Summary statistics - Year 7 Mathematics

What you will learn

calculate the mean (average) of a data set,
find the median (middle value when sorted),
find the mode (most common value),
find the range (max $-$ min) as a simple measure of spread,
choose the most appropriate measure for a situation.

Worked example 0 Real-world example: which 'average' house price?

Five houses on a street sold for $420 000, $440 000, $450 000, $460 000, and $1 800 000 (a mansion).

Mean: $(420 + 440 + 450 + 460 + 1800) \div 5 = 3570 \div 5 = 714$ (thousands).
Median (middle value when sorted): $450$ (thousands) — that’s $450 000.
The mansion pulls the mean up to $714 000 — far above what four of the five houses actually cost.
A news headline saying “average house price $714 000” is technically correct but misleading. The median ($450 000) better represents a typical house on this street.

Key idea: one outlier can drag the mean far from the centre. That is why property reports use the median — it resists extreme values.

1. Mean

The mean adds all the values and divides by how many there are.

Mean

\text{mean} \;=\; \frac{\text{sum of all values}}{\text{number of values}}.

Worked example 1 Mean of five numbers

Find the mean of $7, 9, 11, 10, 8$ .

\text{mean} = \frac{7 + 9 + 11 + 10 + 8}{5} = \frac{45}{5} = 9.

2. Median

The median is the middle value after sorting the data from smallest to largest. If the count is even, the median is the mean of the two middle values.

Worked example 2 Median - odd count

Find the median of $5, 8, 3, 9, 7$ .

Sort: $3, 5, 7, 8, 9$ .
The middle (3rd of 5) is $7$ .

\text{median} = 7.

Worked example 3 Median - even count

Find the median of $2, 4, 6, 8$ .

Sort: $2, 4, 6, 8$ . The two middle values are $4$ and $6$ .

\text{median} = \frac{4 + 6}{2} = 5.

3. Mode

The mode is the value that appears most often. A data set can have one mode, two modes (bimodal), or no mode (all values unique).

$1, 2, 2, 3, 4, 4, 4, 5 \;\to\; mode = 4$ .

$1, 2, 2, 3, 3, 4 \;\to\; modes = 2 and 3$ (bimodal).

4. Range

The range is a simple measure of spread.

Range

\text{range} = \text{largest} - \text{smallest}.

5. Which measure to use?

Choosing a measure

Mean

Uses every value. Sensitive to outliers (extreme values can pull it away from the middle).

Median

Only position matters. Not sensitive to outliers - useful when the data has extreme values.

Mode

The typical or most-common value. Useful for categorical data, where mean and median do not apply.

Worked example 4 Effect of an outlier

A small business has five salaries (in thousands): $40, $45, $50, $55, $300 (the owner).

\text{Mean} = \frac{40 + 45 + 50 + 55 + 300}{5} = \frac{490}{5} = 98.

So the mean salary is $98 000. The median (middle after sorting) is $50 000.

The mean is pulled up sharply by the owner’s high salary; the median gives a better picture of “typical” pay.

Symmetric vs skewed data. When data is symmetric, mean ≈ median. When skewed, the mean is pulled toward the tail — use the median instead.

Practice

Fluency

Tier 1: basic skills

Find the mean of $4, 6, 8, 10, 12$ .
Find the mean of $3, 7, 5, 9, 6$ .
Find the mean of $12, 14, 18, 20$ .
Find the median of $1, 3, 5, 7, 9$ .
Find the median of $4, 8, 6, 2, 10$ .
Find the median of $3, 5, 8, 10$ .
Find the median of $11, 13, 14, 17, 18, 20$ .
Find the mode of $2, 3, 3, 5, 7, 3, 9$ .
Find the mode of $4, 4, 6, 6, 7, 8$ .
Find the range of $2, 7, 5, 9, 3$ .
Find the range of $45, 60, 72, 38, 55$ .
A data set: $10, 12, 14, 16, 18, 20$ . Find the mean.
For the set $5, 5, 6, 8, 11$ , find the mean, median, mode, range.

Reasoning

Tier 2: mixed practice

Use this data for questions 1-5: $12, 15, 18, 15, 14, 20, 17, 15, 16, 18$ .

Sort the data and find the median.
Find the mean.
Find the mode.
Find the range.
If you added the value $100$ to this data set, which of mean, median, mode, range would change most? Explain briefly.
Five students scored an average of $70$ on a test. Four of the scores are $65, 72, 68, 75$ . Find the fifth score.
The mean of $6$ numbers is $18$ . Five of them are $15, 20, 12, 18, 22$ . Find the sixth.
A data set has mean $15$ and $10$ values. If one value is wrongly recorded as $8$ but should be $18$ , what is the correct mean?
Give an example of a data set with mean $10$ , median $10$ , mode $10$ and range $0$ .
Give an example of a data set with $5$ values where mean $>$ median.

Reasoning

Tier 3: explain and spot the mistake

A student writes: “the mode is $10$ because $10$ is the biggest number in the list”. Explain the confusion and give the correct definition.
Explain why the median is usually a better measure than the mean when a data set has a single extreme value.
Can a data set have a mode but no mean? Can it have a mean but no mode? Explain both.
Is the median always in the data set? Give an example where it is not.

Problem solving

Tier 4: real-world problems

A cricket batter’s last $7$ scores are $23, 45, 12, 38, 62, 51, 29$ . Find the mean and median.
A family has children aged $4, 7, 10, 13$ . Another child aged $16$ joins. Find the new mean age.
The daily temperatures ( degC) for a week: $22, 25, 21, 28, 30, 24, 23$ . Find the mean temperature and the range.
Seven students scored an average of $60$ marks. Adding a new student with a score of $76$ changes the class size to $8$ . What is the new mean?
A data set of $10$ values has a mean of $50$ . The smallest value is $20$ and the largest is $90$ . What is the mean of the middle $8$ values (the $10$ values with the min and max removed)?

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1: basic skills

Fluency

$8$
$6$
$16$
$5$
$6$ . Sorted: $2, 4, 6, 8, 10$ .
$6.5$ . Sorted: $3, 5, 8, 10$ ; mean of $5$ and $8$ .
$15.5$ . Mean of middle two $14$ and $17$ .
$3$ (appears three times)
$4$ and $6$ (bimodal)
$7$ . Method: $9 - 2$ .
$34$ . Method: $72 - 38$ .
$15$
Mean $7$ ; median $6$ ; mode $5$ ; range $6$ .

Tier 2: mixed practice

Reasoning

Mixed practice

Sorted: $12, 14, 15, 15, 15, 16, 17, 18, 18, 20$ ; median $= \dfrac{15 + 16}{2} = 15.5$ .
$16$ . Method: sum $= 160$ ; $160 \div 10$ .
$15$ (appears three times).
$8$ . Method: $20 - 12$ .
Range changes most (from $8$ to $88$ ). The mode is unaffected. The median shifts only slightly; the mean goes up by about $7.6$ .
$70$ . Method: total required $= 5 \times 70 = 350$ ; subtract $65 + 72 + 68 + 75 = 280$ .
$21$ . Method: total $= 6 \times 18 = 108$ ; sum of five given $= 87$ ; $108 - 87 = 21$ .
$16$ . Method: original total $= 15 \times 10 = 150$ ; correction $+10$ gives $160$ ; new mean $160 \div 10$ .
Any data set with all values equal to $10$ , e.g. $\{10, 10, 10, 10\}$ .
Many possible. Example: $1, 2, 3, 4, 40$ . Mean $= 10$ , median $= 3$ .

Tier 3: explain and spot the mistake

Reasoning

Explain and spot the mistake

The mode is the most frequently occurring value, not the largest one. The student has confused mode with maximum (the upper end of the range). Correct: the mode is whichever value appears most often; a data set can have no mode, one mode, or multiple modes.
The mean uses every value, so a single extreme number can pull it noticeably up or down. The median depends only on position in the sorted list, so one outlier only shifts the middle by one rank at most - hence the median remains close to the bulk of the data when there are extreme values.
Yes to both. Categorical data (e.g. eye colours) can have a mode (most common colour) but no mean - you cannot average “red”, “blue”, “green”. A data set with all distinct numerical values (e.g. $1, 2, 3, 4$ ) has a mean ( $2.5$ ) but no mode, since no value repeats.
Not always. For an even-count numerical set, the median is the average of the two middle values, which may not itself be in the data. Example: median of $\{2, 4\}$ is $3$ , which is not in the set.

Tier 4: real-world problems

Problem solving

Real-world problems

Mean $\approx 37.1$ ; median $38$ . Method: sum $= 260$ ; mean $= 260 / 7$ ; sort and take the $4$ th value ( $38$ ).
Mean age $= 10$ . Method: sum $= 4 + 7 + 10 + 13 + 16 = 50$ ; $50 \div 5$ .
Mean $\approx 24.7$ degC (exactly $173/7$ ); range $9$ degC. Method: $173 \div 7 \approx 24.71$ ; $30 - 21$ .
New mean $= 62$ . Method: previous total $= 7 \times 60 = 420$ ; new total $= 420 + 76 = 496$ ; $496 \div 8 = 62$ .
Mean of middle $8$ is $48.75$ . Method: total of all $10$ is $500$ ; remove $20 + 90 = 110$ ; remaining total $390$ ; $390 \div 8$ .

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