Mean, median and mode – Statistics with Ungrouped Data (Senior 1 Mathematics) | Ganzaa

Statistics with Ungrouped Data

Lessons

Senior 1 · Mathematics

Statistics with Ungrouped Data

5 lessons

Lesson 4 of 5

Mean, median and mode

25 min

One number for a whole class

Nine learners took a quiz marked out of 20 and scored 12, 15, 9, 18, 15, 11, 14, 15, 17. If the head teacher asks "How did they do?", you cannot read out nine numbers. You need one number that is typical of all of them: an average. Statistics has three averages, and each answers the question in a different way.

Definition

Measure of central tendency (average): a single value that represents the centre of a data set. The three measures are the mean, the median and the mode.

Notation: the Greek capital letter sigma, $\sum$ , means "add up". So $\sum x$ is the sum of all the values of $x$ , and $n$ is the number of values.

Definition

Arithmetic mean: the sum of the values divided by the number of values. The mean of $x$ is written $\bar{x}$ ("x bar"):

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

Example: the mean of the quiz marks is $\bar{x} = \frac{126}{9} = 14$ .

The mean is the balance point: the total distance of the values below it equals the total distance of the values above it.

Why does the mean balance? If every learner had scored the mean, the total would still be $9 \times 14 = 126$ . The mean shares the total out equally, so the amounts above it exactly make up for the amounts below it.

Definition

Median: the middle value when the data are arranged in order of size. For $n$ values in order, the median is the value in position

\frac{n + 1}{2}.

If $n$ is even, this position is a half, such as 3.5; the median is then the mean of the two middle values.

With $n = 9$ , the median is in position $\frac{9 + 1}{2} = 5$ : four values on each side.

Definition

Mode: the value that occurs most often, the value with the highest frequency. A data set can have one mode, two modes (bimodal), or no mode if every value occurs equally often.

Example: in the quiz marks, 15 occurs three times and every other mark once, so the mode is 15.

Worked example 1: the three averages of raw data

Find the mean, median and mode of the quiz marks: 12, 15, 9, 18, 15, 11, 14, 15, 17.

Solution

Mean: $\sum x = 12 + 15 + 9 + 18 + 15 + 11 + 14 + 15 + 17 = 126$ , so $\bar{x} = \frac{126}{9} = 14.$
Median: in order, 9, 11, 12, 14, 15, 15, 15, 17, 18. With $n = 9$ , the median is the $\frac{9+1}{2} = 5$ th value, which is 15.
Mode: 15 occurs most often (3 times), so the mode is 15.

Worked example 2: an even number of values

The numbers of buses leaving Nyabugogo for Huye between 6:00 and 7:00 on six mornings were 8, 11, 9, 12, 10 and 14. Find the mean and the median.

Solution

Mean: $\bar{x} = \frac{8 + 11 + 9 + 12 + 10 + 14}{6} = \frac{64}{6} \approx 10.67 \text{ buses.}$
Median: in order, 8, 9, 10, 11, 12, 14. The position is $\frac{6+1}{2} = 3.5$ , halfway between the 3rd and 4th values: $\frac{10 + 11}{2} = 10.5$ .

Neither average needs to be a whole number, even though every day had a whole number of buses.

Averages from a frequency table

When the data are in a frequency table, a value $x$ with frequency $f$ contributes $f \times x$ to the total. So

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

The median is found by position, using the cumulative frequency, and the mode is the value with the highest frequency.

Worked example 3: the households of the umudugudu

Find the mean, median and mode of the household sizes from Lesson 2.

Solution Add an $f x$ column and a cumulative frequency column.

People $(x)$	Frequency $(f)$	$f x$	Cumulative frequency
2	3	6	3
3	5	15	8
4	8	32	16
5	7	35	23
6	4	24	27
7	2	14	29
8	1	8	30
Total	$\sum f = 30$	$\sum fx = 134$

Mean: $\bar{x} = \frac{\sum fx}{\sum f} = \frac{134}{30} \approx 4.47$ people per household.
Median: position $\frac{30 + 1}{2} = 15.5$ , between the 15th and 16th households. The cumulative frequency of 3 is only 8, and that of 4 is 16, so the 9th to the 16th values are all 4. Both the 15th and 16th values are 4, so the median is 4.
Mode: the highest frequency is 8, for $x = 4$ , so the mode is 4.

The mode is the value under the tallest bar: 4 people.

Worked example 4: a missing mark

Umutoni's mean mark over five tests is 72. Her first four marks were 68, 75, 70 and 71. What did she score in the fifth test?

Solution Turn the mean back into a total: $\sum x = n \times \bar{x} = 5 \times 72 = 360$ . The first four marks add up to $68 + 75 + 70 + 71 = 284$ , so the fifth mark is

360 - 284 = 76.

Which average should you use?

Six workers in a small furniture workshop are paid, per month: 60,000; 60,000; 65,000; 70,000; 75,000; and the owner, 450,000 Frw.

Mean: $\frac{780\,000}{6} = 130\,000$ Frw. Yet five of the six people earn much less than this, because the one very large value pulls the mean up.
Median: $\frac{65\,000 + 70\,000}{2} = 67\,500$ Frw, which describes a typical worker far better.
Mode: 60,000 Frw, the lowest pay, earned by only two people.

Average	Strength	Weakness
mean	uses every value	pulled by extreme values (outliers)
median	not affected by extreme values	ignores the size of most values
mode	the only average for qualitative data, such as favourite sport	may not exist, or there may be several

Warning

Common mistakes

Forgetting to sort. The median is the middle of the ordered data. The middle of the list as written means nothing.
Giving the position as the median. $\frac{n+1}{2}$ tells you where the median is (for example, the 5th value), not what it is.
Dividing by the number of rows. In a frequency table, divide $\sum fx$ by $\sum f$ , the number of data values, not by the number of rows (7 in Worked example 3).
Giving the frequency as the mode. The mode is the value (4 people), not how often it occurs (8).

Tip

Mode is the most common; median is in the middle, like the median strip in the middle of a road; the mean is the "fair share" you get when the total is shared out equally.

Addvance Maths: Frequency Tables: Mean, Median and Mode (7 min). Watch how an fx column gives the mean, and how the median's position is found by counting down the frequencies.

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Practice

Show your working. Questions 5 to 7 need more than one step.

Find the mean, median and mode of: 7, 3, 9, 5, 7, 8, 4.
Find the median of: 21, 17, 25, 30, 19, 23.
The table shows the number of children in 20 families. Find the mean, median and mode.
Children 0 1 2 3 4
Families 2 5 7 4 2
The mean of 8 numbers is 15. What is their sum?
The mean of four numbers is 10. Three of them are 6, 9 and 13. Find the fourth.
Five learners have a mean height of $150\,\text{cm}$ . A sixth learner, $162\,\text{cm}$ tall, joins them. Find the new mean.
The daily sales of a small shop in one week were (in thousands of Frw) 20, 22, 19, 21, 23, 20 and 115 (a festival day). Which average best describes a normal day's sales? Explain.

Show answersHide answers

Mean $\frac{43}{7} \approx 6.14$ ; in order 3, 4, 5, 7, 7, 8, 9, so the median is 7; mode 7.
In order 17, 19, 21, 23, 25, 30: median $\frac{21 + 23}{2} = 22$ .
$\sum fx = 0 + 5 + 14 + 12 + 8 = 39$ , mean $\frac{39}{20} = 1.95$ ; median position 10.5, and the 8th to 14th values are all 2, so the median is 2; mode 2.
$8 \times 15 = 120$ .
$4 \times 10 - (6 + 9 + 13) = 40 - 28 = 12$ .
Total $5 \times 150 + 162 = 912$ , new mean $\frac{912}{6} = 152\,\text{cm}$ .
The median, 21 thousand Frw. The festival day (115) pulls the mean up to about 34.3 thousand, higher than every normal day.

Sources and credits

8.3: Mean, Median and Mode, from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 26 September 2026). Adapted by Ganzaa.
Video: Frequency Tables: Mean, Median & Mode, Addvance Maths, embedded from YouTube.

The balance-point diagram, the Rwandan data, the worked examples, the practice items and the quiz questions were written by Ganzaa.

1One number for a whole class

2Averages from a frequency table

3Which average should you use?

One number for a whole class

Averages from a frequency table

Which average should you use?