Linear Interpolation (Edexcel GCSE Statistics: Higher): Revision Note

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.

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:

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

Answer:

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

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

So the median is in the third class interval

The median is in the $20<x\le 30$ 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 $20\le x\le 30$ 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

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

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

Answer:

$\frac{60}{2}=30$, 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  $\mathrm{frequency}\mathrm{density}=\frac{\mathrm{frequency}}{\mathrm{class}\mathrm{width}}$  as  $\mathrm{frequency}=\mathrm{frequency}\mathrm{density}\times \mathrm{class}\mathrm{width}$

 
$\begin{array}{l}4-8\mathrm{kg}:\mathrm{frequency}=1\times 4=4\\ 8-10\mathrm{kg}:\mathrm{frequency}=8\times 2=16\\ 10-12\mathrm{kg}:\mathrm{frequency}=9.5\times 2=19\end{array}$
 

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  $\mathrm{frequency}\mathrm{density}=\frac{\mathrm{frequency}}{\mathrm{class}\mathrm{width}}$   and solve for width
 

$\begin{array}{l}9.5=\frac{10}{\mathrm{width}}\\ \\ 9.5\times \mathrm{width}=10\\ \\ \mathrm{width}=\frac{10}{9.5}=\frac{20}{19}=1.0526...=1.05\left(2\mathrm{d}.\mathrm{p}.\right)\end{array}$

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

$10+1.05=11.05$

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