Maths · 10 Statistics

10.4 Averages and measures of spread

Before you start

Putting whole numbers and decimals in order from smallest to largest.
The difference between discrete and continuous data (10.3).
Inequality notation for class intervals: $10 < t \leq 20$ means $t$ is more than $10$ and up to and including $20$ .
Rounding to $3$ significant figures.

Syllabus objectives covered (3)

Calculate the mean, median, mode, quartiles, range and interquartile range for individual data and distinguish between the purposes for which these are used.
Calculate an estimate of the mean for grouped discrete or grouped continuous data.
Identify the modal class from a grouped frequency distribution.

Part 1 of 7

The mean and the mode

An average is one number that stands for a whole set of data. There are three averages in this objective, because 'typical' can mean three different things.
The mean shares the total out equally. Add every value, then divide by how many values there are.
$mean = \frac{sum of the values}{number of values}$
For $4, 7, 2, 9, 8$ the total is $30$ and there are $5$ values, so the mean is $30 \div 5 = 6$ . Notice that $6$ is not one of the values. A mean does not have to be.
Because the mean is a total divided by a count, you can run it backwards. If you know the mean and the number of values, you know the total. This one line answers every 'missing value' question.
$sum of the values = mean \times number of values$
The mode is the value that occurs most often. In $3, 5, 5, 6, 8, 8, 8, 10$ the mode is $8$ , because it occurs three times and no other value occurs more.
A set can have two modes: $1, 1, 4, 6, 6$ has modes $1$ and $6$ . A set where every value occurs once has no mode.
The mode is the only one of the three averages you can find for data that are not numbers, such as favourite colours. You cannot add 'red' and 'blue', but you can count which one appears most.
No statistics formula is printed on the List of formulas, so every method in this lesson must be known, not looked up.

Part 2 of 7

The median and the quartiles

The median is the middle value once the data are in order. Ordering comes first. The middle of an unordered list means nothing, because it depends on the order the values happened to be written in.
For $n$ values in order, the median is the value in this position.
$position of the median = \frac{n + 1}{2}$
That gives a position, not the answer. For $11$ values, $\frac{11 + 1}{2} = 6$ tells you to count along to the 6th value. The median is whatever number sits there.
With an even number of values the position falls between two values. For $8$ values, $\frac{8 + 1}{2} = 4.5$ , so the median is halfway between the 4th and 5th values: add those two and halve.
The quartiles cut the ordered data into four equal parts. A quarter of the data lie below the lower quartile (LQ), and a quarter lie above the upper quartile (UQ). The median is the cut in the middle.
To find the quartiles, split the ordered list at the median into a lower half and an upper half. When $n$ is odd, the median itself belongs to neither half, so leave it out.
The LQ is the median of the lower half. The UQ is the median of the upper half. You already know how to find a median, so the quartiles need no new skill.
Take $2, 3, 4, 5, 6, 7, 7, 7, 9, 10, 12$ , where $n = 11$ . The median is the 6th value, $7$ . Leave it out: the lower half $2, 3, 4, 5, 6$ gives LQ $= 4$ , and the upper half $7, 7, 9, 10, 12$ gives UQ $= 9$ .
When $n$ is even, the list splits cleanly into two halves of the same size. For $1, 3, 4, 6, 8, 9, 11, 14$ the halves are $1, 3, 4, 6$ and $8, 9, 11, 14$ . Each half has an even number of values, so LQ $= \frac{3 + 4}{2} = 3.5$ and UQ $= \frac{9 + 11}{2} = 10$ .

Part 3 of 7

Range and interquartile range

An average tells you where the data sit. A measure of spread tells you how far apart the values are. Two classes can have the same mean mark when in one everyone scored about the same and in the other the marks ran from very low to very high.
The range is the largest value minus the smallest value.
$range = largest value - smallest value$
The interquartile range (IQR) is the upper quartile minus the lower quartile. It is the width of the middle half of the data.
$IQR = UQ - LQ$
For $2, 3, 4, 5, 6, 7, 7, 7, 9, 10, 12$ : the range is $12 - 2 = 10$ , and the IQR is $9 - 4 = 5$ .
Both are single numbers. 'From $2$ to $12$ ' says where the data lie. It is not the range. The range is $10$ .
A larger spread means the values are less consistent. A smaller spread means they are bunched more closely together.

Part 4 of 7

Which measure to use, and why

The objective asks you to distinguish between the purposes of these measures. So you need a reason for each choice, not only a calculation.
The mean uses every value. That is its strength and its weakness. One extreme value, a value far away from the rest, drags the mean towards it.
The weekly earnings of eight café workers, in dollars, are $290, 290, 300, 310, 320, 330, 340$ and $1200$ . The mean is $3380 \div 8 = 422.50$ dollars, which is more than seven of the eight people earn. The median is $315$ dollars, which sits among the typical earnings.
graph diagram — not drawn yet
A horizontal number line from 250 to 1250 dollars with eight dots above it: two dots stacked at 290 and one dot each at 300, 310, 320, 330, 340 and 1200; a vertical marker at 315 labelled 'median 315' and a vertical marker at 422.50 labelled 'mean 422.50'.
The one extreme value at $1200$ pulls the mean away from the cluster. The median stays among the typical values.
The median depends only on the middle of the ordered list, so the size of an extreme value makes no difference to it. Change the $1200$ to $5000$ and the median is still $315$ dollars, while the mean jumps to $897.50$ dollars. Use the median when the data contain extreme values. Incomes and house prices are standard examples, because a few very large values sit far above the rest.
Use the mean when there are no extreme values and you want every value to count, for example the average of a student's test marks.
Use the mode when you want the most common item, such as the shoe size a shop should stock most of, or when the data are not numbers. The mode can be a poor summary when it sits at one end of the data. In the earnings above the mode is $290$ , the lowest value.
The same reasoning applies to spread. The range uses only the two most extreme values, so a single extreme value changes it a lot. The IQR ignores the top quarter and the bottom quarter of the data, so the size of an extreme value makes no difference to it.
That is why the median and the IQR belong together. Both come from positions in the ordered list, and the size of an extreme value changes neither of them.
When you explain a choice, name the feature of the data and its effect. 'The $1200$ is an extreme value, which makes the mean higher than seven of the eight earnings; the median is not affected by it' is a reason. 'The median is more accurate' is not.

Part 5 of 7

The mean from a frequency table

When values repeat, the data are often given in a frequency table. The frequency $f$ is how many times each value $x$ occurs.
A team's goals in $14$ matches: $0$ goals in $3$ matches, $1$ goal in $5$ matches, $2$ goals in $4$ matches and $3$ goals in $2$ matches.
$\begin{matrix} x (goals) & 0 & 1 & 2 & 3 \\ f (matches) & 3 & 5 & 4 & 2 \end{matrix}$
Multiply each value by its frequency to get the total for that column. Five matches with $1$ goal give $5$ goals, four matches with $2$ goals give $8$ goals, and so on.
$f x : 0 \times 3 = 0, 1 \times 5 = 5, 2 \times 4 = 8, 3 \times 2 = 6$
The symbol $\sum$ means 'add up'. $\sum f x$ is the total of all the values, here $19$ goals. $\sum f$ is how many values there are, here $14$ matches. Then divide, exactly as before.
$mean = \frac{\sum f x}{\sum f} = \frac{19}{14} = 1.36 (3 s.f.)$
Divide by $\sum f = 14$ , the number of matches, not by $4$ , the number of columns. The columns are not the data. The matches are.

Part 6 of 7

Estimating the mean from grouped data

In a grouped frequency table the values are collected into classes, such as $10 < t \leq 20$ . You know how many values are in each class, but not what they are.
So you cannot find the exact total. Instead, let the midpoint of each class stand for every value in that class. The midpoint is halfway between the ends of the class.
$midpoint = \frac{lower end + upper end}{2}$
For the continuous class $10 < t \leq 20$ the midpoint is $\frac{10 + 20}{2} = 15$ .
For grouped discrete data, such as $6$ – $10$ books, use the smallest and largest values in the class: the midpoint is $\frac{6 + 10}{2} = 8$ .
Then work exactly as with a frequency table, with the midpoint as $x$ : multiply each midpoint by its frequency, add, and divide by the total frequency.
$estimate of the mean = \frac{\sum f x}{\sum f}, x = midpoint of each class$
The answer is an estimate because each midpoint is only a stand-in for the values in its class. It is exact only if the values in every class average out to the midpoint, and you cannot check that without the original data.
Check your answer against the table. An estimate of the mean must lie between the lowest and the highest class ends. A mean of $85$ minutes from classes that stop at $50$ minutes means you have made a mistake, for example dividing by the number of classes instead of the total frequency.
Keep the full value of $\sum f x$ for the division, and round only the final answer, to $3$ significant figures if it is not exact.

Part 7 of 7

The modal class

With grouped data you cannot name a single most common value, because the individual values are not listed. Instead you name the modal class: the class with the highest frequency.
Write the class itself, in the same form as the table, for example $10 < t \leq 20$ . The highest frequency is how you find the modal class, but it is not the answer, and neither is the midpoint.
If two classes share the highest frequency, both are modal classes.

Worked examples

Example 1. The numbers of text messages sent by $11$ students in one hour are $4, 7, 2, 9, 7, 12, 5, 7, 3, 10, 6$ . Find the mean, the median, the mode, the lower and upper quartiles, the range and the interquartile range.

Show the working

Put the values in order first. The median, quartiles and range all need the ordered list, it makes the mode easy to spot, and writing it out once means you do not sort again for each part.
$2, 3, 4, 5, 6, 7, 7, 7, 9, 10, 12$
Count the ordered values: $11$ . This matches the original list, so no value was dropped while ordering.
Mean: add all the values and divide by $11$ . The answer is not exact, so give it to $3$ significant figures.
$\frac{72}{11} = 6.5454 \dots = 6.55 (3 s.f.)$
Median: the position is $\frac{11 + 1}{2} = 6$ , and the 6th value is $7$ .
Mode: $7$ occurs three times, more than any other value.
Quartiles: $n$ is odd, so leave the median out. The lower half $2, 3, 4, 5, 6$ has median $4$ , so LQ $= 4$ . The upper half $7, 7, 9, 10, 12$ has median $9$ , so UQ $= 9$ .
Spread: the range is $12 - 2 = 10$ , and the IQR is $9 - 4 = 5$ .

Answer: Mean $6.55$ (3 s.f.), median $7$ , mode $7$ , LQ $4$ , UQ $9$ , range $10$ , interquartile range $5$ .

Example 2. The weekly earnings, in dollars, of the $8$ people who work in a café are $310, 290, 340, 300, 1200, 320, 290, 330$ . (a) Find the mean and the median. (b) Explain which of these two averages better describes what a typical worker earns.

Show the working

Order the values.
$290, 290, 300, 310, 320, 330, 340, 1200$
(a) Mean: the total is $3380$ and there are $8$ values.
$\frac{3380}{8} = 422.5$
Median: the position is $\frac{8 + 1}{2} = 4.5$ , so the median is halfway between the 4th value, $310$ , and the 5th value, $320$ .
$\frac{310 + 320}{2} = 315$
(b) Compare the mean with the data. Seven of the eight earnings are below $422.50$ dollars. A 'typical' figure that almost nobody reaches is not typical.
The cause is the $1200$ dollars. It is an extreme value. The mean includes it, but the median depends only on the middle two values, so the $1200$ does not affect it.

Answer: (a) Mean $422.50$ dollars, median $315$ dollars. (b) The median: $1200$ dollars is an extreme value that pulls the mean above seven of the eight earnings, but it does not affect the median.

Example 3. The table shows the time, $t$ minutes, that $40$ patients waited at a clinic. $\begin{matrix} Time (t minutes) & Frequency \\ 0 < t \leq 10 & 6 \\ 10 < t \leq 20 & 13 \\ 20 < t \leq 30 & 11 \\ 30 < t \leq 40 & 7 \\ 40 < t \leq 50 & 3 \end{matrix}$ (a) Write down the modal class. (b) Calculate an estimate of the mean waiting time.

Show the working

(a) The highest frequency is $13$ , so the modal class is $10 < t \leq 20$ . Write the class, not the $13$ .
(b) Find the midpoint of each class: halfway between its two ends.
$5, 15, 25, 35, 45$
Multiply each midpoint by its frequency. Each product estimates the total waiting time for that class.
$5 \times 6 = 30, 15 \times 13 = 195, 25 \times 11 = 275, 35 \times 7 = 245, 45 \times 3 = 135$
Add the products to estimate the total waiting time for everybody. Add the frequencies to confirm there are $40$ patients.
$\sum f x = 880, \sum f = 40$
Divide the estimated total by the number of patients.
$\frac{880}{40} = 22$
Check: $22$ minutes is between $0$ and $50$ , and it is in the class next to the modal class, so it is a sensible value.

Answer: (a) $10 < t \leq 20$ . (b) $22$ minutes.

Example 4. The mean height of the $6$ players in a basketball squad is $1.75$ m. A seventh player joins the squad, and the mean height of all $7$ players becomes $1.77$ m. Work out the height of the new player.

Show the working

You cannot subtract one mean from the other, because they belong to groups of different sizes. Change each mean into a total: total $=$ mean $\times$ number of values.
Total height of the first $6$ players, in metres:
$6 \times 1.75 = 10.5$
Total height of all $7$ players, in metres:
$7 \times 1.77 = 12.39$
The only difference between the two totals is the new player, so subtract.
$12.39 - 10.5 = 1.89$
Check: the new player is taller than the old mean of $1.75$ m, which is why the mean went up. The difference of the means, $0.02$ m, is not a height of a person.

Answer: $1.89$ m.

Where people go wrong

Mistake: Finding the median of the list in the order it is given, without sorting it first.

Instead: Put the values in order, then count along to position $\frac{n + 1}{2}$ .

The median is the middle of the ordered data. The middle of an unsorted list depends only on the order in which the values were written down.

Mistake: Giving the position as the median, for example answering $6$ for a list of $11$ values.

Instead: $\frac{n + 1}{2}$ tells you which value to look at. The median is the value in that position.

The position depends only on how many values there are. Two completely different lists of $11$ values both have their median in position $6$ .

Mistake: Writing the range or the interquartile range as two numbers, such as 'from $2$ to $12$ '.

Instead: Subtract: the range is $12 - 2 = 10$ , and the IQR is UQ $-$ LQ.

A measure of spread is a width, so it is one number. The two end values say where the data lie, not how spread out they are.

Mistake: In a grouped table, dividing $\sum f x$ by the number of classes, or adding the midpoints and dividing by the number of classes.

Instead: Divide $\sum f x$ by $\sum f$ , the total frequency.

In the clinic table there are $40$ patients, not $5$ . Each class holds $f$ values, all represented by its midpoint, so the estimated total is $\sum f x$ and the number of values is $\sum f$ .

Mistake: Giving the highest frequency, or the midpoint of the class, as the modal class.

Instead: The modal class is a class interval, such as $10 < t \leq 20$ .

The question asks which class. The frequency is only how you find it.

Check yourself

Write down the mode and the median of $5, 8, 3, 8, 6, 1, 9$ .
Answer
In order: $1, 3, 5, 6, 8, 8, 9$ . The mode is $8$ and the median (the 4th value) is $6$ .
Find the median of $12, 4, 9, 15, 7, 10$ .
Answer
In order: $4, 7, 9, 10, 12, 15$ . The position is $3.5$ , so the median is $\frac{9 + 10}{2} = 9.5$ .
Find the lower quartile, the upper quartile, the interquartile range and the range of $3, 5, 6, 8, 11, 13, 14$ .
Answer
The median is $8$ . The lower half $3, 5, 6$ gives LQ $= 5$ ; the upper half $11, 13, 14$ gives UQ $= 13$ . IQR $= 13 - 5 = 8$ . Range $= 14 - 3 = 11$ .
The mean of $4$ numbers is $7.5$ . What is the total of the $4$ numbers?
Answer
$4 \times 7.5 = 30$ .
A data set has one value that is far larger than all the others. Which average, mean or median, is affected by that value? Which measure of spread, range or IQR, is not affected by it?
Answer
The mean is affected, because it uses every value. The IQR is not affected by the size of that value, because it uses only the middle half of the ordered data.
Write down the midpoint of the class $20 < m \leq 35$ , and the midpoint of the discrete class $11$ – $15$ .
Answer
$\frac{20 + 35}{2} = 27.5$ and $\frac{11 + 15}{2} = 13$ .
The numbers of books read in a term by $20$ students are grouped: $0$ – $4$ : frequency $5$ ; $5$ – $9$ : frequency $8$ ; $10$ – $14$ : frequency $4$ ; $15$ – $19$ : frequency $3$ . Write down the modal class and calculate an estimate of the mean.
Answer
Modal class $5$ – $9$ . Midpoints $2, 7, 12, 17$ ; $\sum f x = 10 + 56 + 48 + 51 = 165$ ; estimate of the mean $= 165 \div 20 = 8.25$ books.
Why is a mean calculated from a grouped frequency table only an estimate?
Answer
The actual values in each class are not known, so the midpoint of each class is used to stand for every value in it.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The marks, out of $20$ , scored by $15$ students in a test are $14, 9, 17, 11, 19, 8, 13, 11, 16, 6, 14, 10, 18, 11, 15$ . Do not use a calculator in this question.

(a) Write down the mode. 1 mark No calculator

Mark scheme

B1 11

(b) Find the median. 1 mark No calculator

Mark scheme

B1 13

(c) Find the interquartile range. 2 marks No calculator

Mark scheme

M1 LQ = 10 and UQ = 16 seen, or 16 − 10 seen
A1 6

(d) Calculate the mean mark. 2 marks No calculator

Mark scheme

M1 192 ÷ 15 oe, or the sum of the 15 marks divided by 15
A1 12.8 oe, e.g. $\frac{64}{5}$ or $12 \frac{4}{5}$

(e) One more student takes the test later and scores $20$ . Show that the mean of all $16$ marks is $13.25$ . 2 marks No calculator

Mark scheme

M1 (192 + 20) ÷ 16 oe, or 212 ÷ 16 seen
A1 reaches 13.25 with correct working; nfww

Question 2. The masses, $m$ grams, of $50$ apples are recorded in the table. $\begin{matrix} Mass (m grams) & Frequency \\ 80 < m \leq 100 & 4 \\ 100 < m \leq 120 & 11 \\ 120 < m \leq 140 & 18 \\ 140 < m \leq 160 & 12 \\ 160 < m \leq 180 & 5 \end{matrix}$

(a) Write down the modal class. 1 mark

Mark scheme

B1 120 < m ≤ 140

(b) Calculate an estimate of the mean mass of these apples. 4 marks

Mark scheme

M1 midpoints 90, 110, 130, 150, 170 soi, at least four correct
M1 Σfx with their midpoints: 4 × 90 + 11 × 110 + 18 × 130 + 12 × 150 + 5 × 170 oe, 6560 if correct
M1 their Σfx ÷ 50; dep on the previous M1
A1 131.2 cao

(c) Explain why your answer to part (b) is only an estimate. 1 mark

Mark scheme

B1 the actual masses are not known, so midpoints are used to stand for the values in each class oe

Question 3. The prices, in dollars, of the seven houses sold in a village last month were $185 000$ , $172 000$ , $198 000$ , $640 000$ , $190 000$ , $176 000$ and $181 000$ .

(a) Calculate the mean price. 2 marks

Mark scheme

M1 1 742 000 ÷ 7 oe
A1 249 000 or better, e.g. 248 857

(b) Find the median price. 1 mark

Mark scheme

B1 185 000

(c) An estate agent wants one figure to describe the price of a typical house in the village. Explain why the median is a better choice than the mean. 2 marks

Mark scheme

B1 identifies 640 000 as an extreme value, or as much larger than the other prices oe
B1 the mean is higher than six of the seven prices, or the median is not affected by the extreme value oe

How this is examined

Estimating the mean from a grouped frequency table with midpoints: a Calculate on a frequency table of class intervals such as $100 < m \leq 120$ , with the word 'estimate' in the question M1 for the midpoints, e.g. $110$ for $100 < m \leq 120$ ; M1 for $\sum f x$ , the midpoints multiplied by the frequencies and added; M1 for dividing that total by $\sum f$ , the total frequency; A1 for the answer, exact or to $3$ significant figures. Write the $f x$ products down: they show the method, which is what the M marks are for.
Reading the median, quartiles, IQR, range and mode from a list of individual values: a list of raw values followed by short Write down, Find and Calculate parts, one measure in each part, which can be set on the non-calculator paper The ordered list written out; the median as the value in position $\frac{n + 1}{2}$ , not the position itself; for the IQR, both quartiles seen (M1) and then their difference as one number (A1); the range as one number, largest minus smallest.
Working backwards from a mean to a missing value or a new mean: 'The mean of $n$ numbers is …' followed by a Find, Work out or Show (that) for a missing value, or for the mean after a value is added Total $=$ mean $\times$ number of values, written for each group; then the difference of the totals, or the new total divided by the new number of values. Means of groups of different sizes are never added or subtracted directly.
Saying which average suits the data, with a reason: a one- or two-mark Explain after you have calculated both the mean and the median of data that contain an extreme value Name the extreme value and state its effect: 'the $640 000$ is much larger than the other prices, so it pulls the mean above six of the seven prices; the median is not affected by it.' 'The median is more accurate' gives no reason.
Naming the modal class of a grouped table: a one-mark Write down on the same grouped table as an estimate of the mean The class interval exactly as the table writes it, e.g. $120 < m \leq 140$ . The frequency $18$ or the midpoint $130$ is not the class.

Words to know

Mean: The sum of the values divided by the number of values.
Median: The middle value when the data are in order; the value in position $\frac{n + 1}{2}$ .
Mode: The value that occurs most often. A set can have more than one mode, or none.
Lower quartile (LQ): The median of the lower half of the ordered data. A quarter of the data lie below it.
Upper quartile (UQ): The median of the upper half of the ordered data. A quarter of the data lie above it.
Range: The largest value minus the smallest value.
Interquartile range (IQR): The upper quartile minus the lower quartile: the width of the middle half of the data.
Extreme value: A value far away from the rest of the data. Its size changes the mean and the range, but not the median or the IQR.
Class interval: One group in a grouped frequency table, such as $10 < t \leq 20$ or $6$ – $10$ .
Midpoint: The value halfway between the two ends of a class, used to stand for every value in that class.
Modal class: The class in a grouped frequency table with the highest frequency.

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