Linear Interpolation

In statistics, linear interpolation is often used to find the estimated median, quartiles or percentiles of a set of data and particularly when the data is presented in a group frequency table with class intervals. In this article we will look at how to do a linear interpolation calculation with the use of a table and graph to find the median, 1^st quartile and 3^rd quartile.

\[y = y_1 + (x-x_1) \frac{(y_2-y_1)}{(x_2-x_1)}\]

where:

x₂ and y₂ are the second coordinates.

x is the point to perform the interpolation.

y is the interpolated value.

Solved example for linear interpolation

How to do linear interpolation

There are a few useful steps that will help you compute the desired value such as the median, 1^st quartile and 3^rd quartile. We will go through each step with the use of an example so that it is clear.

Finding the median

The median is the value in the middle of the data.

The position of the median is at the \(\Big( \frac{n}{2} \Big)^{th}\) value, where n is the total cumulative frequency

In this example, n = 68

Step 1: Solve for the position of the median \(\frac{68}{2} = 34^{th} \space position\)

Step 2: Look for where the 34^th position lies in the data using the cumulative frequency.

According to the cumulative frequency, the 34^th value lies in the 41-50 class interval.

Step 3: Given the graph, use linear interpolation to find the specific median value.

We treat the segment of the graph where the class interval lies as a straight line and use the gradient formula to assist.

\(\text{Gradient} = \frac{(\text{Median cf - previous cf})}{(\text{upper bound - lower bound})} =\frac{(42-24)}{(50-41)} = 2\)

We can manipulate this formula and substitute the value of the median (m) as the upper bound and the position of the median as the median cf which is also equal to the gradient.

\(\text{Gradient} = \frac{(34-24)}{(m-41)}\)

So it follows that,

\(2 = \frac{(34-24)}{(m-41)} \quad 2 = \frac{10}{m-41} \quad m-41 = \frac{10}{2} \quad m-41 = 5 \quad m = 46\)

So the median is 46.

Finding the first quartile

The 1^st quartile is also known as the lower quartile. This is where the first 25% of the data lies.

The position of the 1^st quartile is the \(\Big(\frac{n}{4} \Big)^{th}\) value.

The steps to find the 1^st quartile are very similar to the steps to find the median.

Step 1: solve for the position of the 1^st quartile \(\frac{68}{4} = 17^{th} \text{ position}\)

Step 2: Look for where the 17^th position lies in the data using the cumulative frequency.

According to the cumulative frequency, the 17^th value lies in the 31-40 class interval.

Step 3: Given the graph, use linear interpolation to find the specific 1^st quartile value.

We treat the segment of the graph where the class interval lies as a straight line and use the gradient formula to assist.

\(\text{Gradient} = \frac{(1^{st}\text{quartile cf - previous cf})}{(\text{upper bound - lower bound})} =\frac{(24-16)}{(40-31)} = \frac{8}{9}\)

We can manipulate this formula and substitute the value of the 1^st quartile (Q₁) as the upper bound and the position of the 1^st quartile as the 1^st quartile cf which is also equal to the gradient.

\(\text{Gradient} = \frac{(17-16)}{(Q_1-31)}\)

It follows that,

\(\frac{8}{9} = \frac{(17-16)}{(Q_1 - 31)} \quad \frac{8}{9} = \frac{1}{Q_1 - 31} \quad Q_1 - 31 = \frac{9}{8} \quad Q_1 = 32.125\)

So the 1^st quartile is 32.125.

Finding the third quartile

The 1^st quartile is also known as the lower quartile. This is where the first 25% of the data lies.

The position of the 3^rd quartile is the \(\Big(\frac{3n}{4} \Big)^{th}\) value.

Step 1: solve for the position of the 3^rd quartile \(\frac{3(68)}{4} = 51^{st} \text{ position}\)

Step 2: Look for where the 51^st position lies in the data using the cumulative frequency.

According to the cumulative frequency, the 51^st value lies in the 61-70 class interval.

Step 3: Given the graph, use linear interpolation to find the specific 3^rd quartile value.

We treat the segment of the graph where the class interval lies as a straight line and use the gradient formula to assist.

\(\text{Gradient} = \frac{3^{rd} \text{quartile cf - previous cf}}{\text{upper bound - lower bound}} = \frac{(68-48)}{(70-61)} = \frac{20}{9}\)

We can manipulate this formula and substitute the value of the 3^rd quartile (Q₃) as the upper bound and the position of the 3^rd quartile as the 3^rd quartile cf which is also equal to the gradient.

\(\text{Gradient} = \frac{(51-48)}{(Q_3 -61)}\)

It follows that, \(\frac{20}{9} = \frac{(51-48)}{(Q_3 - 61)} \quad \frac{20}{9} = \frac{3}{Q_3 - 61} \quad Q_3 - 61 = \frac{27}{20} \quad Q_3 = 62.35\)

So the 3^rd quartile is 32.125.