Linear Interpolation (Edexcel GCSE Statistics)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Median from Grouped Data

How do I find the median for grouped data?

Grouped data doesn’t contain the individual data values
- So we can’t find the exact median in the usual way
We can estimate the median for grouped data using linear interpolation
- This assumes the data is evenly spread across the class containing the median
STEP 1
Identify the class interval containing the median
- This is the class containing the ‘ $\frac{n}{2}$ ^th’ data value
  - Divide the total number of values, n, by 2
- e.g. if there are 80 data values, $\frac{n}{2} = \frac{80}{2} = 40$
  - Consider cumulative frequencies until you find the class interval containing the 40^th value
STEP 2
Find ‘how far into’ that class interval the $\frac{n}{2}$ ^th data value is
- e.g. if the interval with the median contains the 36^th through 43^rd data values
  - then the 40^th value is ‘5 values in’ (36, 37, 38, 39, 40)
  - and there are 8 values in the interval
  - so the 40^th value is $\frac{5}{8}$ of the way into the interval
STEP 3
Multiply the class width of the class interval containing the median by the fraction found in Step 2
- e.g. if the interval with the median is $50 \leq x \leq 70$
  - the class width is $70 - 50 = 20$
  - $20 \times \frac{5}{8} = 12.5$
STEP 4
Add the result from Step 3 to the lower boundary of the class interval containing the median
- The result is the estimated median
- e.g. lower bound of $50 \leq x \leq 70$ is 50
  - The estimated median is $50 + 12.5 = 62.5$
The estimated median can also be found using the following formula:
- $estimated median = L + \frac{\frac{n}{2} - C}{f} \times w$
  - L is the lower boundary of the class interval containing the median
  - n is the total number of data values
  - C is the cumulative frequency of all the class intervals before the one containing the median
  - f is the frequency of (i.e. the number of values in) the class interval containing the median
  - w is the width of the class interval containing the median (upper boundary minus lower boundary)
- This formula combines all four steps of the process
  - but it is not on the exam formula sheet
  - So if you want to use it you’ll need to remember it

Examiner Tips and Tricks

The formula can be tricky to remember correctly
- It’s better to understand how the method works
- Then you don’t need to remember the formula
Remember that the median found this way is an estimate
- You can’t find the exact median without knowing all the data values

Worked Example

A student collected data about the length of time (x hours) students in his school spent listening to music in a given week. He collected data from 50 students in total. The following table summarises the data:

Time spent, x (hours)	Number of students
0 ≤ x ≤ 10	3
10 < x ≤ 20	19
20 < x ≤ 30	12
30 < x ≤ 40	10
40 < x ≤ 50	5
50 < x ≤ 60	1

Work out an estimate for the median amount of time spent listening to music by the students.

STEP 1: Identify the class interval containing the median

Divide the total number of values, n, by 2

Here n = 50

$\frac{50}{2} = 25$

So we’re looking for the interval with the 25^th value

Note that this is different from finding the median from a set of data values

In that case we would be looking for the value halfway between the 25^th and 26^th values

With linear interpolation we don’t have to worry about that!

Add a cumulative frequency column to the table and work out the cumulative frequencies

Time spent, x (hours)	Number of students	Cumulative Frequency
0 ≤ x ≤ 10	3	3
10 < x ≤ 20	19	22
20 < x ≤ 30	12	34
30 < x ≤ 40	10	44
40 < x ≤ 50	5	49
50 < x ≤ 60	1	50

The second class interval goes up to the 22^nd data value and the third class interval goes up to the 34^thdata value

So the median is in the third class interval

The median is in the $20 < x \leq 30$ class interval

STEP 2: Find how far into the class interval the $\frac{n}{2}$ ^th data value is

The interval with the median contains the 23^rd through 34^th data values

The 25^th data value is ‘3 values in’ to the interval (23, 24, 25)

And there are 12 data values in the interval

$\frac{3}{12} = \frac{1}{4}$

So the median is 1/4 of the way into the interval

STEP 3: Multiply the class width by the fraction found in Step 2

Subtract the lower boundary from the upper boundary to find the class width of the $20 \leq x \leq 30$ interval

30-20=10

Multiply by the fraction in Step 2

$10 \times \frac{1}{4} = 2.5$

STEP 4: Add the result from Step 3 to the lower boundary of the class interval

20 + 2.5 = 22.5

Don’t forget the units in your answer!

Estimated median = 22.5 hours

How do I find the median for data on a histogram?

A histogram is a way of representing grouped data as a diagram
- See the 'Histograms & Frequency Polygons' revision note for full details
The connection between the frequency density shown on the histogram and the frequency that would be shown in a grouped data table is given by the formula
- $frequency density = \frac{frequency}{class width}$
If you are asked to estimate a median for data in a histogram there are two options:
- You can recreate the grouped data table using the frequency density formula, and then follow the method given above
- Or you can work out the estimated median directly from the histogram
  - See the following Worked Example for how to do this

Worked Example

The histogram shows the weight, in kg, of 60 newborn bottlenose dolphins.

A histogram for the weights of newborn bottlenose dolphins

Find an estimate for the median weight of the dolphins in the sample, giving your answer correct to two decimal places.

$\frac{60}{2} = 30$ , so to estimate the median we need to find the weight of the 30th dolphin

To find the frequencies represented by the different bars, rearrange $frequency density = \frac{frequency}{class width}$ as $frequency = frequency density \times class width$

$4 - 8 kg : frequency = 1 \times 4 = 4 8 - 10 kg : frequency = 8 \times 2 = 16 10 - 12 kg : frequency = 9.5 \times 2 = 19$

The first two classes have a cumulative frequency of $4 + 16 = 20$
So the median is going to be '10 dolphins into' the 10-12 kg class

The height (frequency density) of the 10-12 kg bar is 9.5
We need to find what width would give a frequency of 10
Use $frequency density = \frac{frequency}{class width}$ and solve for width

$9.5 = \frac{10}{width} 9.5 \times width = 10 width = \frac{10}{9.5} = \frac{20}{19} = 1.0526 . . . = 1.05 (2 d . p .)$

That means that the median lies 1.05 kg into the 10-12 kg class interval

$10 + 1.05 = 11.05$

A histogram showing the part of the 10-12 kg class interval calculated in the answer

Estimated median = 11.05 kg (2 d.p.)

Last updated: 7 May 2024

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Previous:Mode & Mean from Grouped DataNext:Transforming Data

Linear Interpolation (Edexcel GCSE Statistics)

Revision Note

Median from Grouped Data

How do I find the median for grouped data?

Examiner Tips and Tricks

Worked Example

How do I find the median for data on a histogram?

Worked Example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Author: Dan Finlay