They are indicating that they scored lower than only 10% of the other test-takers. That obviously sounds very good, but the 10th percentile is well below the mean, right? Well, they are not really saying that they are in the tenth percentile. Let's say that someone were to report that they scored in the top 10th percentile of a test. This can be helpful when you want to find percentiles that may be presented differently. The same is true for the 10th and 90th percentiles. The 25th percentile and the 75th percentile are both 25 percentile points away from the mean, so their z-scores are both 0.675, with the only difference being the negative to show that the 25th percentile is below the mean. Notice that these percentiles are symmetric, just like the standard deviations. Standard normal distribution with z-scores for common percentiles. The graph below shows a standard normal distribution curve with a few common percentiles marked with their corresponding z-scores.įig. In order to do so, we recall the following definition of z-score. In fact, sometimes you will work with values that fall somewhere between the standard deviations, or you may be interested in a specific percentile that does not correspond to one of the standard deviations mentioned above, nor the mean.Īnd this is where the need of a normal distribution percentile formula arises. When working with a normal distribution, you will not just be interested in the percentile of the standard deviations, or the mean's percentile. Standard normal distribution showing the percentage of data below each standard deviation. The following normal distribution graph shows the corresponding percentage that lie below each standard deviation.įig. You can subtract the next standard deviation percentage to find the percentile of 2 standard deviations below the mean, 15.87% - 13.59% is 2.28%, or about the 2nd percentile. Standard Normal Distribution with standard deviation percentages.įor finding the percentile of a standard deviation below the mean, that is to the left of the mean, subtract the standard deviation's percentage from 50%.įor 1 standard deviation below the mean, find the percentile by subtracting 34.13% from 50% to get 15.87%, or about the 16th percentile. This is sometimes called the "68-95-99.7 Rule".
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