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 **1.0**60) 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 1.0**60**). 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 **0.6**25), and following the row for the last two digits (in this case the second and third decimals in 0.6**25**). 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 iq-brain.com, we automatically calculate percentiles for your IQ test scores. Take the test **here**.