One of the best things about a normal distribution of data is that, well, it’s normal! Because you know what to expect from it, you can figure out a lot of things about the data it’s describing, since a standard normal distribution having a mean of 0 and a standard deviation of 1, is proportional to the data set that it is describing.
So, for any data set, you can know what percentage of the data is in a particular section of the graph. In particular, the percentage you will care the most about is the percentage of the data that is below your desired value, commonly known as the percentile.
In this article, we will learn more about percentages and percentiles from a normal distribution.
Normal Distribution Percentile Meaning
A normal distribution is a probability distribution where the data is distributed about the mean symmetrically to look like a bell-shaped curve, which is sometimes called a density curve.
Normal distributions are generally more suitable for large data sets. Many naturally occurring data, like test scores or organisms’ mass, tend to pattern themselves close to a normal distribution.
The normal distribution curve shown in the graph below, shows that the majority of the data is clustered around the middle of the graph, right where the mean is located.
The graph then tapers off towards the left and the right ends, to show smaller portion of the data far from the mean. Half of the data falls below the mean, and half of the data falls above the mean and thus, the mean is also the median of the data. The highest point on the graph is located at the middle of the graph as well, therefore this is where the mode is.
So, for a normal distribution, the mean, median, and mode are all equal.
Furthermore, the curve is divided into pieces by the standard deviations. The area under the normal distribution curve represents 100% of the data. For a standard normal distribution, this means that the area under the curve is equal to 1.
A specific percentage of the data is assigned to each standard deviation away from the mean on a normal distribution. These specific percentages are called the Empirical Rule of Normal Distribution,
- About 68% of the data falls within 1 standard deviation of the mean.
- About 95% of the data falls within 2 standard deviations of the mean.
- About 99.7% (almost all of teh data!) falls within 3 standard deviations of the mean.
This is sometimes called the "68-95-99.7 Rule".
Standard Normal Distribution with standard deviation percentages.
Those percentages are very helpful in knowing information about the repartition of the data. But one of the most important pieces of information to know about a data value in a normal distribution, is how much of the data it is greater than or less than a specific value, called the percentile.
The percentile for a normal distribution is a value that has a specific percentage of the observed data below it.
For a standardized test like the GRE test, you would receive both your score on the test as well as what percentage of test takers tested below your score. This tells you where a certain data value, here your score, lies relative to the rest of the data, comapring to the scores of the test takers.
Your score is called the percentile.
Percentile is a cumulative measurement, it is the sum of all the sections of percentages below that value. Many times, a value’s percentile is reported alongside the value itself.
Normal Distribution Percentile of Mean
As stated earlier in the above paragraph, the mean in the normal distribution curve lies right in its middle. The curve distributes thus the data symmetrically about the mean, that is 50% of the data are above the mean and 50% of the data are below the mean. This means that the mean is the 50th percentile of the data.
For a normal distribution probability, the normal distribution percentile of mean, is the 50th percentile.
We take the following example to understand this better.
If you were to score the average test score on a standardized test, your score report would say that you fall in the 50th percentile. That can sound bad at first, since it sounds like you got a 50% on the test, but it is simply telling you where you fall relative to all the other test-takers.
The 50th percentile would make your score perfectly average.
Does the Standard deviation has a percentile of its own as well? Let's figure this out in the next paragraph!
Normal Distribution Percentile of Standard Deviation
A very good question that one could have is the following, what is the percentile for each standard deviation?
Well, knowing that the mean is the 50th percentile, and recalling what does each percentage represent in every section of the normal distribution graph, you can figure out the percentile at each standard deviation.
For 1 standard deviation above the mean, that is to the right of the mean, find the percentile by adding the 34.13% above the mean to the 50% to get 84.13%. Usually for percentile, you round to the nearest whole number.
So, 1 standard deviation is about the 84th percentile.
If you wanted to find the percentile of 2 standard deviations, you would continue to add the percentages to the right of the mean to 50%. Therefore, the second standard deviation's percentile is 13.59% and 34.13% added to 50%, that gives you 97.72%, or about the 98th percentile.
And thus, 2 standard deviations are about the 98% percentile.
For 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%.
For 1 standard deviation below the mean, find the percentile by subtracting 34.13% from 50% to get 15.87%, or about the 16th percentile.
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.
The following normal distribution graph shows the corresponding percentage that lie below each standard deviation.
Fig. 1. Standard normal distribution showing the percentage of data below each standard deviation.
Normal Distribution Percentile Formula
When working with a normal distribution, you will not just be interested in the percentile of the standard deviations, or the mean's percentile. 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.
And this is where the need of a normal distribution percentile formula arises. In order to do so, we recall the following definition of z-score.
For further explanation on how z-scores are found, see the Z-score article.
The z-score inidicates how much a given value differs from a standard deviation.
For a normal distribution with a mean of \(\mu\) and a standard deviation of \(\sigma\), the z-score of any data value \(x\) is given by, \[Z=\frac{x-\mu}{\sigma}.\]
The above formula recenters the data around a mean of 0 and a standard deviation of 1, so that we can compare all normal distributions.
The importance of the z-score is that not only it tells you about the value itself, but where it is located on the distribution.
Conversely, in order to find a value based on a given percentile, the z-score formula can be reformulated into \[x=\mu+Z\sigma.\]
Luckily, you probably won't have to calculate the percentile every time for the z-score you want, that would be rather burdensome! Instead, you can use a z-score table, like the ones below.
A z-score table has the proportion of the data that falls below each z-score so that you can find the percentile directly.
Fig. 2. Negative z-score table for a normal distribution
Fig. 3. Positive z-score table for a normal distribution.
How to read a z-score table in order to find the percentile?
Once you have found your z-score, follow these steps for using the z-score to find the corresponding percentile. Most z-score tables show z-scores out to the hundredths place, but you can find more precise tables if needed.
Reading a z-score table can be done using the following steps,
Step 1. Look at the z-score you are given or have found.
Step 2. Look along the left side of the table, which shows the ones and the tenths places of your z-score. Find the row that matches your first two digits.
Step 3. Look along the top of the table, which shows the hundredths place. Find the column that matches your third digit.
Step 4. Find the intersection of the row and the column that matches your ones, tenths, and hundredths places. This is the proportion of data below your z-score, which is equal to the percentage of data below your z-score.
Step 5. Multiply by 100 to get a percentage. Generally, you round to the nearest whole number to get a percentile.
For a standard normal distribution, what is the percentile of 0.47?
Solution:
Step 1. For the standard normal distribution, this value is the same thing as the z-score. It is the number of standard deviations away from the mean. It is also to the right of the mean, so it should be a percentile higher than the 50th.
Step 2. Using the z-score table, the ones and tenths places are 0 and 4, so look at the entire row next to 0.4.
Step 3. The hundredths place is 7, or 0.07. Look at the column below 0.07.
Step 4. The intersection of the 0.4 row and the 0.07 column is 0.6808.
Step 5. So 68.08% of the data is below 0.47. Therefore, 0.47 is about the 68th percentile of a standard normal distribution.
Normal Distribution Percentile Graph
The graph below shows a standard normal distribution curve with a few common percentiles marked with their corresponding z-scores.
Fig. 4. Standard normal distribution with z-scores for common percentiles.
Notice that these percentiles are symmetric, just like the standard deviations. 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. The same is true for the 10th and 90th percentiles.
This can be helpful when you want to find percentiles that may be presented differently.
Let's say that someone were to report that they scored in the top 10th percentile of a test. 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. They are indicating that they scored lower than only 10% of the other test-takers. This is equivalent to saying they scored higher than 90% of the test-takers, or rather scored in the 90th percentile.
Knowing that normal distribution is symmetric allows flexibility in how we view the data.
The graphs above and the z-score tables all are based on the standard normal distribution that has a mean of 0 and a standard deviation of 1. This is used as the standard so that it is scalable for any data set.
But, obviously, most data sets do not have a mean of zero or a standard deviation of 1. That is what the z-score formulas can help with.
Examples of Normal Distribution Percentile
Growth charts, test scores, and probability problems are common problems you will see when working with normal distributions.
A farmer has a new calf on his ranch, and he needs to weigh it for his records. The calf weighs \(46.2\) kg. He consults his Angus calf growth chart and notes that the average weight of a newborn calf is \(41.9\) kg with a standard deviation of \(6.7\) kg. In what percentile is his calf's weight?
Solution:
You need to start by finding the z-score of the calf's weight. For this, you will need the formula \[Z=\frac{x-\mu}{\sigma}.\]
For this breed's growth chart, the mean is \(\mu =41.9\), the standard deviation is \(\sigma =6.7\), and the value \(x=46.2\). Substitute these values into the formula to get, \[Z=\frac{46.2-41.9}{6.7}=\frac{4.3}{6.7} \approx 0.64.\]
Now turn to your z-score table. Find the row for \(0.6\) and the column for \(0.04.\)
Fig. 5. Finding percentile from a z-score table for a normal distribution.
The row and column intersect at \(0.73891\). So, multiply by \(100\) to find that a proportion of 73.891% of the population falls below the z-score \(0.64.\) Therefore, the calf's weight is in about the 74th percentile.
You may also need to find a value based on a certain percentile. For the most part, that will involve doing the steps above in reverse.
Mary is taking the GRE test in order to apply for graduate school. She wants to have a strong chance of getting into the school of her dreams and decides to try and score in the 95th percentile. She does some research and finds that the average GRE score is \(302\) with a standard deviation of \(15.2.\) What score should she be aiming for?
Solution:
For this problem, you start with the z-score table. Find the cell that contains the value closest to 95%, which will be about \(0.95\) in the table.
Fig. 6 Finding z-score from percentile.
The first value that is at least \(0.95\) is the cell shown above with \(0.95053\) in it. Look at the label for its row, \(1.6\), and its column, \(0.05\), to find the z-score for the 95th percentile. The z-score will be \(1.65.\) This means that Mary needs to score about \(1.65\) standard deviations above the mean of \(302\). To find the corresponding test score, use the formula \[x=\mu+Z\sigma.\]
Substitute in the values for \(\mu\), \(Z\), and \(\sigma\) to get, \[x=302+1.65(15.2)\approx 327.\]
So, Mary needs to score at least a 327 on the GRE to meet her goal.
Normal Distribution Proportion
Normal distributions are so useful because they are proportional to each other via the z-score and percentiles.
Each normal distribution may have its own mean and standard deviation, which can affect the spread of the data. But the proportion of the data that lies within each standard deviation is the same across all normal distributions. Each area under the curve represents a proportion of the data set or the population.
This means that you can find the percentile for any value in any normal distribution as long as you know the mean and standard deviation.
Let's look at the two following examples of standardized tests to compare.
Two teachers gave the same group of students their final exams and are comparing their students' results. The math teacher reports a mean score of \(81\) with a standard deviation of \(10\). The history teacher reports a mean score of \(86\) with a standard deviation of \(6.\)
The graph below shows both exams' normal distributions.
Fig. 7. Comparing Normal Distributions with different means and standard deviations.
Both graphs represent normal distributions of the students' scores. But they look different side-by-side.Because the students scored higher on average on their history exam, the center of the history exam graph is farther to the right. And because the students had a higher standard deviation, which is basically a greater range of scores, on their math exam, the graph is lower and more spread out. This is because both graphs represent the same number of students.For both graphs, the center represents the 50th percentile, and thus the "typical" exam score.
By the empirical rule of normal distributions, about 68% of the students scored within 1 standard deviation of the mean. So for the two exams, this 68% would represent the same number of students. But for the math exam, the middle 68% of students scored between \(71\) and \(91\), whereas the middle 68% of students scored between \(80\) and \(92\) on the history exam. Same number of students covering different data values. A student who scored in the 90th percentile on the math exam and another student who scored in the 90th percentile on the history exam both performed the same relative to the rest of the students, even though their scores differed.The data represented by the graphs is proportional to each other, even though the graphs look different.
Comparing Data Using Normal Distribution
Because all normal distributions are proportional, you are able to compare the data from two different sets, with different means and standard deviations, as long as both are normally distributed.
Mary took the GRE test , but she has also been thinking about going to law school, for which she needed to take the LSAT test.
Now she wants to compare her scores, and maybe her chances of getting into the program of her choice, but the two tests are scored differently.
Her GRE score was \(321\) with the mean of \(302\) and the standard deviation of \(15.2\). And her LSAT score was \(164\) with a mean of \(151\) and with a standard deviation of \(9.5\).
Which test did she perform better on? What percentile did she fall in for each test?
Solution:
Start with the GRE score and the formula \[Z=\frac{x-\mu}{\sigma}.\] Substitute in the mean, standard deviation, and her score for the GRE, to get \[Z=\frac{321-302}{15.2}=1.25.\]
Look at the z-score table above to find the proportion for the z-score \(1.25.\) The proportion of data below \(1.25\) is \(0.89435\). This represents a percentage of 89.435%, or about the 89th percentile.
Now look at her LSAT score, and substitute its mean, standard deviation, and score into the formula, \[Z=\frac{164-151}{9.5}\approx 1.37.\]
You can tell just from the z-scores that she performed better on the LSAT since \(1.37\) standard deviations is farther to the right than \(1.25\) standard deviations.
But the question also asks for the percentile she achieved on each test. So, once again, consult the z-score table above and find the proportion corresponding to \(1.37\), which is \(0.91466.\) This is a percentage of 91.466% or about the 91st percentile.
So, she performed better than 89% of the other GRE test-takers and better than 91% of the other LSAT test-takers.
Normal distribution Percentile - Key takeaways
- For a normal distribution, the z-score is the number of standard deviation away from the mean a value is, and the percentile is the percentage of data that lies below that z-score.
- For a z-score \(Z\) within a normal distribution, a data value \(x\), a mean \(\mu\), and a standard deviation \(\sigma\), you can use either formula: \[Z=\frac{x-\mu}{\sigma}.\] \[x=\mu+Z\sigma.\]
- You need a z-score table to find the proportion of the data that corresponds to each z-score so you can find the percentile.
- For a normal distribution, the mean is the 50% percentile.
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