Mean

AS · S1 · 9 min

The mean is the everyday "average": add up all the values and divide by how many there are. It is the most used measure of central tendency because it uses every value and feeds directly into the standard deviation. On Paper 5 you calculate it from raw data, from frequency tables, from grouped data (as an estimate), and from summary totals such as $\sum x$ , often for two data sets that are combined or compared.

The mean of raw data

Definition

The mean of $n$ values $x_1, x_2, \dots, x_n$ is

\bar{x} = \frac{\sum x}{n},

where $\sum x$ (read "sigma $x$ ") is the sum of all the values.

For $4, 7, 7, 9, 13$ : $\sum x = 40$ , $n = 5$ , so $\bar{x} = 8$ .

A helpful picture: the mean is the balance point of the data. If each value were a unit weight on a ruler, the ruler would balance at $\bar{x}$ . The deviations above the mean exactly cancel those below: $\sum (x - \bar{x}) = 0$ . This is why one extreme value drags the mean towards it: a weight far out along the ruler moves the balance point a long way.

Frequency tables

When values repeat, a frequency table lists each value $x$ with its frequency $f$ . Each value contributes $f \times x$ to the total.

Key result

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

$\sum f$ is the total number of values, $n$ .

Grouped data: estimating the mean

When the data are grouped into classes, the individual values are lost. To estimate the mean, assume every value in a class sits at the class mid-point, then use the frequency-table formula with $x$ as the mid-point.

Method

Find each class's boundaries, then its mid-point $x = \tfrac{1}{2}(\text{lower boundary} + \text{upper boundary})$ .
Multiply each mid-point by its frequency to get $fx$ .
Estimate $\bar{x} = \dfrac{\sum fx}{\sum f}$ , and call it an estimate.

Mid-points depend on the boundaries, so the class boundary rules matter here. For lengths " $10$ – $19$ " to the nearest cm, the boundaries are $9.5$ and $19.5$ and the mid-point is $14.5$ . For ages " $10$ – $19$ " the boundaries are $10$ and $20$ and the mid-point is $15$ . For " $10 \le x < 20$ " it is $15$ .

The result is only an estimate because the values in each class are not really all at the mid-point. If values are spread fairly evenly within classes, the errors largely cancel and the estimate is good.

Working from totals

Many questions never give you the data; they give summary totals such as $n = 20$ and $\sum x = 364$ . Then $\bar{x} = \tfrac{364}{20} = 18.2$ immediately.

Turned around, $\sum x = n\bar{x}$ . This one fact solves most "adding, removing or combining" problems:

Key result

Total of the values: $\sum x = n \bar{x}$ .
Combined mean of two data sets: $\bar{x} = \dfrac{\sum x_1 + \sum x_2}{n_1 + n_2} = \dfrac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$ .
Adding or removing a value: change the total and the count, then divide again.

The combined mean is not the average of the two means unless the sets are the same size. A class of $10$ with mean $80$ and a class of $30$ with mean $60$ have combined mean $\tfrac{800 + 1800}{40} = 65$ , not $70$ .

Properties of the mean

It uses every data value.
It is affected by extreme values (outliers), so for skewed data it is pulled towards the tail.
It may not be one of the data values, and for discrete data it may not be a possible value ( $2.18$ children).
It is the measure used with the standard deviation, and it is the natural choice for data that are roughly symmetrical with no outliers.

For a full comparison with the median and mode, see Interpreting and comparing distributions.

Worked examples

Mean from a frequency table (routine)

The numbers of children in $50$ families are recorded.

Number of children $x$	$0$	$1$	$2$	$3$	$4$	$5$
Frequency $f$	$6$	$11$	$14$	$9$	$7$	$3$

Find the mean number of children per family.

Solution

$\sum f = 6 + 11 + 14 + 9 + 7 + 3 = 50$ .

$\sum fx = 0 + 11 + 28 + 27 + 28 + 15 = 109$ .

$\bar{x} = \tfrac{109}{50} = 2.18$ children.

This is an exact mean (not an estimate): the data are not grouped.

Estimated mean from grouped data

The journey times of $120$ people are summarised.

Time $t$ (min)	$0 \le t < 10$	$10 \le t < 20$	$20 \le t < 30$	$30 \le t < 40$	$40 \le t < 60$	$60 \le t < 80$
Frequency	$6$	$18$	$36$	$32$	$20$	$8$

Calculate an estimate of the mean journey time.

Solution

Mid-point $x$	$5$	$15$	$25$	$35$	$50$	$70$
$fx$	$30$	$270$	$900$	$1120$	$1000$	$560$

$\sum fx = 3880$ , $\sum f = 120$ .

\bar{x} \approx \frac{3880}{120} = 32.3\ \text{minutes (3 s.f.)}.

The estimated median of these data is $30$ minutes (see Cumulative frequency). The mean is larger because the long journeys in the upper classes pull it up: the data are positively skewed.

A missing frequency

The scores on a $5$ -point scale given by some customers are shown.

Score	$1$	$2$	$3$	$4$	$5$
Frequency	$3$	$7$	$k$	$8$	$2$

The mean score is $2.96$ . Find $k$ .

Solution

$\sum f = 20 + k$ and $\sum fx = 3 + 14 + 3k + 32 + 10 = 59 + 3k$ .

\frac{59 + 3k}{20 + k} = 2.96 \quad\Rightarrow\quad 59 + 3k = 59.2 + 2.96k \quad\Rightarrow\quad 0.04k = 0.2 \quad\Rightarrow\quad k = 5.

Combining two groups (exam style)

In a class of $30$ students, the $12$ boys have mean mark $58$ . The mean mark of the whole class is $61$ . Find the mean mark of the girls.

Solution

Total for the class: $30 \times 61 = 1830$ . Total for the boys: $12 \times 58 = 696$ .

Total for the $18$ girls: $1830 - 696 = 1134$ , so the girls' mean is $\tfrac{1134}{18} = 63$ .

Removing and adding values (exam-hard)

The mean of $20$ values is $14.5$ . One value is removed, and the mean of the remaining $19$ values is $14.2$ .

(a) Find the value that was removed. (b) A further $15$ values, with mean $16$ , are then added to the $19$ values. Find the mean of all $34$ values.

Solution

(a) Original total: $20 \times 14.5 = 290$ . New total: $19 \times 14.2 = 269.8$ . Removed value: $290 - 269.8 = 20.2$ .

(b) Total of the new values: $15 \times 16 = 240$ . Overall mean:

\frac{269.8 + 240}{34} = \frac{509.8}{34} = 14.99\ (\text{4 s.f.}) \approx 15.0.

Note how working with totals avoids ever needing the individual values.

Watch out

Averaging the averages. The mean of two groups combined is the total over the total count. Only when the groups are equal in size does it equal the mean of the two means.

Watch out

Using class widths or upper boundaries instead of mid-points. For grouped data the representative value is the mid-point, found from the boundaries. " $20$ – $29$ " (rounded) has mid-point $24.5$ , not $25$ and not $29$ ; ages " $20$ – $29$ " have mid-point $25$ .

Watch out

Dividing by the number of classes. In $\tfrac{\sum fx}{\sum f}$ , the denominator is the total frequency, not the number of rows in the table.

Exam tip

Show the $fx$ column (or the sum $\sum fx$ written out). A bare final answer from a calculator's statistics mode risks losing all the marks if it is wrong.
Say "estimate" for grouped data, and if asked why: the actual values within each class are unknown, so the mid-points are used.
Give answers to $3$ s.f. unless exact; keep full accuracy in intermediate totals.
Questions on two data sets almost always need totals: write $\sum x = n\bar{x}$ for each set before you do anything else.

Summary

$\bar{x} = \dfrac{\sum x}{n}$ for raw data; $\bar{x} = \dfrac{\sum fx}{\sum f}$ for a frequency table.
Grouped data: use class mid-points; the result is an estimate.
Mid-points come from class boundaries; take care with rounded data and ages.
$\sum x = n\bar{x}$ : the key to combining sets and adding or removing values.
The combined mean is $\dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$ , not the mean of the means.
The mean uses all the data but is sensitive to extreme values.

Practice questions

Question

Find the mean of $3.2, 4.8, 5.1, 2.7, 6.0, 4.4$ .
The numbers of goals scored in $40$ matches: $0$ goals ( $8$ matches), $1$ ( $12$ ), $2$ ( $10$ ), $3$ ( $6$ ), $4$ ( $3$ ), $5$ ( $1$ ). Find the mean number of goals per match.
The ages of $40$ people at an event are grouped as $10$ – $19$ ( $8$ ), $20$ – $29$ ( $15$ ), $30$ – $49$ ( $12$ ), $50$ – $79$ ( $5$ ). Estimate the mean age.
For a set of $18$ values, $\sum x = 432$ . Find the mean. Another value, $43$ , is added. Find the new mean.
Class A has $25$ students with mean height $162\ \text{cm}$ and class B has $15$ students with mean height $170\ \text{cm}$ . Find the mean height of all $40$ students.
The mean of eight numbers is $12$ . Seven of them are $9, 15, 11, 14, 8, 13, 12$ . Find the eighth.
The times taken, $t$ minutes, are grouped as $0 \le t < 10$ ( $5$ people), $10 \le t < 20$ ( $x$ people) and $20 \le t < 40$ ( $10$ people). The estimated mean time is $17$ minutes. Find $x$ .
A student calculates the mean of $25$ values to be $40.4$ . She then discovers that one value, which should have been $38$ , was recorded as $83$ . Find the correct mean.

Answers

$\sum x = 26.2$ , $n = 6$ , $\bar{x} = 4.37$ (3 s.f.).
$\sum fx = 0 + 12 + 20 + 18 + 12 + 5 = 67$ ; $\sum f = 40$ ; $\bar{x} = 1.675$ goals.
Ages: boundaries $10, 20, 30, 50, 80$ ; mid-points $15, 25, 40, 65$ . $\sum fx = 120 + 375 + 480 + 325 = 1300$ . Estimated mean $= \tfrac{1300}{40} = 32.5$ years.
$\bar{x} = \tfrac{432}{18} = 24$ . New mean $= \tfrac{432 + 43}{19} = \tfrac{475}{19} = 25$ .
$\tfrac{25 \times 162 + 15 \times 170}{40} = \tfrac{4050 + 2550}{40} = \tfrac{6600}{40} = 165\ \text{cm}$ .
Total $= 8 \times 12 = 96$ . The seven sum to $82$ , so the eighth is $14$ .
Mid-points $5, 15, 30$ . $\tfrac{25 + 15x + 300}{15 + x} = 17$ , so $325 + 15x = 255 + 17x$ , giving $2x = 70$ and $x = 35$ .
Incorrect total $= 25 \times 40.4 = 1010$ . Correct total $= 1010 - 83 + 38 = 965$ . Correct mean $= \tfrac{965}{25} = 38.6$ .

The mean of raw data

Frequency tables

Grouped data: estimating the mean

Working from totals

Properties of the mean

Worked examples

Practice questions

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