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, 1st quartile and 3rd quartile.
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Team Linear Interpolation Teachers
The linear interpolation formula is the simplest method used to estimate the value of a function between any two known points. This formula is also useful for curve fitting using linear polynomials. This formula is often used for data forecasting, data prediction and other mathematical and scientific applications. The linear interpolation equation is given by:
\[y = y_1 + (x-x_1) \frac{(y_2-y_1)}{(x_2-x_1)}\]
where:
x1 and y1 are the first coordinates.
x2 and y2 are the second coordinates.
x is the point to perform the interpolation.
y is the interpolated value.
The best way to understand linear interpolation is through the use of an example.
Find the value of y if x = 5 and some set of value given are (3,2), (7,9).
Step 1: First assign each coordinate the right value
x = 5 (note that this is given)
x1 = 3 and y1 = 2
x2 = 7 and y2 = 9
Step 2: Substitute these values into the equations, then get the answer for y.
\(y = 2 +(5-3)\frac{(9-2)}{(7-3)} \quad y = \frac{11}{2}\)
There are a few useful steps that will help you compute the desired value such as the median, 1st quartile and 3rd quartile. We will go through each step with the use of an example so that it is clear.
In this example, we will look at grouped data with class intervals.
| Class | Frequency |
| 0-10 | 5 |
| 11-20 | 10 |
| 21-30 | 1 |
| 31-40 | 8 |
| 41-50 | 18 |
| 51-60 | 6 |
| 61-70 | 20 |
Frequency is how often a value in a specific class appears in the data.
Step 1: Given the class and the frequency, you have to create another column called the cumulative frequency (also known as CF).
Cumulative frequency is therefore defined as the running total of frequencies.
| Class | Frequency | CF |
| 0-10 | 5 | 5 |
| 11-20 | 10 | 15 |
| 21-30 | 1 | 16 |
| 31-40 | 8 | 24 |
| 41-50 | 18 | 42 |
| 51-60 | 6 | 48 |
| 61-70 | 20 | 68 |
Step 2: Plot the cumulative frequency graph. To do this, you plot the upper boundary of the class against the cumulative frequency.
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 34th position lies in the data using the cumulative frequency.
According to the cumulative frequency, the 34th 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.
The 1st quartile is also known as the lower quartile. This is where the first 25% of the data lies.
The position of the 1st quartile is the \(\Big(\frac{n}{4} \Big)^{th}\) value.
The steps to find the 1st quartile are very similar to the steps to find the median.
Step 1: solve for the position of the 1st quartile \(\frac{68}{4} = 17^{th} \text{ position}\)
Step 2: Look for where the 17th position lies in the data using the cumulative frequency.
According to the cumulative frequency, the 17th value lies in the 31-40 class interval.
Step 3: Given the graph, use linear interpolation to find the specific 1st 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 1st quartile (Q1) as the upper bound and the position of the 1st quartile as the 1st 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 1st quartile is 32.125.
The 1st quartile is also known as the lower quartile. This is where the first 25% of the data lies.
The position of the 3rd quartile is the \(\Big(\frac{3n}{4} \Big)^{th}\) value.
Step 1: solve for the position of the 3rd quartile \(\frac{3(68)}{4} = 51^{st} \text{ position}\)
Step 2: Look for where the 51st position lies in the data using the cumulative frequency.
According to the cumulative frequency, the 51st value lies in the 61-70 class interval.
Step 3: Given the graph, use linear interpolation to find the specific 3rd 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 3rd quartile (Q3) as the upper bound and the position of the 3rd quartile as the 3rd 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 3rd quartile is 32.125.
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What is linear interpolation?
Linear interpolation is a method to fit a curve using linear polynomials.
How do you calculate linear interpolation?
How to calculate linear interpolation: Linear interpolation can be calculated using the formula
y=y1+(x-x1)(y2-y1)/(x2-x1)
where,
x1 and y1 are the first coordinates.
x2 and y2 are the second coordinates.
x is the point to perform the interpolation.
y is the interpolated value.
How do you use linear interpolation?
How to use linear interpolation: Linear interpolation can be be used by substituting the values of x1, x2, y1 and y2 in the below formula
y=y1+(x-x1)(y2-y1)/(x2-x1)
where,
x1 and y1 are the first coordinates.
x2 and y2 are the second coordinates.
x is the point to perform the interpolation.
y is the interpolated value.
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