Percentiles and standard deviation | Statistics and data interpretation | Quantitative reasoning

Percentiles

Percentiles are an essential part of the GRE. You’ll interpret percentiles to understand how your score compares to other test takers. For example, if you scored in the 81st percentile, it means you scored higher than 81% of the test takers.

If you scored exactly in the middle of the group, what would your percentile be? It would be the 50th percentile. Scoring at the 50th percentile means:

You scored higher than 50% of test takers.
50% of test takers scored higher than you.

Percentiles can be tricky, especially when they show up in standard deviation problems. For now, remember this key idea: a larger percentile means a better relative score.

Box and whisker plot, range, interquartile range (IQR)

A box and whisker plot is a way to illustrate how measurements are distributed in a data set. Box and whisker plots connect closely to percentiles. Here’s an example labeled box and whisker plot:

Example box and whisker plot

Imagine the box and whisker plot above represents a set of 100 observations of how many times people use public transit in a week.

The whiskers show the extreme values in the data set.
The minimum observation here is 0.
The maximum observation here is 8.

The second quartile $Q_{2}$ is one of the easiest points to interpret. A quartile divides the data into four equal parts (each representing 25% of the data). The second quartile reaches 50% of the data, so $Q_{2}$ is the median. The median always sits at the 50th percentile.

In this example, the people who rode public transport 4 times a week are exactly in the middle of the distribution.

Can you guess what percentile $Q_{1}$ represents?

(spoiler)

If you guessed the 25th percentile, nice job!

The people who rode 2 times a week used public transport more often than 25% of the others, but less often than the other 75% of people.

Similarly, $Q_{3}$ represents the 75th percentile.

The range of this set is the maximum value minus the minimum value, which is $8 - 0 = 8$ .

The interquartile range (IQR) is different from the full range. The interquartile range is the difference between $Q_{3}$ and $Q_{1}$ . In this case, the interquartile range is $6 - 2 = 4$ .

How to calculate quartiles and interquartile range (IQR)

Calculating quartiles is straightforward once you know the process. Let’s walk through an example.

List $L$ consists of $11$ adjacent positive numbers, the largest being $33$ . What is the IQR of $L$ ?

To find the IQR, you need $Q_{1}$ and $Q_{3}$ , since:

$I QR = Q_{3} - Q_{1}$

For a sorted list:

$Q_{2}$ is the median (the 50th percentile).
$Q_{1}$ is the median of the lower half of the list.
$Q_{3}$ is the median of the upper half of the list.

Start by finding $Q_{2}$ (the median). Since there are 11 numbers, the median is the 6th value.

$L = [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33]$

The median of the list is $Q_{2} = 28$ .

Now split the original list into two halves (excluding the median), and find the median of each half.

$L_{1} L_{3} = [23, 24, 25, 26, 27] = [29, 30, 31, 32, 33]$

So $Q_{1} = 25$ and $Q_{3} = 31$ .

Now compute the IQR.

$I QR = Q_{3} - Q 1 = 31 - 25 = 6$

Now it’s your turn. Try a similar question:

List $M$ consists of $11$ adjacent positive numbers, the largest being $25$ . What is the IQR of $M$ ?

Try solving it, and then check your answer below.

(spoiler)

Answer: 6

$L = [15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]$

The median of the list is $Q_{2} = 20$ .

Now we split the original list into two parts, and find the median of each.

$L_{1} L_{3} = [15, 16, 17, 18, 19] = [21, 22, 23, 24, 25]$

So that means $Q_{1} = 17$ and $Q_{3} = 23$ .

$I QR = Q_{3} - Q 1 = 23 - 17 = 6$

Notice that the IQR for $M$ is the same as for $L$ in the walkthrough above. That’s because the lists have the same length and the same spacing between each number. They’re essentially the same shape of distribution - $M$ is just shifted so that it ends at $25$ instead of $33$ . Shifting all values by the same amount doesn’t change the IQR.

Let’s try one more to reinforce the core idea.

Quantity A: The IQR of a list of integers from 1 to 10
Quantity B: The IQR of a list of integers from 1 to 9

Try solving it, and then check your answer!

(spoiler)

The two quantities are equal.

The IQR for Quantity A $[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ :

The list has an even number of elements, so we split it in half to calculate Q1 and Q3
Q1 is the median of $[1, 2, 3, 4, 5] = 3$
Q3 is the median of $[6, 7, 8, 9, 10] = 8$
So the IQR is $8 - 3 = 5$

The IQR for Quantity B $[1, 2, 3, 4, 5, 6, 7, 8, 9]$ :

The list has an odd number of elements, so we drop the middle number, and then split it in half to calculate Q1 and Q3
Q1 is the median of $[1, 2, 3, 4] = 2.5$
Q3 is the median of $[6, 7, 8, 9] = 7.5$
So the IQR is $7.5 - 2.5 = 5$

Other important percentiles to know

There are two other “special” percentiles you should know:

84th percentile
16th percentile

These are special because they are one “standard deviation” (i.e. 34%) away from the center (50%) of the distribution.

84th percentile ( $50 + 34 = 84$ )
16th percentile ( $50 - 34 = 16$ )

Sometimes a question will give you an 84th percentile measurement and the standard deviation. From that information, you can calculate the mean.

For example, in a normal distribution with an 84th percentile measurement of 26 and a standard deviation of 3, the mean is $26 - 3 = 23$ .

Standard deviation and the normal distribution

The normal distribution is a well-known distribution pattern, commonly called the bell curve. Any time you see the term “normal distribution,” you should think of this figure.

Standard distribution

A normal distribution is perfectly symmetric: the left and right sides are mirror images.

At the center is 0, which means 0 standard deviations from the mean. The mean is the center of the distribution, so having 0 deviation from the mean puts you at the center.

The 34% represents the portion of the distribution between 0 and 1 standard deviation from the mean. A standard deviation is essentially the “average distance from the mean” (the technical definition is more precise, but this description is enough for GRE problems). You won’t be asked to compute a standard deviation on the GRE.

Notice that the percentages on either side of the 0 standard deviation line add up to 50: $2 + 14 + 34 = 50$ . That’s because 50% of values are above the mean and 50% are below the mean.

Some questions ask for the percentage of values above 1 standard deviation (not just above the mean). That’s the area to the right of the +1 line, which is $14 + 2 = 16$ .

The ETS uses approximations for the standard deviations on the GRE: 34%, 14%, and 2%.

The actual values are closer to 34.1%, 13.6%, and 2.3%.

If you’d like to learn more, check out Wikipedia’s normal distribution page.

Many problems will give you a mean and a standard deviation. With that information, you can label the values along the bottom of the distribution.

For example, if the mean is 82 and the standard deviation is 6, then one standard deviation (6) below the mean (82) is 76. The distribution would look like the figure below.

Standard distribution with example values

Now connect this back to percentiles. If you scored 2 standard deviations above the mean, your percentile would be 98. That’s because only 2% of the population is above +2 standard deviations, so 98% is below it.

Remember: the higher the percentile, the better.

Just like how you’re aiming for a top-percentile score on the GRE 🚀

Let’s try an example question using a normal distribution.

A car tire manufacturer is testing the average number of miles their tires can be used before the treads are fully stripped. The experiment involves running 3,600 tires on a treadmill that simulates the conditions a tire may experience on the road. The experiment found that the mean miles driven before the treads were stripped was 60,000 miles. The distances were normally distributed with a standard deviation of 750 miles. How many tires survived beyond 58,500 miles in the experiment?

Using what you’ve learned so far, you should be able to find the exact number.

(spoiler)

Answer: 3,528

When solving normal distribution / standard deviation questions, a good first step is to sketch the distribution and label the values you’re given.

Standard distribution with example values

The mean is 60,000 and the standard deviation is 750, so the center of the chart is 60,000 and each tick mark is 750 miles apart.

The value 58,500 is exactly two standard deviations below the mean:

$60, 000 - 2 (750) = 58, 500$

In a normal distribution, 2% of values lie below −2 standard deviations. That means 98% of values are above 58,500 miles:

$14 + 34 + 34 + 14 + 2 = 98%$

So the number of tires that survived beyond 58,500 miles is 98% of 3,600:

$(0.98) 3600 = 3528$

Of the 3,600 tires total, 3,528 survived beyond 58,500 miles.

This topic takes practice because you’re combining percentiles with the normal distribution chart. The main routine is always the same: locate the value in standard deviations, convert that to a percentage using the chart, and then apply the percentage to the total.

Common themes

Look out for questions that reference the $16$ th or $84$ th percentiles. These are the $- 1$ and $1$ standard deviation points respectively.
For a normal distribution the mean, median, and mode are all the same value, and they all represent the $50$ th percentile.
Do not assume that a distribution is the bell-shaped curve unless the question specifically states that it is a normal distribution.

Percentiles

If you scored exactly in the middle of the group, what would your percentile be? It would be the 50th percentile. Scoring at the 50th percentile means:

You scored higher than 50% of test takers.
50% of test takers scored higher than you.

Percentiles can be tricky, especially when they show up in standard deviation problems. For now, remember this key idea: a larger percentile means a better relative score.

Box and whisker plot, range, interquartile range (IQR)

A box and whisker plot is a way to illustrate how measurements are distributed in a data set. Box and whisker plots connect closely to percentiles. Here’s an example labeled box and whisker plot:

Example box and whisker plot

Imagine the box and whisker plot above represents a set of 100 observations of how many times people use public transit in a week.

The whiskers show the extreme values in the data set.
The minimum observation here is 0.
The maximum observation here is 8.

In this example, the people who rode public transport 4 times a week are exactly in the middle of the distribution.

Can you guess what percentile $Q_{1}$ represents?

(spoiler)

If you guessed the 25th percentile, nice job!

The people who rode 2 times a week used public transport more often than 25% of the others, but less often than the other 75% of people.

Similarly, $Q_{3}$ represents the 75th percentile.

The range of this set is the maximum value minus the minimum value, which is $8 - 0 = 8$ .

The interquartile range (IQR) is different from the full range. The interquartile range is the difference between $Q_{3}$ and $Q_{1}$ . In this case, the interquartile range is $6 - 2 = 4$ .

How to calculate quartiles and interquartile range (IQR)

Calculating quartiles is straightforward once you know the process. Let’s walk through an example.

List $L$ consists of $11$ adjacent positive numbers, the largest being $33$ . What is the IQR of $L$ ?

To find the IQR, you need $Q_{1}$ and $Q_{3}$ , since:

$I QR = Q_{3} - Q_{1}$

For a sorted list:

$Q_{2}$ is the median (the 50th percentile).
$Q_{1}$ is the median of the lower half of the list.
$Q_{3}$ is the median of the upper half of the list.

Start by finding $Q_{2}$ (the median). Since there are 11 numbers, the median is the 6th value.

$L = [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33]$

The median of the list is $Q_{2} = 28$ .

Now split the original list into two halves (excluding the median), and find the median of each half.

$L_{1} L_{3} = [23, 24, 25, 26, 27] = [29, 30, 31, 32, 33]$

So $Q_{1} = 25$ and $Q_{3} = 31$ .

Now compute the IQR.

$I QR = Q_{3} - Q 1 = 31 - 25 = 6$

Now it’s your turn. Try a similar question:

List $M$ consists of $11$ adjacent positive numbers, the largest being $25$ . What is the IQR of $M$ ?

Try solving it, and then check your answer below.

(spoiler)

Answer: 6

$L = [15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]$

The median of the list is $Q_{2} = 20$ .

Now we split the original list into two parts, and find the median of each.

$L_{1} L_{3} = [15, 16, 17, 18, 19] = [21, 22, 23, 24, 25]$

So that means $Q_{1} = 17$ and $Q_{3} = 23$ .

$I QR = Q_{3} - Q 1 = 23 - 17 = 6$

Let’s try one more to reinforce the core idea.

Quantity A: The IQR of a list of integers from 1 to 10
Quantity B: The IQR of a list of integers from 1 to 9

Try solving it, and then check your answer!

(spoiler)

The two quantities are equal.

The IQR for Quantity A $[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ :

The list has an even number of elements, so we split it in half to calculate Q1 and Q3
Q1 is the median of $[1, 2, 3, 4, 5] = 3$
Q3 is the median of $[6, 7, 8, 9, 10] = 8$
So the IQR is $8 - 3 = 5$

The IQR for Quantity B $[1, 2, 3, 4, 5, 6, 7, 8, 9]$ :

The list has an odd number of elements, so we drop the middle number, and then split it in half to calculate Q1 and Q3
Q1 is the median of $[1, 2, 3, 4] = 2.5$
Q3 is the median of $[6, 7, 8, 9] = 7.5$
So the IQR is $7.5 - 2.5 = 5$

Other important percentiles to know

There are two other “special” percentiles you should know:

84th percentile
16th percentile

These are special because they are one “standard deviation” (i.e. 34%) away from the center (50%) of the distribution.

84th percentile ( $50 + 34 = 84$ )
16th percentile ( $50 - 34 = 16$ )

Sometimes a question will give you an 84th percentile measurement and the standard deviation. From that information, you can calculate the mean.

For example, in a normal distribution with an 84th percentile measurement of 26 and a standard deviation of 3, the mean is $26 - 3 = 23$ .

Standard deviation and the normal distribution

The normal distribution is a well-known distribution pattern, commonly called the bell curve. Any time you see the term “normal distribution,” you should think of this figure.

Standard distribution

A normal distribution is perfectly symmetric: the left and right sides are mirror images.

At the center is 0, which means 0 standard deviations from the mean. The mean is the center of the distribution, so having 0 deviation from the mean puts you at the center.

Notice that the percentages on either side of the 0 standard deviation line add up to 50: $2 + 14 + 34 = 50$ . That’s because 50% of values are above the mean and 50% are below the mean.

Some questions ask for the percentage of values above 1 standard deviation (not just above the mean). That’s the area to the right of the +1 line, which is $14 + 2 = 16$ .

The ETS uses approximations for the standard deviations on the GRE: 34%, 14%, and 2%.

The actual values are closer to 34.1%, 13.6%, and 2.3%.

If you’d like to learn more, check out Wikipedia’s normal distribution page.

Many problems will give you a mean and a standard deviation. With that information, you can label the values along the bottom of the distribution.

For example, if the mean is 82 and the standard deviation is 6, then one standard deviation (6) below the mean (82) is 76. The distribution would look like the figure below.

Standard distribution with example values

Remember: the higher the percentile, the better.

Just like how you’re aiming for a top-percentile score on the GRE 🚀

Let’s try an example question using a normal distribution.

A car tire manufacturer is testing the average number of miles their tires can be used before the treads are fully stripped. The experiment involves running 3,600 tires on a treadmill that simulates the conditions a tire may experience on the road. The experiment found that the mean miles driven before the treads were stripped was 60,000 miles. The distances were normally distributed with a standard deviation of 750 miles. How many tires survived beyond 58,500 miles in the experiment?

Using what you’ve learned so far, you should be able to find the exact number.

(spoiler)

Answer: 3,528

When solving normal distribution / standard deviation questions, a good first step is to sketch the distribution and label the values you’re given.

Standard distribution with example values

The mean is 60,000 and the standard deviation is 750, so the center of the chart is 60,000 and each tick mark is 750 miles apart.

The value 58,500 is exactly two standard deviations below the mean:

$60, 000 - 2 (750) = 58, 500$

In a normal distribution, 2% of values lie below −2 standard deviations. That means 98% of values are above 58,500 miles:

$14 + 34 + 34 + 14 + 2 = 98%$

So the number of tires that survived beyond 58,500 miles is 98% of 3,600:

$(0.98) 3600 = 3528$

Of the 3,600 tires total, 3,528 survived beyond 58,500 miles.

Common themes

Look out for questions that reference the $16$ th or $84$ th percentiles. These are the $- 1$ and $1$ standard deviation points respectively.
For a normal distribution the mean, median, and mode are all the same value, and they all represent the $50$ th percentile.
Do not assume that a distribution is the bell-shaped curve unless the question specifically states that it is a normal distribution.