The Mean

The mean of a set of data, the sum of the values divided by their number: from a list, a frequency table, a dot diagram and a stem-and-leaf diagram, with missing values, new values and combined groups.

What You Should Already Know

From Primary school: the average of some numbers is their total divided by how many there are. The average of 4, 7 and 10 is $21 \div 3 = 7$ .
A frequency table, a dot diagram and a stem-and-leaf diagram (lessons 12.1 to 12.3) show every value of a set of data and how often it occurs.

Averages and the Mean

A set of data can hold many values. An average is one value that stands for the whole set: its centre, or a typical value. It lets us describe the data in one number and compare one set with another.

This chapter has three averages, each finding the centre in its own way: the mean (this lesson), the median (lesson 12.8) and the mode (lesson 12.9, met already in lesson 12.3 as the value with the highest frequency). Lesson 12.11 compares them.

The mean

The mean of a set of data is the sum of the values divided by the number of values:

Mean $= \dfrac{\text{sum of the values}}{\text{number of values}}$

The mean is written $\bar{x}$ , read "x bar". It has the same unit as the data.

The mean evens things out

Five pupils have 3, 4, 4, 9 and 10 books. Together they have 30 books. If the 30 books were shared equally, each pupil would get $30 \div 5 = 6$ . The mean, 6 books, is that equal share.

Example 1: The mean of a list

Five pupils scored these marks in a Mathematics test: 65, 72, 80, 75, 88. Find the mean mark.

Solution

Sum of the marks $= 65 + 72 + 80 + 75 + 88 = 380\text{.}$

Number of marks $= 5$ .

Mean $= 380 \div 5 = 76$ .

Check that the answer is sensible: the mean lies between the smallest mark, 65, and the largest, 88.

The mean mark is 76 marks.

Mean × Number = Sum

Multiplying both sides of the rule by the number of values gives the sum. This is the key to every question where a value is missing, added or taken away.

The sum from the mean

Sum of the values $=$ mean $\times$ number of values

Example 2: A missing value

Ravi read 10, 15, 12, 20 and $x$ pages on five evenings. He read a mean of 13 pages an evening. Find $x$ .

Solution

Sum of the five values $= 13 \times 5 = 65$ .

The four known values add up to $10 + 15 + 12 + 20 = 57\text{.}$

So $57 + x = 65$ , and $x = 65 - 57 = 8$ .

Check: $(10 + 15 + 12 + 20 + 8) \div 5 = 65 \div 5 = 13\text{.}$

$x = 8$ : Ravi read 8 pages on the fifth evening.

Example 3: A value added, a value taken away

(a) The mean mark of 24 pupils in a test is 68. A 25th pupil, who was absent, takes the test later and scores 93. Find the new mean mark of the 25 pupils. (b) Five friends get a mean of $12 of pocket money a day. Sean, who gets $20 a day, moves to another school. Find the mean pocket money of the other four.

Solution

(a) The sum of the 24 marks $= 68 \times 24 = 1632$ .

With the 25th mark, the sum is $1632 + 93 = 1725$ .

New mean $= 1725 \div 25 = 69$ .

The new mark, 93, is above the old mean, so it has pulled the mean up, from 68 to 69.

(b) The five friends get $12 × 5 = $60 a day altogether.

Without Sean: $60 − $20 = $40, shared by 4 friends.

New mean = $40 ÷ 4 = $10.

(a) The new mean is 69 marks. (b) The mean of the other four is $10 a day.

Example 4: Combining two groups

In a test, the 20 pupils of Class 2D scored a mean of 75 marks and the 30 pupils of Class 2F a mean of 80 marks. Find the mean mark of all 50 pupils.

Solution

The mean of the two means, $(75 + 80) \div 2 = 77.5\text{,}$ is wrong: it counts the 20 pupils of 2D as if they were as many as the 30 of 2F.

Find the total marks of each class: Class 2D $20 \times 75 = 1500$ ; Class 2F $30 \times 80 = 2400$ .

All the marks: $1500 + 2400 = 3900$ , scored by $20 + 30 = 50$ pupils.

Mean $= 3900 \div 50 = 78$ .

The answer is nearer to 80 than to 75, because more pupils are in Class 2F, whose mean is 80.

The mean mark of all 50 pupils is 78 marks.

The Mean from a Frequency Table

In a frequency table, a value $x$ with frequency $f$ occurs $f$ times. Those $f$ values add up to $f \times x$ , written $fx$ . So the sum of all the values is the sum of the $fx$ , and the number of values is the sum of the frequencies.

The mean of a frequency table

The Greek capital letter $\Sigma$ (sigma) means "the sum of". $\Sigma f$ is the sum of the frequencies, the number of values. $\Sigma fx$ is the sum of the products $fx$ , the sum of the values.

$\bar{x} = \dfrac{\Sigma fx}{\Sigma f}$

Add an $fx$ column to the table, and a total row with $\Sigma f$ and $\Sigma fx$ .

Example 5: The mean of a frequency table

The table shows the number of goals a school hockey team scored in each of its matches. Find the mean number of goals per match.

Goals	Number of matches
$0$	$2$
$1$	$5$
$2$	$7$
$3$	$4$
$4$	$2$
Total	$20$

Solution

Add the column $fx$ (goals $\times$ number of matches) and the totals.

Goals, $x$	Matches, $f$	$fx$
$0$	$2$	$0 \times 2 = 0$
$1$	$5$	$1 \times 5 = 5$
$2$	$7$	$2 \times 7 = 14$
$3$	$4$	$3 \times 4 = 12$
$4$	$2$	$4 \times 2 = 8$
Total	$20$	$39$

$\Sigma f = 2 + 5 + 7 + 4 + 2 = 20$ matches, and $\Sigma fx = 0 + 5 + 14 + 12 + 8 = 39$ goals.

$\bar{x} = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{39}{20} = 1.95$

The mean is 1.95 goals per match.

The mean need not be a value of the data

No match ended with 1.95 goals: every number of goals is a whole number. The mean is an equal share (39 goals shared over 20 matches), so it can be a decimal. Do not round it to a whole number just because the data are whole numbers.

Example 6: A missing frequency

The table shows the number of people in each car that entered a shopping centre's car park in ten minutes. The mean number of people per car is 2.6. Find $k$ .

People in the car	Number of cars
$1$	$6$
$2$	$9$
$3$	$k$
$4$	$5$
$5$	$2$

Solution

$\Sigma f = 6 + 9 + k + 5 + 2 = 22 + k\text{.}$

$\Sigma fx = 1 \times 6 + 2 \times 9 + 3k + 4 \times 5 + 5 \times 2 = 54 + 3k\text{.}$

The mean is 2.6, so $\dfrac{54 + 3k}{22 + k} = 2.6\text{.}$

Multiply both sides by $22 + k$ : $54 + 3k = 2.6(22 + k) = 57.2 + 2.6k\text{.}$

$3k - 2.6k = 57.2 - 54\text{,}$ so $0.4k = 3.2$ and $k = 8$ .

Check: with $k = 8$ the table has 30 cars and $54 + 24 = 78$ people, and $78 \div 30 = 2.6$ .

People, $x$	Cars, $f$	$fx$
$1$	$6$	$1 \times 6 = 6$
$2$	$9$	$2 \times 9 = 18$
$3$	$8$	$3 \times 8 = 24$
$4$	$5$	$4 \times 5 = 20$
$5$	$2$	$5 \times 2 = 10$
Total	$30$	$78$

$k = 8$ : 8 cars carried 3 people.

The Mean from a Dot Diagram or a Stem-and-Leaf Diagram

A dot diagram is a frequency table drawn: the number of dots over a value is its frequency. Count the dots in each column, multiply by the value, and add.

A stem-and-leaf diagram keeps every value. Read each value with the key, then add. Adding one row at a time keeps the working tidy.

Example 7: The mean from a dot diagram

The dot diagram shows how many hours each pupil of a CCA group spent reading last week. Find the mean number of hours.

Solution

Count the dots in each column: 1, 3, 5, 4, 4, 2, 1 over the values 2 to 8. Number of pupils $= 1 + 3 + 5 + 4 + 4 + 2 + 1 = 20\text{.}$

Sum of the hours $= 2 \times 1 + 3 \times 3 + 4 \times 5 + 5 \times 4 + 6 \times 4 + 7 \times 2 + 8 \times 1$

$= 2 + 9 + 20 + 20 + 24 + 14 + 8 = 97\text{.}$

Mean $= 97 \div 20 = 4.85$ .

The mean is 4.85 hours.

Example 8: The mean from a stem-and-leaf diagram

The stem-and-leaf diagram shows the times, in seconds, that 15 pupils took to fold a paper crane. Find the mean time.

Solution

Read each row with the key (3 | 4 means 34 seconds) and add it:

Stem 3: $34 + 38 = 72$

Stem 4: $41 + 42 + 45 + 47 + 47 = 222$

Stem 5: $50 + 53 + 56 + 58 = 217$

Stem 6: $61 + 63 + 67 = 191$

Stem 7: $72$

Sum $= 72 + 222 + 217 + 191 + 72 = 774\text{,}$ and there are $2 + 5 + 4 + 3 + 1 = 15$ leaves.

Mean $= 774 \div 15 = 51.6$ .

The mean time is 51.6 seconds.

Common Mistakes

Dividing $\Sigma fx$ by the number of rows of the table instead of by $\Sigma f$ , the number of values. In Example 5, $39 \div 5 = 7.8$ goals per match is more than any match had.
Adding the values of a table without their frequencies: $(0 + 1 + 2 + 3 + 4) \div 5 = 2$ treats every number of goals as if it happened once.
Taking the mean of two means when the groups are of different sizes. Find each group's total first (Example 4).
Rounding a mean to a whole number because the data are whole numbers: 1.95 goals per match is the right answer.
Reading a stem-and-leaf diagram without its key: 3 | 4 is 34 seconds there, not 3.4.

Practice Questions

Try each question first, then open the solution to check your working. Give an answer that is not exact to 3 significant figures.

Question 1

The heights, in cm, of seven pupils are 154, 167, 149, 154, 176, 169 and 158. Find their mean height.

ShowHide solution

Sum $= 154 + 167 + 149 + 154 + 176 + 169 + 158 = 1127$ cm.

Mean $= 1127 \div 7 = 161$ cm.

The mean height is 161 cm.

Question 2

Jun Hao's scores in six spelling tests, each marked out of 15, were 8, 11, 7, $y$ , 14 and 9. His mean score was 10. Find $y$ .

ShowHide solution

Sum of the six scores $= 10 \times 6 = 60$ .

The five known scores add up to $8 + 11 + 7 + 14 + 9 = 49\text{.}$

$y = 60 - 49 = 11$ .

$y = 11$

Question 3

Nurul's mean mark in four tests is 72. What must she score in the fifth test to raise her mean mark to 75?

ShowHide solution

Sum of the first four marks $= 72 \times 4 = 288$ .

For a mean of 75 in five tests, the sum must be $75 \times 5 = 375$ .

Fifth mark $= 375 - 288 = 87$ .

She must score 87 marks.

Question 4

The table shows how much each of 50 part-time helpers at a food fair earned in a day. (a) Find the mean amount earned. (b) How many helpers earned more than the mean?

Amount earned ($)	Number of helpers
$100$	$20$
$110$	$8$
$120$	$10$
$210$	$7$
$250$	$5$
Total	$50$

ShowHide solution

(a)

Amount ($), $x$	Helpers, $f$	$fx$
$100$	$20$	$100 \times 20 = 2000$
$110$	$8$	$110 \times 8 = 880$
$120$	$10$	$120 \times 10 = 1200$
$210$	$7$	$210 \times 7 = 1470$
$250$	$5$	$250 \times 5 = 1250$
Total	$50$	$6800$

$\bar{x} = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{6800}{50} = 136$

(b) More than $136: the helpers who earned $210 or $250, $7 + 5 = 12$ helpers.

(a) $136 (b) 12 helpers; the other 38 earned less than the mean.

Question 5

The dot diagram shows the number of people living in each of the 16 flats on one floor of an HDB block. (a) Find the mean number of people per flat. (b) No flat has that number of people. Explain why the mean can still be right.

ShowHide solution

(a) The columns hold 1, 3, 4, 5, 2 and 1 dots: 16 flats.

Sum $= 1 \times 1 + 2 \times 3 + 3 \times 4 + 4 \times 5 + 5 \times 2 + 6 \times 1 = 1 + 6 + 12 + 20 + 10 + 6 = 55$ people.

Mean $= 55 \div 16 = 3.4375 \approx 3.44$ (to 3 significant figures).

(b) The mean is the equal share: 55 people shared equally among 16 flats. It need not be one of the values.

(a) 3.44 people per flat (b) The mean is the share each flat would have if the 55 people were spread equally; it need not be a whole number.

Question 6

The stem-and-leaf diagram shows the times, in seconds, of the 12 runners in a 100 m heat. Find the mean time.

ShowHide solution

The key says 12 | 8 means 12.8 seconds: the leaf is the tenths.

Stem 12: $12.8$

Stem 13: $13.1 + 13.4 + 13.4 + 13.9 = 53.8$

Stem 14: $14.0 + 14.2 + 14.5 + 14.7 = 57.4$

Stem 15: $15.0 + 15.3 + 15.8 = 46.1$

Sum $= 12.8 + 53.8 + 57.4 + 46.1 = 170.1\text{,}$ from 12 runners.

Mean $= 170.1 \div 12 = 14.175 \approx 14.2$ s (to 3 significant figures).

The mean time is 14.2 seconds.

Question 7

Pupils took a quiz marked out of 5. The table shows their scores. The mean score is 3.2. Find $k$ .

Score	Number of pupils
$1$	$2$
$2$	$4$
$3$	$k$
$4$	$7$
$5$	$3$

ShowHide solution

$\Sigma f = 2 + 4 + k + 7 + 3 = 16 + k\text{.}$

$\Sigma fx = 2 + 8 + 3k + 28 + 15 = 53 + 3k\text{.}$

$\dfrac{53 + 3k}{16 + k} = 3.2\text{,}$ so $53 + 3k = 51.2 + 3.2k\text{.}$

$53 - 51.2 = 3.2k - 3k\text{,}$ so $1.8 = 0.2k$ and $k = 9$ .

Score, $x$	Pupils, $f$	$fx$
$1$	$2$	$1 \times 2 = 2$
$2$	$4$	$2 \times 4 = 8$
$3$	$9$	$3 \times 9 = 27$
$4$	$7$	$4 \times 7 = 28$
$5$	$3$	$5 \times 3 = 15$
Total	$25$	$80$

Check: $80 \div 25 = 3.2$ .

$k = 9$

Question 8

A school choir has 15 boys, of mean age 13.4 years, and 25 girls, of mean age 14.2 years. Find the mean age of the whole choir.

ShowHide solution

Sum of the boys' ages $= 15 \times 13.4 = 201$ years.

Sum of the girls' ages $= 25 \times 14.2 = 355$ years.

Mean $= (201 + 355) \div (15 + 25) = 556 \div 40 = 13.9$ years.

The mean age of the choir is 13.9 years.

Question 9

The mean mass of 5 parcels is 3.2 kg. One parcel is taken away, and the mean mass of the other 4 is 2.9 kg. Find the mass of the parcel taken away.

ShowHide solution

Sum of the 5 masses $= 3.2 \times 5 = 16$ kg.

Sum of the other 4 $= 2.9 \times 4 = 11.6$ kg.

Mass taken away $= 16 - 11.6 = 4.4$ kg.

The parcel taken away has a mass of 4.4 kg.

Question 10

The table shows the number of hours that each pupil of a class spent at CCA in one week. Wei Ling worked out the mean as $104 \div 4 = 26$ hours. Explain her mistake and find the mean.

Hours	Number of pupils
$2$	$6$
$3$	$10$
$4$	$8$
$5$	$6$
Total	$30$

ShowHide solution

She found $\Sigma fx = 2 \times 6 + 3 \times 10 + 4 \times 8 + 5 \times 6 = 12 + 30 + 32 + 30 = 104$ correctly, but divided by 4, the number of rows.

A mean of 26 hours is more than any pupil spent: a sign that something is wrong. Divide by the number of pupils, $\Sigma f = 6 + 10 + 8 + 6 = 30\text{.}$

Hours, $x$	Pupils, $f$	$fx$
$2$	$6$	$2 \times 6 = 12$
$3$	$10$	$3 \times 10 = 30$
$4$	$8$	$4 \times 8 = 32$
$5$	$6$	$5 \times 6 = 30$
Total	$30$	$104$

$\bar{x} = 104 \div 30 = 3.466\ldots \approx 3.47$ hours (to 3 significant figures).

She divided by the number of rows (4) instead of the number of pupils (30). The mean is 3.47 hours.

Question 11

Sarah's mean mark in her first 3 tests was 80, and in her next 2 tests 70. She says her mean mark in the 5 tests is $(80 + 70) \div 2 = 75$ . Is she right? Find her mean mark.

ShowHide solution

She is not right: the two means come from different numbers of tests (3 and 2).

Sum of the first 3 marks $= 80 \times 3 = 240$ ; sum of the next 2 $= 70 \times 2 = 140$ .

Mean $= (240 + 140) \div 5 = 380 \div 5 = 76\text{.}$

No. Her mean mark in the 5 tests is 76.

Key Points

An average is one value that stands for a set of data; the mean, median and mode are the three averages of this chapter.
Mean $=$ sum of the values $\div$ number of values, written $\bar{x}$ , in the unit of the data.
Sum $=$ mean $\times$ number of values: use it to find a missing value, or the new mean when a value is added or taken away.
To combine groups, add their totals and divide by the total number; do not take the mean of the means.
From a frequency table, $\bar{x} = \dfrac{\Sigma fx}{\Sigma f}$ ; divide by $\Sigma f$ , not by the number of rows.
A dot diagram is read as a frequency table; a stem-and-leaf diagram is read value by value with its key.
The mean need not be a value of the data, and need not be a whole number.