Choosing the Mean, Median or Mode

What each average uses, how an extreme value affects it, which average suits categories, and how to choose the average that fits a purpose, with a reason.

What You Should Already Know

The mean is the sum of the values divided by their number (lesson 12.7).
The median is the middle value of the data in order (lesson 12.8).
The mode is the value with the highest frequency (lesson 12.9).

What Each Average Uses

Each average describes the centre of the data in its own way, so the three can give very different answers for the same data. Choosing well starts from what each one uses.

The three averages compared

Mean: uses the size of every value. One extreme value (far from the rest) can pull it a long way. It needs numbers. There is always exactly one mean.

Median: uses only the order of the values. An extreme value hardly moves it. It needs values that can be put in order. There is always exactly one median.

Mode: uses only the frequencies. An extreme value does not change it unless it is repeated. It is the only average for categories with no order, such as colours or flavours. There can be one mode, two modes, or none: if every value occurs the same number of times, there is no mode (5, 7, 9; or 3, 3, 5, 5); otherwise two values sharing the highest frequency are both modes, and more than two sharing it give no mode.

Extreme Values

Example 1: A company with one large salary

The nine employees of a small company earn these monthly salaries: $2000, $2100, $2200, $2400, $2600, $2700, $2800, $2800 and $10 000 (the director). (a) Find the mean, median and mode. (b) Which average best describes what a typical employee earns?

Solution

(a) Sum $= 2000 + 2100 + 2200 + 2400 + 2600 + 2700 + 2800 + 2800 + 10\,000 = 29\,600\text{.}$

Mean $= 29\,600 \div 9 = 3288.88\ldots\text{,}$ which is $3289 to the nearest dollar.

The salaries are in order, and $n = 9$ : the median is the 5th, $2600.

$2800 occurs twice and every other salary once: the mode is $2800.

Drawn to scale, with the mean and the median marked:

(b) Eight of the nine employees earn less than the mean. The one large salary has pulled the mean up, away from the cluster of the others. The median, $2600, sits in the middle of that cluster, and the mode, $2800, is inside it too.

The mean still has a use: mean $\times$ 9 gives the total the company pays each month, $29 600.

(a) Mean $3289, median $2600, mode $2800. (b) The median, $2600: the mean is pulled up by the director's salary, and 8 of the 9 employees earn less than it.

Example 2: A company with no extreme values

The five employees of another company earn $5500, $5700, $5900, $6100 and $6300 a month. Find the three averages and say which describes the salaries well.

Solution

Sum $= 5500 + 5700 + 5900 + 6100 + 6300 = 29\,500\text{,}$ so the mean $= 29\,500 \div 5 = 5900\text{.}$

The median is the 3rd salary, $5900.

Every salary occurs once: there is no mode.

With no extreme value, the mean and the median agree. Either describes the salaries; the mean has the advantage of using every one of them.

Mean $5900, median $5900, no mode. The mean (or the median) describes them well.

Choosing for a Purpose

Which average?

Categories with no order (colours, flavours, CCAs): only the mode can be found. Categories in an order, such as the letter sizes S, M, L, XL of T-shirts (letters, not numbers), have a median as well as a mode, but no mean.

Numbers with an extreme value: the median describes a typical value; the mean is pulled towards the extreme value.

Numbers with no extreme value: the mean, which uses every value; the median will be close to it.

The most common value is wanted (what to stock, what most often happens): the mode.

Always give a reason for the choice, using the data.

Example 3: A property agent

A property agent lists five HDB flats sold in one town last month for $480 000, $510 000, $495 000, $530 000 and $1 250 000. Which average should she quote as the typical price, and why?

Solution

Mean $= (480\,000 + 510\,000 + 495\,000 + 530\,000 + 1\,250\,000) \div 5 = 3\,265\,000 \div 5 = 653\,000\text{.}$

In order: 480 000, 495 000, 510 000, 530 000, 1 250 000. The median is the 3rd, 510 000.

Every price occurs once, so there is no mode.

The flat sold for $1 250 000 is an extreme value. Four of the five flats cost less than the mean of $653 000, so the mean would give buyers a wrong idea of a typical price.

The median, $510 000: the one very expensive flat pulls the mean up to $653 000, more than four of the five prices.

Example 4: A school shop

The histogram shows the sizes of the canvas shoes bought from a school shop last term. (a) Find the mean, median and mode of the sizes. (b) The shop is ordering new stock. Which average is the most useful?

Solution

(a) $\Sigma f = 30$ pairs, and $\Sigma fx = 4 \times 2 + 5 \times 4 + 6 \times 7 + 7 \times 9 + 8 \times 5 + 9 \times 3 = 8 + 20 + 42 + 63 + 40 + 27 = 200\text{.}$

Mean $= 200 \div 30 = 6.66\ldots \approx 6.67$ (to 3 significant figures).

Shoe sizes are numbers that go up in equal steps, so a mean can be worked out, unlike the letter sizes S, M, L of lesson 12.9, which are ordered categories with no mean.

Median: the 15th and 16th values. Running totals 2, 6, 13, 22: both are size 7.

Mode: the tallest bar, size 7.

(b) The mode. The shop needs to know which size is bought most often, so that it orders more of it. A mean of 6.67 is not a size the shop can stock.

(a) Mean 6.67, median 7, mode 7. (b) The mode, size 7: it is the size bought most often.

Example 5: The average that makes a claim look best

A shop advertises, "Our members save an average of $45 a visit!" The savings of 10 members on one visit were $5, $5, $5, $8, $10, $10, $10, $10, $12 and $375. Which average did the shop use? Is the claim fair?

Solution

Mean $= (5 + 5 + 5 + 8 + 10 + 10 + 10 + 10 + 12 + 375) \div 10 = 450 \div 10 = 45\text{.}$ The shop used the mean.

Median: the 5th and 6th values are both 10, so the median is $10. The mode is also $10 (4 times).

One member saved $375; without that visit the other nine saved $75 altogether. Nine of the ten saved $12 or less.

The claim is true about the mean, but it misleads: a typical member saves about $10. When an advertisement says "average", ask which average, and why that one was chosen.

The mean. It is not fair: one saving of $375 pulls the mean up; the median and mode, $10, describe a typical member.

How a Change Affects Each Average

Example 6: Changing and adding a value

Seven pupils did 3, 5, 5, 6, 8, 9 and 13 hours of volunteer work in a month. Find the mean, median and mode. Then find them again if (a) the 13 is changed to 34, (b) instead, an eighth pupil who did 6 hours is added, (c) instead, an eighth pupil who did 20 hours is added.

Solution

Original: sum 49, mean $49 \div 7 = 7$ ; median the 4th value, 6; mode 5.

(a) Sum $49 - 13 + 34 = 70$ , mean 10. The 34 is still the largest, so the median is still 6; the mode is still 5.

(b) 3, 5, 5, 6, 6, 8, 9, 13: sum 55, mean $55 \div 8 = 6.875$ ; median $\frac{6 + 6}{2} = 6$ ; 5 and 6 now occur twice each, so there are two modes, 5 and 6.

Making the largest value larger moves only the mean. Adding a value moves the median no further than halfway to the next value in order (here from 6 to 7 at most), while an extreme new value pulls the mean a long way. The mode changes only if the new value makes another value as common as the mode, or more.

Original: 7, 6, 5. (a) 10, 6, 5 (b) 6.875, 6, modes 5 and 6 (c) 8.625, 7, 5 (mean, median, mode in each).

Common Mistakes

Using the mean to describe a typical value when one value is far from the rest.
Finding a mean of categories (or a median of categories with no order), or of numbers given to categories as labels.
Choosing an average without a reason. Say what the data are like (an extreme value, categories) and what the average is for.
Saying "the average" without saying which: the three can differ a great deal.
Calling one average the "true average". All three are averages; each answers a different question.
Thinking the median never changes. Making an extreme value larger does not move it, but adding or removing a value can.

Practice Questions

Try each question first, then open the solution to check your working. Give a reason for every choice.

Question 1

Five condominium units in an estate were sold for $1 200 000, $1 280 000, $1 240 000, $1 320 000 and $3 800 000 (a penthouse). (a) Find the mean and the median price. (b) Which should a property agent use to describe a typical unit in the estate?

ShowHide solution

(a) Sum $= 8\,840\,000$ , so the mean $= 8\,840\,000 \div 5 = 1\,768\,000\text{.}$

In order: 1 200 000, 1 240 000, 1 280 000, 1 320 000, 3 800 000. Median $=$ 1 280 000.

(b) The median. The penthouse is an extreme value that pulls the mean above four of the five prices.

(a) Mean $1 768 000, median $1 280 000. (b) The median, because of the one extreme price.

Question 2

The bar graph shows the scoops of each flavour sold by an ice cream stall in a week. The stall wants to know which flavour to order more of. (a) Which average should it use, and what is it? (b) Ravi works out $(45 + 38 + 52 + 30) \div 4 = 41.25\text{.}$ Why does this not help?

ShowHide solution

(a) Flavours are categories with no order, so only the mode can be found: Strawberry, with the tallest bar (52 scoops).

(b) 41.25 is the mean number of scoops per flavour, not an average flavour. It does not say which flavour to order.

(a) The mode: Strawberry. (b) It averages the frequencies; it names no flavour.

Question 3

A teacher wants to report how her class did in a test. The marks were 65, 70, 72, 75, 78, 80, 82, 85, 88, 90. (a) Find the mean and the median. (b) Which should she report?

ShowHide solution

(a) Sum $= 785$ ; mean $= 785 \div 10 = 78.5$ . Median $= \frac{78 + 80}{2} = 79$ .

(b) There is no extreme mark, and the mean and median are close. The mean uses every pupil's mark, so it is a good choice; the median would also be acceptable.

(a) Mean 78.5, median 79 (b) The mean, 78.5: no extreme marks, and it uses every mark.

Question 4

The dot diagram shows how many of 20 training sessions each of the 15 members of a CCA attended this term. (a) Find the mean, median and mode. (b) Which average best describes the attendance of a typical member?

ShowHide solution

(a) Sum $= 2 + 3 + 16 + 17 + 18 \times 2 + 19 \times 4 + 20 \times 5 = 250\text{.}$ Mean $= 250 \div 15 = 16.66\ldots \approx 16.7\text{.}$

Median: the 8th value. Counting from the left: 2, 3, 16, 17, 18, 18, then 19 is the 7th to 10th. Median 19.

Mode: 20 (5 members).

(b) The median, 19. Two members who came only 2 and 3 times pull the mean down to 16.7, below the attendance of 12 of the 15 members.

(a) Mean 16.7, median 19, mode 20 (b) The median: two very low values pull the mean down.

Question 5

For each set of data, say which averages can be found, and which is the most suitable. (a) The blood types of 30 blood donors. (b) The heights of the pupils in a class. (c) The monthly allowances of 30 pupils, one of whom gets $500 while the rest get $40 to $80.

ShowHide solution

(a) Blood types are categories with no order: only the mode.

(b) Heights are numbers with no extreme values to expect: the mean and the median can always be found, and the mode if one height is the most common; the mean is suitable.

(c) The mean and the median can be found (and the mode, if one allowance is the most common), but the $500 is an extreme value: the median.

(a) Only the mode. (b) The mean. (c) The median.

Question 6

Seven numbers have a mean of 12, a median of 11 and a mode of 10. The largest number is increased by 14. Find the new mean, median and mode.

ShowHide solution

Sum $= 12 \times 7 = 84$ ; new sum $= 84 + 14 = 98$ ; new mean $= 98 \div 7 = 14$ .

The largest number stays the largest, so the order and the middle value do not change: the median is still 11.

The largest number is at least 11, so it is not one of the 10s; the frequencies of the others do not change. The mode is still 10.

Mean 14, median 11, mode 10.

Question 7

The data 4, 6, 6, 7, 9 have mean 6.4, median 6 and mode 6. A value of 30 is added. Find the three averages again. Which changed most?

ShowHide solution

4, 6, 6, 7, 9, 30: sum $= 62$ , mean $= 62 \div 6 = 10.33\ldots \approx 10.3$ (to 3 significant figures).

Median $= \frac{6 + 7}{2} = 6.5$ . Mode: still 6.

The mean rose by about 3.9, the median by 0.5, and the mode did not change.

Mean 10.3, median 6.5, mode 6: the mean changed most.

Question 8

A tuition centre says its students' marks "improved by 7 marks on average". The improvements of its 8 students were 2, 3, 3, 3, 4, 5, 6 and 30 marks. (a) Find the mean, median and mode. (b) Which did the centre quote, and which would describe a typical student fairly?

ShowHide solution

(a) Mean $= 56 \div 8 = 7$ . Median $= \frac{3 + 4}{2} = 3.5$ . Mode $= 3$ .

(b) It quoted the mean, the largest of the three, pulled up by one student who improved by 30. Seven of the 8 improved by 6 marks or less. The median, 3.5 marks, or the mode, 3 marks, describes a typical student fairly.

(a) Mean 7, median 3.5, mode 3 (b) The mean. The median (3.5) or the mode (3) describes a typical student fairly.

Question 9

The seven workers of a small firm earn $2400, $2400, $2600, $2800, $3000, $3200 and $9600 a month. (a) Find the mean (to the nearest dollar), median and mode. (b) The boss says, "The average salary here is $3714." A worker says, "The average salary is $2400." Who is right, and which average best describes a typical salary?

ShowHide solution

(a) Sum $= 26\,000$ ; mean $= 26\,000 \div 7 = 3714.28\ldots\text{,}$ which is $3714. Median: the 4th, $2800. Mode: $2400.

(b) Both are right about an average: the boss quotes the mean and the worker the mode. Six of the seven earn less than the mean, which the $9600 pulls up. The median, $2800, best describes a typical salary.

(a) Mean $3714, median $2800, mode $2400. (b) Both, for different averages; the median, $2800, is the best description.

Key Points

The mean uses every value; the median uses their order; the mode uses their frequencies.
An extreme value pulls the mean towards it, hardly moves the median, and does not change the mode unless it repeats.
With an extreme value, the median describes a typical value best; with none, the mean (and the median is close to it).
For categories with no order (colours, flavours, CCAs), only the mode can be found; ordered categories (letter sizes S, M, L) also have a median, but no mean; numbers given to categories are labels. Shoe sizes in equal steps are numbers and have a mean.
There is no mode if every value occurs the same number of times (5, 7, 9; or 3, 3, 5, 5); otherwise two values sharing the highest frequency are both modes, and more than two sharing it give no mode.
When the most common value matters (stock, sizes, favourites), use the mode.
Choose an average for the data and the purpose, and give a reason.
An "average" in a claim may be the one that looks best: ask which average was used.