Sec 2 Statistics: Mean, Median, Mode & Grouped Data

Secondary 2 G3 Mathematics · Study guide

Connect an average to the question being asked: use totals for the mean, positions for the median and frequencies for the mode.

Concept → worked examples → practice14 min read · practice linked below

An average answers a particular question

A large collection of values is easier to discuss when we describe its centre. The mean shares the total equally, the median locates the middle observation, and the mode identifies the most frequent value. They need not be equal, and no one measure suits every situation.

Before you begin: revise ordering numbers, fractions and frequency tables. A frequency counts observations; it is not itself the value being measured. Keep the units of the data in your final average.

Three different meanings of average
Measure	How to find it	Useful when…
Mean	$\text{total of values}\div\text{number of values}$	Every value should contribute, such as average cost or total output per person.
Median	Order the values and find the middle; average the two central values for even-sized data.	Extreme values would make the mean unrepresentative.
Mode	Find the value with the greatest frequency.	You need the most common category or size.

There can be several modes when values tie for the greatest frequency. In school descriptions, if all values occur equally often, we say there is no mode rather than select one arbitrarily.

Raw data: totals, positions and frequencies

Worked example 1 — three averages from the same data

Five pupils spend 7, 4, 12, 6 and 6 minutes on a puzzle. Find the mean, median and mode.

For the mean, add all five observations: $7+4+12+6+6=35$ . Divide by five: $\bar{x}=35/5=7\text{ minutes}$ .
For the median, first order the values: $4,6,6,7,12$ . There are five values, so the middle position is $(5+1)/2=3$ . The third value is 6, giving a median of 6 minutes.
The value 6 appears twice, more often than any other value. The mode is 6 minutes, not 2; two is its frequency.

The total is 35 minutes. The 12-minute value has 5 minutes above the mean; the shortfalls of 3, 1 and 1 minutes below the mean also total 5. Equal sharing gives 7 minutes to each of the five observations.

Check: the mean lies between the smallest and largest observations. If the 12-minute observation were corrected to 32 minutes, the total would increase by 20, so the mean would increase by $20/5=4$ to 11 minutes. The median and mode would still be 6 minutes: they are less affected by this extreme value.

For an even number of observations, the median may not appear in the data. In $2,4,7,9$ , the middle positions are 2 and 3, so the median is $(4+7)/2=5.5$ , not position 2.5 as a data value.

Frequency tables: weight each value by its count

Worked example 2 — goals scored in twenty matches

A team records the following numbers of goals per match. Find all three averages.

Goals per match and their frequencies
Goals, $x$	Matches, $f$	Total goals, $fx$	Cumulative frequency
0	2	0	2
1	5	5	7
2	8	16	15
3	4	12	19
4	1	4	20
Total	20	37	—

Each two-goal match contributes 2 goals, and there are eight such matches, contributing $2\times8=16$ goals. This explains the product $fx$ .

$\text{Mean}=\frac{\sum fx}{\sum f}=\frac{37}{20}=1.85\text{ goals per match}.$

With 20 observations, the median is the average of the 10th and 11th values. The cumulative frequencies show that positions 8 through 15 are all 2 goals. Both central values are 2, so the median is 2 goals. The largest frequency is 8, corresponding to 2 goals; the mode is also 2 goals.

Check: include the two zero-goal matches in the denominator. An average of 1.85 is valid even though no match contains 1.85 goals. Dividing by five would count categories rather than matches.

What if a frequency is missing? Suppose the frequency for 2 goals and the table totals were erased, but the mean 1.85 was known. Let that missing frequency be $n$ . The remaining frequencies add to $2+5+4+1=12$ , and their goal total is $0+5+12+4=21$ . Each of the $n$ two-goal matches contributes 2 goals, so

$\frac{21+2n}{12+n}=1.85=\frac{37}{20}.$

Multiply by $20(12+n)$ : $20(21+2n)=37(12+n)$ . Expand to obtain $420+40n=444+37n$ , so $3n=24$ and $n=8$ . The answer is a non-negative integer, as a frequency must be. Substitution gives $(21+16)/(12+8)=37/20=1.85$ , recovering the original table.

Grouped data: use midpoints and state the limitation

Worked example 3 — an estimated mean travel time

Twenty students report their journey times $t$ , in minutes, in the following intervals. Estimate the mean journey time and identify the median class.

Grouped journey times
Time, minutes	Frequency $f$	Midpoint $m$	$fm$
$0\le t<10$	4	5	20
$10\le t<20$	10	15	150
$20\le t<30$	6	25	150
Total	20	—	320

The exact times within each class are unknown. Represent every observation in a class by its midpoint, for example $(10+20)/2=15$ . This gives

$\text{Estimated mean}=\frac{\sum fm}{\sum f}=\frac{320}{20}=16\text{ minutes}.$

The first class contains positions 1–4 and the second contains positions 5–14. Therefore both the 10th and 11th observations lie in $10\le t<20$ , which is the median class. The exact median cannot be recovered from the grouped table. This class also has the highest frequency, so it is the modal class.

Check: a time of exactly 10 minutes belongs to the second class, and exactly 20 belongs to the third. The intervals neither overlap nor leave a gap. The midpoint estimate does not mean every student took exactly 5, 15 or 25 minutes.

A table with exact values, as in the goals example, gives an exact mean. A table with intervals usually gives only an estimate. State “estimated mean” when you have substituted midpoints for unknown observations.

Recover totals before combining averages

Worked example 4 — unequal groups and a missing observation

Twelve pupils have a mean journey time of 14 minutes. Another eight pupils have a mean of 19 minutes. Find the mean for all 20 pupils. One more pupil joins, making the overall mean 16.5 minutes. Find that pupil’s journey time. Treat the stated means as exact.

Use $\text{total}=\text{mean}\times\text{count}$ . The first group’s total is $14\times12=168$ minutes and the second group’s total is $19\times8=152$ minutes.
The combined mean is $(168+152)/(12+8)=320/20=16$ minutes. The simple average $(14+19)/2=16.5$ would give both groups equal weight even though their sizes differ.
After the extra pupil joins, there are 21 pupils and the new total is $16.5\times21=346.5$ minutes.
The new pupil’s time is the increase in the total: $346.5-320=26.5$ minutes.

Check: $(320+26.5)/21=16.5$ . The extra time exceeds the previous mean of 16, as expected when the mean rises. A journey time may be fractional; a number of pupils must be a whole number.

Explore an extreme value’s effect

Move one observation

Four observations stay at 2, 3, 3 and 4 minutes. Change the fifth value and compare the mean and median. Each dot represents one observation; stacked dots represent repeated values.

All observations stay on the same labelled axis, so the movement of the fifth value is visible.

Total = 20 minutes across 5 observations. Mean = 4 minutes; median = 3 minutes; mode = 3 minutes. Increasing only the largest value pulls the mean upward while the median stays at 3.

Choose a measure and explain it in context

A shop deciding which shoe size to stock wants the mode, because the most common size matters. A report on typical income may use the median, because a few very high incomes can raise the mean. A calculation of total cost per person needs the mean, because it uses every cost.

An average alone does not describe the whole data set. The sets $4,5,6$ and $0,5,10$ both have mean and median 5, but their values are distributed very differently. Examine a suitable diagram as well.

Graphs can mislead through an altered axis, unequal class widths or a missing key. Read the scale and what is counted before interpreting the apparent pattern. See statistical diagrams for dot, stem-and-leaf and histogram methods.

Common mistakes and checking strategies

Not ordering before finding the median: write the sorted list or use cumulative frequencies.
Dividing by the number of rows: the mean uses the total frequency, including observations whose value is zero.
Confusing a position with a value: “the 10th observation” tells you where to look, not the median’s numerical value.
Treating a grouped estimate as exact: class midpoints replace unknown values; use “estimated”.
Averaging unequal groups’ means: recover each total first, then divide the combined total by the combined count.

Practise choosing and calculating averages

Use the Sec 2 statistics practice questions for raw data, missing values, frequency tables and grouped estimates. Explain what each answer means, not only how it was calculated.