Averages from Tables

Averages from Tables & Charts

How do we find averages if there are lots of values?

In reality there will be far more data to work with than just a few numbers
In these cases the data is usually organised in such a way to make it easier to follow and understand
- For example in a table or chart
We can still find the mean, median and mode but have to ensure we understand what the table or chart is telling us

How do I find averages from a table or chart?

Finding the median and mode from tables/charts is fairly straightforward once you understand what the table/chart is telling you
Tables allow data to be summarised neatly
- and quite importantly, they put the data into order

Finding the mean from (discrete) data presented in tables

The mean can be found as you long as understand what a table is telling you
- Tables tell us the data value
  - e.g. the number of pets per household
- and the frequency of that data value
  - e.g. the number of households with that number of pets

   STEP 1
   Add a column to the table and work out "data value" × "frequency"
   (This is effectively doing the 'adding up' part of finding the mean in stages)

   STEP 2
   Find the total of the extra column to give the overall total of the data values

   STEP 3
   Find the mean by dividing this total by the total of the frequency column
   i.e. divide the total of the data values by the number of data values

Finding the median from (discrete) data presented in tables

The median is the middle value when the data is in order
The position of the median can be found by using $\frac{n + 1}{2}$ , where $n$ is the number of data values
Use the table to deduce where the ${(\frac{n + 1}{2})}^{th}$ value lies
- e.g. if the median is the 7 ^th value and the frequency of the first two rows are 4 and 7
  - the median will be one of the 7 values the second row of that table

Finding the mode (or modal value)

The mode (or modal value) is simple to identify
- Look for the highest frequency
  - and thus find the corresponding data value
- Make sure you do not confuse the data value with the frequency!
  - The frequency (in a table) tells us the row the mode is in

Worked example

The bar chart shows data about the shoe sizes of pupils in class 11A.

Find the mean shoe size for the class,

Although the data is given in a bar chart, this is essentially the same as a table.
Rewrite it as a table but add an extra column to help find the total of all the shoe sizes.

Shoe size ( x )	Frequency ( f )	x f
6	1	6 × 1 = 6
6.5	1	6.5 × 1 = 6.5
7	3	7 × 3 = 21
7.5	2	7.5 × 2 = 15
8	4	8 × 4 = 32
9	6	9 × 6 = 54
10	11	10 × 11 = 110
11	2	11 × 2 = 22
12	1	12 × 1 = 12
Total	31	278.5

Mean $= \frac{278.5}{31} = 8.983 870 . . .$

Mean = 8.98 (3 s.f.)

Note that the mean does not have to be an actual shoe size.

Find the median shoe size,

The bar chart/table has the data in order already so find the position of the median.

$\frac{n + 1}{2} = \frac{31 + 1}{2} = \frac{32}{2} = 16$

The median is the 16 ^th value.
There are 1 + 1 + 3 + 2 + 4 = 11 values in the first five rows of the table.
There are 11 + 6 = 17 values in the first six rows of the table.
Therefore the 16 ^th value must be in the sixth row.

Median shoe size is 9

Suggest a reason the shop owner may wish to know the modal shoe size of their customers.

A shop owner would want to know the modal shoe size of their customers as this size will be more likely to sell than other sizes so the shop owner should order more shoes in the modal size to stock the shop with

Worked example

15 students were timed how long it took them to solve a maths problem. Their times, in seconds, are given below.

12	10	15	14	17
11	12	13	9	21
14	20	19	16	23

Find the mean and median times.

There are a fair amount of numbers so it may be wise to do the adding up in bits - we've used rows.

12 + 10 + 15 + 14 + 17 = 68
11 + 12 + 13 + 9 + 21 = 66
14 + 20 + 19 + 16 + 23 = 92

$\begin{array}{rcl} Mean & = & \frac{68 + 66 + 92}{15} \\ = & \frac{226}{15} \\ = & 15.066 666 . . . \end{array}$

For the median, the data needs to be in order first.

$9 10 11 12 12 13 14 14 15 16 17 19 20 21 23$

Mean = 15.1 seconds (3 s.f.)
Median time = 14 seconds

Comment on the mode of the data.

The mode (or lack of) is easiest to see from the data listed in order in the median question above.

There are two modes (bi-modal) - 12 and 14 seconds

Alternatively we could say there is no mode.

Averages from Grouped Data

What is grouped data and why use it?

Some data for a particular scenario can vary a lot
- For example the heights of people
  - particularly if you include a mixture of children and adults
Data like height is also continuous (data that can be measured)
- Such data is difficult, even using a table, to list every value
- There is also little difference between someone who is, say, 176 cm tall and someone who is 177cm tall
So we often group data into classes but that leads to one crucial point
- When data is grouped, we lose the raw data
- With height, we know that 10 people have a height between 150 cm and 160 cm
  - but we won't know exactly what those 10 heights are
This means we cannot find the actual mean, median and mode from grouped data by their original definitions
- but we can estimate the mean
  - look for the word estimate in questions - it's a big clue to use the method in this note!
- we can talk about the class interval that the median lies in
- and we can talk about the modal class (the class interval the mode lies in)

How do I find/estimate the mean from grouped data?

There is one extra stage to this method compared to finding the mean from tables with discrete data
- We use the class midpoints as our data values
- For example, if heights are split into class intervals 150 ≤ x < 160, 160 ≤ x < 170, etc
  - the midpoints, and so data values, would be 155, 165, etc

STEP 1
Draw an extra two columns on the end of a table of the grouped data
In the first new column write down the midpoint of each class interval

STEP 2
Work out "frequency" × "midpoint" (This is often called fx )

STEP 3
Total the fx column, and if not already done nor mentioned in the question, total the frequency column to find the number of data values involved

STEP 4
Estimate the mean by using its formula; " total of fx " ÷ " no. of data values"

Be careful with midpoints
- not all class intervals will be of equal size
- so there may not be a nice pattern to the midpoints

How do I find the class interval that the median lies in?

Find the position of the median using $\frac{n + 1}{2}$ , where $n$ is the number of data values (total of the frequency column)
Use the table to deduce the class interval containing the ${(\frac{n + 1}{2})}^{th}$ value
- e.g. if the median is the 7 ^th value and the frequency of the first two class intervals are 4 and 7
  - the median will lie in the second class interval of the table
Note the language used when working with the median for grouped data
- Rather than 'the median' we refer to the 'class interval containing the median'

How do I find the modal class interval?

Similar to finding the median we are only interested in the class interval the modal value lies in
Look for the highest frequency in the table
- and then find the corresponding class interval
Take care not to confuse the class interval with the frequency
- The frequency tells us the class interval the modal value lies in

Exam Tip

When presented with data in a table it may not be obvious whether the data is grouped or not
- when you see the phrase “estimate the mean” you know that you are in the world of grouped data
- so use the midpoint technique to answer the question

Worked example

The weights of 20 three-week-old Labrador puppies were recorded at a vet's clinic.
The results are shown in the table below.

Weight, w kg	Frequency
3 ≤ w < 3.5	3
3.5 ≤ w < 4	4
4 ≤ w < 4.5	6
4.5 ≤ w < 5	5
5 ≤ w < 6	2

Estimate the mean weight of these puppies.

First add two columns to the table and complete the first new column with the midpoints of the class intervals.
Complete the second extra column by calculating " fx".
A total row would also be useful.

Weight, w kg	Frequency	Midpoint	" fx"
3 ≤ w < 3.5	3	3.25	3 × 3.25 = 9.75
3.5 ≤ w < 4	4	3.75	4 × 3.75 = 15
4 ≤ w < 4.5	6	4.25	6 × 4.25 = 25.5
4.5 ≤ w < 5	5	4.75	5 × 4.75 = 23.75
5 ≤ w < 6	2	5.5	2 × 5.5 = 11
Total	20		85

Now we can find the mean.

Mean $= \frac{85}{20} = 4.25$

An estimate of the mean weight of the puppies is 4.25 kg

Write down the modal class.

Looking for the highest frequency in the table we can see it is 6.
This corresponds to the interval 4 ≤ w < 4.5.

The modal class is 4 ≤ w < 4.5

A common error is to write down 6 as the mode (modal value).

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