Maths · Statistics
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Averages and Range
Finding the mean, median, mode and range, using frequency tables and grouped data, and choosing the best average.
Learning Objectives
- 1Calculate the mean, median, mode and range of a small data set.
- 2Choose the most suitable average and explain the effect of an outlier.
- 3Find the mean, median and mode from a frequency table.
- 4Estimate the mean from grouped data using mid-points, and find the modal class and median class.
One number to describe the data
An average gives one value that stands for a whole set of data, and the range tells you how spread out the data is. Statistics questions on a non-calculator paper use small sets of numbers, so you need to be accurate rather than fast: write the numbers in order, show your adding up, and always say what the answer means. Frequency tables are where many marks are lost, because the mean needs the frequency multiplied by the value, not just the values added up.
Mean, median, mode and range
The three averages are different ways of finding a typical value. The data here is 2, 3, 3, 4, 5, 5, 6, 6, 6, 10, which gives a mean of 5, a median of 5 and a mode of 6.
The measures
- Mean Add all the values and divide by how many there are: \(\dfrac{50}{10} = 5\).
- Median The middle value of the ordered data. With 10 values, take the average of the 5th and 6th: \(\dfrac{5 + 5}{2} = 5\).
- Mode The most common value, which is 6 here.
- Range Largest minus smallest: \(10 - 2 = 8\).
Choosing an average
Each average has strengths and weaknesses, and exam questions often ask you to choose one.
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Mean
Uses every value, but one extreme value, an outlier, can pull it up or down.
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Median
Not affected by outliers, so it is best when there are some very large or very small values.
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Mode
The only average for non-numerical data, such as the most popular colour, but it may not exist or may not be unique.
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Range
A measure of spread, not an average. A smaller range means the data is more consistent.
Comparing with a mean and range
Sam's five test scores are 12, 15, 15, 18, 20. Tina's five test scores are 8, 14, 15, 20, 23. Compare their scores using the mean and the range.
Show the solutionHide the solution
- 1 Sam's mean \(\dfrac{12 + 15 + 15 + 18 + 20}{5} = \dfrac{80}{5} = 16\).
- 2 Tina's mean \(\dfrac{8 + 14 + 15 + 20 + 23}{5} = \dfrac{80}{5} = 16\).
- 3 Ranges Sam: \(20 - 12 = 8\). Tina: \(23 - 8 = 15\).
- 4 Conclusion The means are equal, so on average they score the same, but Sam's smaller range shows more consistent scores.
AnswerBoth have a mean of 16, but Sam's range of 8 is smaller than Tina's 15, so Sam is more consistent
Frequency tables
When values are repeated, a frequency table saves writing them all out, but the method changes.
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Mean
Multiply each value by its frequency, add up the products, and divide by the total frequency. \(\bar{x} = \dfrac{\sum fx}{\sum f}\).
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Median
The total frequency is \(n\), so the median is the \(\dfrac{n + 1}{2}\)th value. Use the cumulative frequencies to find which value that is.
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Mode
The value with the highest frequency.
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Range
Largest value minus smallest value, not the largest frequency.
Mean from a frequency table
The table shows the number of pets owned by 20 students. Pets 0, 1, 2, 3, 4 have frequencies 4, 4, 6, 4, 2. Work out the mean, median and mode.
Show the solutionHide the solution
- 1 Products \(0 \times 4 = 0\), \(1 \times 4 = 4\), \(2 \times 6 = 12\), \(3 \times 4 = 12\), \(4 \times 2 = 8\), which add up to 36.
- 2 Mean \(\dfrac{36}{20} = 1.8\).
- 3 Median There are 20 values, so the median is halfway between the 10th and 11th. Cumulative frequencies are 4, 8, 14, 18, 20, so the 10th and 11th are both 2, and the median is 2.
- 4 Mode The highest frequency is 6, for 2 pets, so the mode is 2.
AnswerMean 1.8, median 2, mode 2
Grouped data
When data is in groups, you do not know the exact values, so you estimate.
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Mid-points
Use the middle of each class as its value. For \(20 < w \leq 40\) the mid-point is 30.
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Estimated mean
Multiply each mid-point by its frequency, add up, and divide by the total frequency.
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Modal class
The class with the highest frequency.
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Median class
The class that contains the middle value, found with the cumulative frequency.
Estimating a mean from grouped data
The weights of 20 parcels are 5 in the class \(0 < w \leq 20\), 10 in \(20 < w \leq 40\) and 5 in \(40 < w \leq 60\). Estimate the mean weight.
Show the solutionHide the solution
- 1 Mid-points 10, 30 and 50.
- 2 Products \(10 \times 5 = 50\), \(30 \times 10 = 300\) and \(50 \times 5 = 250\).
- 3 Total \(50 + 300 + 250 = 600\).
- 4 Estimated mean \(\dfrac{600}{20} = 30\) kg. It is only an estimate, because the true values are not known.
Answer30 kg
Working backwards from a mean
The total is the mean times the number of values.
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Total from a mean
If 5 numbers have a mean of 8, they total \(5 \times 8 = 40\).
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Adding a value
Adding a sixth number to give a mean of 9 means the new total is \(6 \times 9 = 54\), so the new number is \(54 - 40 = 14\).
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Combined means
To combine groups, find each total, add the totals, and divide by all the values.
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Missing value
A set with a known mean and known values leaves the missing value as total minus the others.
Test yourself
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1
How do you find the median of 10 ordered values?
Show answerHide answer
The mean of the 5th and 6th.
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2
What is the range of 3, 8, 11, 20?
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\(20 - 3 = 17\).
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3
How do you find the mean from a frequency table?
Show answerHide answer
Divide the total of frequency times value by the total frequency.
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4
Which average is least affected by an outlier?
Show answerHide answer
The median.
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5
What is the mid-point of the class \(10 < t \leq 20\)?
Show answerHide answer
15.
Exam technique: averages
Show the working, and say what the answer means.
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Order the data
Write the values in order before finding the median.
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Show the products
In a frequency table, show each \(f \times x\) separately.
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Give a conclusion
"On average" and "more consistent" are needed in comparison questions.
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Compare like with like
Compare an average with an average, and a range with a range.
Summary and exam focus
- The mean, median and mode are averages, and the range is a measure of spread.
- The median is not affected by outliers, and a smaller range means more consistent data.
- With a frequency table, the mean is \(\dfrac{\sum fx}{\sum f}\), and the median is the \(\dfrac{n + 1}{2}\)th value.
- For grouped data use mid-points to estimate the mean.
Exam focus
The mean of five numbers is 8. Four of the numbers are 5, 7, 9 and 10. Work out the fifth number. (2 marks) (2 marks)
The total of five numbers with a mean of 8 is \(5 \times 8 = 40\). The four known numbers add up to 31, so the fifth number is \(40 - 31 = 9\). Do not average the four known numbers.
Key terms
The words this lesson expects you to use. Each one is linked from the first place it appears above.
- Mean
- The sum of the values divided by how many there are.
- Median
- The middle value when the data is in order.
- Mode
- The value that occurs most often.
- Range
- The difference between the largest and smallest values.
- Outlier
- A value that is much larger or smaller than the others.
- Frequency table
- A table showing how often each value occurs.
- Grouped data
- Data sorted into classes, such as \(0 < t \leq 10\).
- Mid-point
- The middle value of a class, used to estimate the mean of grouped data.
- Modal class
- The class with the highest frequency.
Questions and answers
15 questions set on this lesson, with the mark schemes and model answers open.
Here are the ages, in years, of seven children: 14, 11, 15, 11, 13, 17, 11. Work out (a) the mode, (b) the median, (c) the range. [3 marks]
Mark scheme — 3 marks available
- (a) 11 — B1
- (b) 13 — B1
- (c) 6 — B1
Model answer
(a) 11 appears three times, so the mode is 11. (b) In order: 11, 11, 11, 13, 14, 15, 17, so the median is 13. (c) \(17 - 11 = 6\).
The table shows the number of brothers and sisters that each of 20 pupils has. (a) Write down the mode. [1 mark] (b) Work out the mean. [3 marks] (c) Work out the median. [2 marks]
Mark scheme — 6 marks available
- (a) 1 — B1
- (b) Products \(f \times x\) with at least 4 correct — M1
- (b) \(\dfrac{28}{20}\) or total 28 — M1
- (b) 1.4 — A1
- (c) Cumulative frequencies, or the 10th and 11th values — M1
- (c) 1 — A1
Model answer
(a) The highest frequency is 8, so the mode is 1. (b) \(0 \times 4 + 1 \times 8 + 2 \times 5 + 3 \times 2 + 4 \times 1 = 0 + 8 + 10 + 6 + 4 = 28\). The mean is \(\dfrac{28}{20} = 1.4\). (c) The cumulative frequencies are 4, 12, 17, 19, 20, so the 10th and 11th values are both 1. The median is 1.
Team A scored a mean of 24 points in a game and the range was 10. Team B scored a mean of 27 points and the range was 18. Compare the points scored by the two teams. [2 marks]
Mark scheme — 2 marks available
- A comparison of the means, in context — Q1
- A comparison of the ranges, in context — Q1
Model answer
On average Team B scored more points, with a mean of 27 compared with 24. Team A was more consistent, because its range of 10 is smaller than 18.
The mean of six numbers is 9. Five of the numbers are 4, 8, 10, 11 and 13. Work out the sixth number. [3 marks]
Mark scheme — 3 marks available
- \(6 \times 9 = 54\) — M1
- \(54 - 46\) — M1
- 8 — A1
Model answer
The total of six numbers is \(6 \times 9 = 54\). The five known numbers add up to \(4 + 8 + 10 + 11 + 13 = 46\). The sixth number is \(54 - 46 = 8\).
The table shows the heights of 40 plants. For the classes \(0 < h \leq 10\), \(10 < h \leq 20\), \(20 < h \leq 30\) and \(30 < h \leq 40\) the frequencies are 6, 14, 12 and 8. (a) Write down the modal class. [1 mark] (b) Work out an estimate for the mean height. [3 marks]
Mark scheme — 4 marks available
- (a) \(10 < h \leq 20\) — B1
- (b) Mid-points 5, 15, 25, 35 used — M1
- (b) \(\dfrac{820}{40}\) or total 820 — M1
- (b) 20.5 — A1
Model answer
(a) The highest frequency is 14, so the modal class is \(10 < h \leq 20\). (b) The mid-points are 5, 15, 25 and 35. \(5 \times 6 + 15 \times 14 + 25 \times 12 + 35 \times 8 = 30 + 210 + 300 + 280 = 820\). The estimate of the mean is \(\dfrac{820}{40} = 20.5\).
The mean of 3 numbers is 5. The mean of 7 other numbers is 9. Work out the mean of all 10 numbers. [3 marks]
Mark scheme — 3 marks available
- \(3 \times 5 = 15\) or \(7 \times 9 = 63\) — M1
- \(\dfrac{15 + 63}{10}\) — M1
- 7.8 — A1
Model answer
The total of the first 3 numbers is \(3 \times 5 = 15\) and of the other 7 is \(7 \times 9 = 63\). The mean of all 10 is \(\dfrac{15 + 63}{10} = \dfrac{78}{10} = 7.8\).
What is the median of 3, 5, 6, 8, 9, 12?
Why: With 6 values, take the mean of the 3rd and 4th: \(\dfrac{6 + 8}{2} = 7\).
What is the range of 3, 8, 11, 20?
Why: \(20 - 3 = 17\).
What is the mean of 4, 6, 8, 10, 12?
Why: \(\dfrac{40}{5} = 8\).
Which average is least affected by an outlier?
Why: The median depends only on the middle value.
The numbers 0, 1, 2, 3, 4 have frequencies 4, 4, 6, 4, 2. What is the mean?
Why: \(\dfrac{0 + 4 + 12 + 12 + 8}{20} = \dfrac{36}{20} = 1.8\).
What is the mode of the numbers 0, 1, 2, 3, 4 with frequencies 4, 4, 6, 4, 2?
Why: The highest frequency, 6, is for the value 2.
What is the mid-point of the class \(20 < w \leq 40\)?
Why: \(\dfrac{20 + 40}{2} = 30\).
Classes \(0 < w \leq 20\), \(20 < w \leq 40\) and \(40 < w \leq 60\) have frequencies 5, 10 and 5. What is the estimated mean?
Why: Using mid-points: \(\dfrac{10 \times 5 + 30 \times 10 + 50 \times 5}{20} = \dfrac{600}{20} = 30\).
The mean of five numbers is 8. Four of them are 5, 7, 9 and 10. What is the fifth?
Why: The total is \(5 \times 8 = 40\), and \(40 - 31 = 9\).