Averages from Grouped Data (Edexcel IGCSE Maths)

Spec point — 6.2C

Averages from grouped data

What is grouped data?

• Data can be collected into groups or class intervals

  • It is useful for organising data if you have a lot of individual data points
  • You can present grouped data in a grouped frequency table

• Grouped data may be discrete or continuous

  • Discrete data is numerical data that can only take on specific values, it needs to be counted

    • E.g. Shoe size

  • Continuous data can take any value within a range of infinite values, it needs to be measured

    • E.g. Length of a foot in cm

Why can I only find an estimate for the mean from grouped data?

• It is impossible to find the mean for grouped data, because we don't have access to the original data values

  • i.e. there is no way to find the exact sum of all the data values
  • so we can't use the formula, $\mathrm{mean}=\frac{\mathrm{sum} \mathrm{of} \mathrm{values}}{\mathrm{number} \mathrm{of} \mathrm{values}}$

• However we can estimate the mean for grouped data

  • To do this we use the class midpoints as our data values

    • e.g. if a class interval is 150 ≤ x < 160
    • we assume that all the data values are equal to the midpoint, 155

Exam Hint 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!

How do I find an estimate for the mean from grouped data?

• To find an estimate for the mean from grouped data, complete the following steps:

• 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

  • If the midpoint isn't obvious, add the endpoints and divide by 2

    • e.g. if a class interval is 150 ≤ x < 160
    • the midpoint is $\frac{150+160}{2}=\frac{310}{2}=155$

• STEP 2 Calculate "frequency" × "midpoint" (this is often called fx)

  • Write these values in the second column you added to the table

• STEP 3 Find the total for the fx column

  • If the question does not tell you the total number of data values (i.e. the total frequency), find the total of the frequency column also

• STEP 4 Estimate the mean by using the formula

  • $\text{estimated} \text{mean} = \frac{ \text{total} of ( \text{midpoints} \times \text{frequencies} ) }{ \text{total} frequency}$
  • i.e. divide the total of the fx column by the total number of data values

How do I find the modal class?

• For grouped data we talk about the modal class instead of the mode

  • This is the class with the highest frequency

• Find the highest frequency in the table

  • The corresponding class interval tells you the modal class

Exam Hint Be careful not to confuse the modal class with its frequency.

e.g. if the highest frequency in the table is 34, corresponding to the class interval $40\leq x<50$ then the modal class is $40\leq x<50$, not '34'.

This is a common mistake in exams!

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	2
3.5 ≤ w < 4	4
4 ≤ w < 4.5	6
4.5 ≤ w < 5	5
5 ≤ w < 5.5	2
5.5 ≤ w < 6	1

(a) Estimate the mean weight of these puppies.

Answer:

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

Weight, w kg	Frequency	Midpoint	"fx"
3 ≤ w < 3.5	2	3.25	2 × 3.25 = 6.5
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 < 5.5	2	5.25	2 × 5.5 = 10.5
5.5 ≤ w < 6	1	5.75	1 × 5.75 = 5.75
Total	20		87

Now we can find the mean using

$\text{estimated} \text{mean} = \frac{ \text{total} of ( \text{midpoints} \times \text{frequencies} ) }{ \text{total} frequency}$

$\mathrm{estimated} \mathrm{mean}=\frac{87}{20}=4.35$

Final answer: 4.35 kg

(b) Write down the modal class.

Answer:

The highest frequency in the table is 6 This corresponds to the interval 4 ≤ w < 4.5

Final answer: 4 ≤ w < 4.5