1MA1 · S4StatisticsFoundation & HigherGrades 2–5

Averages & Spread

Q: Why is the grouped mean only an estimate?

The exact values in each class are unknown, so the midpoint is used as a best guess for all of them.

Q: Can there be more than one mode?

Yes. Data with two modes are bimodal; if every value appears equally, there is no mode.

Q: Which average should I use?

Median when there are outliers, mean when all values should count, mode for categories or the most common value.

An average gives one typical value for a set of data; a measure of spread says how much the values vary. Together they summarise a data set in two numbers, and comparing two data sets means comparing both. This lesson covers the three averages and the range for lists, frequency tables and grouped data.

45 min Statistics

The big picture

Each average has a job. The mean uses every value, so it is sensitive to extreme values. The median is the middle value, so it ignores extremes. The mode is the most common value, and the only average for non-numerical data. Edexcel often asks which is most appropriate and why.

Grouped data lose the exact values, so the mean can only be estimated using midpoints, and the median and mode can only be given as classes. Recognising "estimate" in a question is the signal to use midpoints.

What you'll learn

Find the mean, median, mode and range of a list
Calculate averages from frequency tables
Estimate the mean and identify the modal and median classes for grouped data
Choose an appropriate average and compare distributions

Averages of a list

Mean $=$ total of the values ÷ number of values.

Median: order the values; the middle one is at position $\tfrac{n + 1}{2}$ . With an even number of values, take the mean of the two middle values.

Mode: the most frequent value. Range: largest minus smallest.

Example 1The mean of five numbers is 12. Four of them are 10, 11, 15 and 9. Find the fifth.

1Total of five numbers:

5 \times 12 = 60

2Known total:

45

AnswerThe fifth number is 15.

Your turnSame method, new numbers — try it before moving on.

The mean of six numbers is 15. Five of them are 12, 18, 20, 9 and 14. Find the sixth.

Your answer

Show worked solution

Solution

1Total $= 6 \times 15 = 90$ .
2Known total $73$ , so the sixth is $17$ .

Answer: $17$

Tip — When a question gives a mean, turn it into a total straight away: mean × number of values.

Frequency tables

For a frequency table, the mean is $\dfrac{\sum fx}{\sum f}$ : multiply each value by its frequency, add the results, and divide by the total frequency.

The median is at position $\tfrac{n + 1}{2}$ ; use running totals to find which value it falls on.

The mode is the value with the highest frequency.

\bar{x} = \dfrac{\sum fx}{\sum f}

Mean from a frequency table.

Example 1The number of pets owned by 25 students: 0 pets: 7, 1 pet: 9, 2 pets: 6, 3 pets: 3. Find the mean, median and mode.

\sum fx = 0 + 9 + 12 + 9 = 30

, so mean

= \tfrac{30}{25} = 1.2

2Median at position 13. Running totals 7, 16 — the 13th value is 1.

3Highest frequency is 9, for 1 pet.

AnswerMean 1.2, median 1, mode 1

Your turnSame method, new numbers — try it before moving on.

Goals scored: 0 goals (frequency 3), 1 goal (5), 2 goals (8), 3 goals (4). Find the mean.

Your answer

Show worked solution

Solution

1 $\sum fx = 0 + 5 + 16 + 12 = 33$ .
2 $33 \div 20 = 1.65$ .

Answer: $1.65$

Grouped data

Estimate the mean using the midpoint of each class as $x$ in $\dfrac{\sum fx}{\sum f}$ . It is an estimate because the exact values are unknown.

The $\textbf{modal class}$ has the highest frequency. The $\textbf{median class}$ contains the middle value.

Example 1Heights of 25 plants (cm):

150 \le h < 160

: 4;

160 \le h < 170

: 11;

170 \le h < 180

: 8;

180 \le h < 190

: 2. Estimate the mean and give the median class.

1Midpoints 155, 165, 175, 185:

\sum fx = 620 + 1815 + 1400 + 370 = 4205

2Estimated mean

= \tfrac{4205}{25} = 168.2

3Median at position 13: running totals 4, 15 — so in

160 \le h < 170

AnswerEstimated mean 168.2 cm; median class

160 \le h < 170

Your turnSame method, new numbers — try it before moving on.

Grouped data: $0 \le x < 10$ (frequency 5), $10 \le x < 20$ (10), $20 \le x < 30$ (5). Estimate the mean.

Your answer

Show worked solution

Solution

1Midpoints 5, 15, 25: $\sum fx = 25 + 150 + 125 = 300$ .
2 $300 \div 20 = 15$ .

Answer: $15$

Why this works

A midpoint is a sensible guess for every value in a class, but some classes will have values bunched at one end. That is why the answer is only an estimate — and why questions say "estimate".

Choosing and comparing

Use the $\textbf{median}$ when there are extreme values (outliers), the $\textbf{mean}$ when the data are fairly even and every value matters, and the $\textbf{mode}$ for non-numerical data or the most popular choice.

To compare two data sets, compare an average $\textit{and}$ a measure of spread, in context: "Class A scored higher on average (mean 68 compared with 62), and their scores were more consistent (range 20 compared with 35)."

Tip — Comparisons need two parts, each referring to the context. Just quoting the numbers earns fewer marks than saying what they mean.

Think like an examiner

Show $fx$ values and their total for frequency table means.
Use midpoints and say "estimate" for grouped data.
Compare averages and spread in context.

Averages

\text{mean} = \tfrac{\text{total}}{n}

Mean of a list.

\tfrac{\sum fx}{\sum f}

Mean from a table.

\tfrac{n + 1}{2}

Position of the median.

Watch out for these

Dividing

\sum fx

by the number of rows.

Divide by the total frequency.

Giving the median of a frequency table as the middle row.

Find the middle data value using running totals.

Using class boundaries instead of midpoints for an estimated mean.

Use midpoints.

Giving the mode as the highest frequency itself.

The mode is the value with that frequency.

Stretch yourself

A class of 12 boys has a mean test score of 64, and 18 girls have a mean of 71. A new student joins and the mean for all 31 students becomes 69. What did the new student score?

Hint — Convert each mean into a total.

Reveal solution

SolutionBoys’ total

= 12 \times 64 = 768

. Girls’ total

= 18 \times 71 = 1278

. Class total

= 2046

, so the mean of the 30 is

68.2

. With 31 students and mean 69, the total is

31 \times 69 = 2139

. The new student scored

2139 - 2046 = 93

marks — well above the average, which is why one student raised the mean by 0.8.

\blacksquare

Questions students ask

Why is the grouped mean only an estimate?

The exact values in each class are unknown, so the midpoint is used as a best guess for all of them.

Can there be more than one mode?

Yes. Data with two modes are bimodal; if every value appears equally, there is no mode.

Which average should I use?

Median when there are outliers, mean when all values should count, mode for categories or the most common value.

Key takeaways

Mean: total ÷ number; median: middle value; mode: most common; range: largest − smallest.
Frequency table mean $= \tfrac{\sum fx}{\sum f}$ .
Grouped data: estimate the mean with midpoints; give modal and median classes.
Choose averages to suit the data.
Compare distributions with an average and a spread, in context.

Exam-style questions

Practise it the way it is examined

3 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 22 marks

Find the median of $8, 3, 11, 6, 9, 4$ .[2]
Show mark scheme
- M1Orders: $3, 4, 6, 8, 9, 11$
- A1 $7$
Answer: $7$

Question 2Grade 42 marks

12 boys have a mean mark of 64. 18 girls have a mean mark of 69. Work out the mean mark of all 30 students.[2]
Show mark scheme
- M1Totals $768$ and $1242$
- A1 $67$
Answer: $67$

Question 3Grade 52 marks

Two machines fill bags. Machine P: mean 502 g, range 12 g. Machine Q: mean 500 g, range 4 g.

Compare the two machines.[2]
Show mark scheme
- B1Q fills closer to 500 g on average, or P fills slightly more on average
- B1Q is more consistent, as its range is smaller

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Next step

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