Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Statistics
Practise Averages and spread. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Finding the mean, median, mode and range from a list or a frequency table, estimating the mean of grouped data, working backwards from a mean, and using an average and a measure of spread to compare two sets of data. This comes up on both tiers, often alongside a chart or table; Higher papers add harder problems that need algebra.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
1
Find the mode and range of a listThe mode is the most common value; the range is the largest value minus the smallest value.
2
Find the median of a listPut the values in order and find the middle one, or halfway between the middle two.
3
Calculate the mean of a listAdd up all the values and divide by how many values there are.
4
Find averages from a frequency tableMean = total of the fx column ÷ total frequency; the mode is the value with the highest frequency.
4
Compare two data sets using average and spreadCompare an average and the range, and say what each comparison means in context.
5
Estimate the mean of grouped dataUse the midpoint of each class as the value of every item in that class.
5
Solve problems with totals and combined meansUse total = mean × number of values to find a missing value or the mean of two groups together.
Notes
Mean, median, mode and range
Mode: the most common value. There can be more than one mode, or none.
Median: the middle value once the data are in order. For n values it is the \(\frac{n+1}{2}\)th value; with an even number of values it is halfway between the middle two.
Mean: the total of the values ÷ the number of values.
Range: largest value − smallest value. It measures how spread out the data are.
An outlier is a value much bigger or smaller than the rest. It affects the mean and the range but hardly moves the median, so with an outlier the median is often the better average.
Use the mode for non-numerical data, such as favourite colour.
Frequency tables
Each value x has a frequency f: the number of times it occurs.
Mean = Σfx ÷ Σf. Add an fx column, total it, then divide by the total frequency (not by the number of rows).
The mode is the value with the highest frequency, not the frequency itself.
For the median, find its position, \(\frac{n+1}{2}\), then add up the frequencies from the top until you reach that position.
Grouped data
In a grouped table you do not know the exact values, so the mean can only be estimated.
Estimated mean = Σ(f × midpoint) ÷ Σf, using the midpoint of each class.
The modal class is the class with the highest frequency. You can also find the class that contains the median.
Working with totals
Total = mean × number of values. For a missing value, work out new total − old total.
To combine two groups, add their totals and divide by the total number of values. Do not just average the two means unless the groups are the same size.
Comparing two sets of data
Make two comparisons: one of an average (median or mean) and one of the spread (range).
Say what each comparison means in context, e.g. 'On average, Class A scored higher' and 'Class B's scores were more consistent (smaller range)'.
Cheatsheet
Mean = total of the values ÷ number of values
Median = the \(\frac{n+1}{2}\)th value once the data are in order
Mode = most common value; modal class = class with the highest frequency
Range = largest value − smallest value
Frequency table: mean = Σfx ÷ Σf
Grouped data: estimated mean = Σ(f × midpoint) ÷ Σf
Total = mean × number of values
Compare with one average and one measure of spread, both in context
How to answer each type of question
Find averages from a frequency table
3 to 4 marks4
Add an fx column: multiply each value by its frequency.
Add up the fx column and the frequency column.
Mean = Σfx ÷ Σf. Check that it lies between the smallest and largest values.
The mode is the value with the highest frequency.
Example. The table shows the number of pets owned by each of 30 students. 0 pets: 7 students 1 pet: 10 students 2 pets: 8 students 3 pets: 4 students 4 pets: 1 student (a) Write down the mode. (b) Work out the mean number of pets.
Show the model answer
(a) 1 pet (B1) (b) fx: 0, 10, 16, 12, 4 (M1 for at least three correct products) Σfx = 42, and 42 ÷ 30 (M1) = 1.4 pets (A1)
Work out an estimate for the mean of grouped data
4 to 5 marks5
Find the midpoint of each class.
Multiply each midpoint by its frequency.
Add these up and divide by the total frequency.
Check that the answer lies inside the classes, and if asked, explain why it is only an estimate.
Example. The table shows information about the times, t minutes, that 40 people took to finish a puzzle. \(0 \lt t \le 10\): 6 \(10 \lt t \le 20\): 14 \(20 \lt t \le 30\): 12 \(30 \lt t \le 40\): 8 (a) Work out an estimate for the mean time. (b) Explain why your answer to (a) is only an estimate.
Show the model answer
(a) Midpoints 5, 15, 25, 35 (M1) 6 × 5 + 14 × 15 + 12 × 25 + 8 × 35 = 30 + 210 + 300 + 280 = 820 (M1) 820 ÷ 40 (M1) = 20.5 minutes (A1) (b) The exact times are not known, so the midpoint of each class was used for every time in that class (C1)
Work out a combined mean or a missing value
2 to 3 marks5
Turn each mean into a total: mean × number of values.
Add or subtract the totals, as the question needs.
Divide by the new number of values.
Example. In a test, the mean mark of 12 boys is 64 and the mean mark of 18 girls is 74. Work out the mean mark of all 30 students.
Show the model answer
12 × 64 = 768 or 18 × 74 = 1332 (M1) (768 + 1332) ÷ 30 = 2100 ÷ 30 (M1) = 70 (A1) (64 + 74) ÷ 2 = 69 is wrong, because the two groups are different sizes.
Compare two distributions
2 marks4
Compare the averages: say which is higher and what that means in context.
Compare the spreads: the smaller range means the data are more consistent.
Write each comparison as a full sentence that uses the numbers.
Example. Two basketball teams each played 10 matches. Team A: median 24 points, range 15 points Team B: median 19 points, range 6 points Compare the numbers of points scored by the two teams.
Show the model answer
On average Team A scored more points, because its median (24) is higher than Team B's (19) (C1) Team B's scores were more consistent, because its range (6) is smaller than Team A's (15) (C1)
Shortcuts and memory tricks
Mode = most, median = middle, mean = share it out equally, range = spread.
Sense check: a mean must lie between the smallest and largest values. For grouped data it must lie inside the classes that contain data.
Mean × number = total. Most 'missing value' problems start here.
On a calculator paper you can check a frequency table mean with your calculator's statistics mode, but still write out Σfx and Σf for the method marks.
Median of 30 values: the \(\frac{30+1}{2}\) = 15.5th value, which is halfway between the 15th and 16th values.
Where marks are lost
Forgetting to put the data in order before finding the median.
Giving the position \(\frac{n+1}{2}\) as the median, instead of the value in that position.
Giving the highest frequency as the mode instead of the value that has that frequency.
Dividing Σfx by the number of rows in the table instead of by the total frequency.
Using the class width or one end of a class instead of the midpoint when estimating a mean.
Averaging two means when the groups are different sizes.
Exam technique
Show your fx column or list of products. Method marks can still be given if you slip later.
In 'compare' questions, give two comparisons (an average and the spread), each with the numbers and in the context of the question.
Use 'on average' when comparing averages and 'more consistent' or 'less spread out' when comparing ranges.
When asked which average is best, name it and give a reason linked to the data, e.g. 'the median, because 95 is an outlier that makes the mean too high'.
Do not round in the middle of a calculation. Round only the final answer, if the question asks you to.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
Here are the annual salaries of the five people who work in a small business. £20 000 £21 000 £22 000 £23 000 £150 000
(a) Write down the median salary.[1]
(b) The owner says, 'The mean salary is £47 200, so our staff are well paid.' Explain why the mean is not a good average to represent these salaries.[1]
Show the answer and mark scheme
(a)Answer: £22 000
B1 for £22 000
Worked solution: The middle value of the five ordered salaries is £22 000.
(b)Answer: The £150 000 salary is an extreme value (outlier) that makes the mean much higher than four of the five salaries; the median is more typical.
C1 for explaining that the £150 000 is an outlier / extreme value that raises the mean above almost every salary
Worked solution: Four of the five people earn between £20 000 and £23 000. The one very large salary pulls the mean up to £47 200, which is not typical of the staff.
Question 2Medium5 marks
The table gives information about the times, \(t\) minutes, that 40 people took to complete a puzzle.
[object Object]
(a) Work out an estimate for the mean time.[3]
(b) Explain why your answer to part (a) is only an estimate.[1]
(c) Mo says, 'The median time is in the class \(20 \lt t \le 30\).' Is Mo correct? Give a reason for your answer.[1]
(b)Answer: The exact times are not known; the midpoint of each class is used instead.
C1 for stating that the actual times are not known (midpoints are used)
Worked solution: The data are grouped, so each person's time is taken to be the midpoint of their class.
(c)Answer: Yes: the median is the 20th/21st time (the 20.5th), and the cumulative frequencies are 4, 15, 31, so it is in \(20 \lt t \le 30\).
C1 for 'yes' with a reason, e.g. the 20th and 21st values lie in this class since 15 people take 20 minutes or less and 31 take 30 minutes or less
Worked solution: The median is between the 20th and 21st times. 15 people took 20 minutes or less and 31 took 30 minutes or less, so both are in \(20 \lt t \le 30\).