell curve accounts for 50% of the population, so given that 117 sits in the right half of the distribution (and we already worked out that 35.4% of the population have an IQ score between 100 and 117), we need to add the 50% of observations in the left tail – i.e. IQ test scores less than 100). Adding these two percentages gives 85.5%. In other words, about 85.5% of the population will obtain an IQ score which is 117 or less. This means that a score of 117 on an IQ test with a standard deviation of 16 points is better than roughly 85.5% of test takers for this test. Flipping it around, a score of 117 on this test would put my score in the top 14.5% of test takers (i.e. 100% minus the percentile of 85.5% = 14.5%).

If my IQ test score were 90 on the other hand, the Z-score would be calculated as follows: 90 – 100 / 16 = – 0.625. Looking at the Z table, in the column (remember that you are looking at the first two digits of the Z-score, in this case 0.625), and following the row for the last two digits (in this case the second and third decimals in 0.625). So the issue here is that we have a Z-score corresponding to a second decimal of 0.2 (i.e. 0.23237 or 23.2%) and a Z-score with a second decimal of 0.3 (i.e. 0.23565 or 23.6%), There is no percentage corresponding to second and third decimals of exactly 25. So the answer will be somewhere which is about half way between the two values, (i..e [0.23237 + 0.23565]/2 = 0.23401 or 23.4%).

Remember that the Z-score gives us the probability of observing a value which is between the mean (i.e. 100) and the score achieved (i.e. 90 in this case). So, the probability of observing an IQ between 90 and 100 is 23.4%, which means that an IQ of 90 is higher than about 26.6% of the population (i.e. 50% in the left hand tail of the bell  curve minus 23.4% = 26.6%). So an IQ score of 90 is said to be at the 26.6th percentile.

At, we automatically calculate percentiles for your IQ test scores. Take the test here.