11.025 Quantiles, deciles and percentiles

Quantiles

When we find the median value of a data set, we are finding a central value with the property that there are as many data points below this value as there are above it. That is, we are finding a value that splits the data set into two equal parts.

We then extended this idea in order to define the quartiles. These are three numbers that split a data set into four equal parts - so the first quartile is the median of the lower half of the set, and the third quartile is the median of the upper half. (The second quartile is, of course, just the median of the whole set.)

Extending this idea further still leads to the concept of quantiles. If there are $N$N numbers in the data set and we want to divide the set into $k$k parts, the associated quantile is a number which has, as nearly as possible, $\frac{N}{k}$Nk​ of the numbers are below it in the ordered set, with the remaining data points above it.

Most often we choose either $k=10$k=10 and the resulting quantiles are called deciles, or we choose $k=100$k=100 and call the resulting quantiles percentiles.

Clearly, the median should be the same as the $50$50th percentile, the first quartile should be the same as the $25$25th percentile and the third quartile should be the same as the $75$75th percentile. Similarly, for example, the $4$4th decile is the same as the $40$40th percentile. Thus, if we know how to calculate the percentiles, we automatically have a way of determining the quartiles and the deciles.

There are different methods for determining the percentiles of a data set, each giving slightly different results. The differences disappear when the data sets are large.

The simplest method is the following: to find the $p$pth percentile of a data set with $N$N elements, calculate $\frac{p}{100}\times N$p100​×N. The smallest integer that is greater than or equal to the result is the rank of the number in the data that will be taken to be the required percentile.

 

Worked example

Example 1

Find the $30$30th percentile of the following set of nine numbers: $14,19,23,24,31,33,40,42,56$14,19,23,24,31,33,40,42,56.

Think: Note that, once again, the data set is already arranged in ascending order. So the $30$30th percentile can be found $\frac{30}{100}$30100​ of the way along the data set. Remember that there are $n=9$n=9 scores.

Do:

$\text{Position }$Position	$=$=	$\frac{30}{100}\times9$30100​×9
	$=$=	$2.7$2.7th score

The nearest integer above $2.7$2.7 is $3$3. So, we take the third score to be the $30$30th percentile.

So, for this data set, the $30$30th percentile is $23$23.

Reflect: Note that the $25$25th percentile would also be $23$23 for this set of scores, which happens because the data set is so small. If the data set was much larger, the two percentiles would likely be different.

 

Practice questions

QUESTION 1

QUESTION 2