| Average | How to find it | Best used when… |
|---|---|---|
| Mean | $\dfrac{\text{total}}{\text{how many}}$ | The data have no extreme values |
| Median | The middle value when ordered | There are outliers or skewed data |
| Mode | The most common value | The data are categories, or you want the most popular |
Find the mean, median, mode and range of: $8, 3, 5, 8, 12, 3, 8, 9$.
Five employees earn £$22\,000$, £$24\,000$, £$25\,000$, £$26\,000$ and £$180\,000$. Find the mean and median, and say which better describes a typical salary.
The median is the better average here, because the one very large salary pulls the mean upwards. The median is unaffected by that outlier.
Six numbers have a mean of $14$. Five of them are $9, 12, 15, 18$ and $11$. Find the sixth.
Find the mean, median and mode number of pets.
| Pets $x$ | Frequency $f$ | $fx$ |
|---|---|---|
| $0$ | $7$ | $0$ |
| $1$ | $12$ | $12$ |
| $2$ | $9$ | $18$ |
| $3$ | $4$ | $12$ |
| $4$ | $3$ | $12$ |
| Total | $35$ | $54$ |
Estimate the mean height and state the modal class.
| Height $h$ (cm) | $f$ | Midpoint $x$ | $fx$ |
|---|---|---|---|
| $150 \leq h \lt 160$ | $8$ | $155$ | $1240$ |
| $160 \leq h \lt 170$ | $17$ | $165$ | $2805$ |
| $170 \leq h \lt 180$ | $12$ | $175$ | $2100$ |
| $180 \leq h \lt 190$ | $3$ | $185$ | $555$ |
| Total | $40$ | $6700$ |
IQR $= Q_3 - Q_1$
Find $Q_1$, the median, $Q_3$ and the IQR of: $4, 7, 9, 11, 12, 15, 18, 21, 24, 28, 31$.
Find $Q_1$ and $Q_3$ of: $2, 5, 6, 8, 11, 13, 14, 20$.
A data set has $Q_1 = 20$ and $Q_3 = 32$. Is a value of $55$ an outlier?
| Measure | Affected by outliers? |
|---|---|
| Mean | Yes — strongly |
| Median | No |
| Mode | No |
| Range | Yes — very strongly |
| IQR | No |
- Draw a horizontal scale that covers all the data.
- Mark the five values above it.
- Draw a box from $Q_1$ to $Q_3$, with a vertical line at the median.
- Draw whiskers from the box out to the minimum and maximum.
- Label the axis clearly.
• Box width $=$ the IQR
• Whisker to whisker $=$ the range
• $25\%$ of the data lies in each of the four sections
• A median off-centre in the box shows the data are skewed
A data set has minimum $12$, $Q_1 = 20$, median $27$, $Q_3 = 34$ and maximum $50$. Find the range and IQR, and comment on the shape.
The data are positively skewed — stretched out towards the higher values.
| If… | Then say… |
|---|---|
| Higher median | "On average, group A scored higher" |
| Smaller IQR or range | "Group A's results are more consistent" |
| Larger IQR or range | "Group B's results are more varied" |
Two athletes record their long jump distances.
| Median (m) | IQR (m) | Range (m) | |
|---|---|---|---|
| Amy | $5.4$ | $0.3$ | $0.9$ |
| Beth | $5.6$ | $0.8$ | $2.1$ |
Compare their performances, and say who you would pick for a team event where consistency matters.
Mean
Total $\div$ how many. Distorted by outliers.
Median
Middle of the ordered list; position $\dfrac{n+1}{2}$.
Mode
Most common value — state the value, not the frequency.
Missing value
Mean $\times$ how many $=$ total.
Table mean
$\dfrac{\sum fx}{\sum f}$; use midpoints for grouped data.
Grouped is an estimate
Midpoints assume even spread within each class.
IQR
$Q_3 - Q_1$ — the middle $50\%$, unaffected by outliers.
Outlier test
Beyond $Q_1 - 1.5\,\text{IQR}$ or $Q_3 + 1.5\,\text{IQR}$.
Box plot
Five values: min, $Q_1$, median, $Q_3$, max.
Comparing
Average $+$ spread $+$ context. All three.
Find the mean, median, mode and range of: $6, 2, 9, 6, 4, 11, 6$.
▶ Show solution
Ordered: $2, 4, 6, 6, 6, 9, 11$
Mean $= \dfrac{44}{7} = 6.29$ (2 d.p.)
Median: the $4$th value $= 6$
Mode $= 6$
Range $= 11 - 2 = 9$
Seven numbers have a mean of $12$. Six of them are $8, 15, 10, 14, 9$ and $16$. Find the seventh.
▶ Show solution
Total $= 12 \times 7 = 84$
Sum of the six $= 8 + 15 + 10 + 14 + 9 + 16 = 72$
Seventh $= 84 - 72 = 12$
Find the mean number of goals from this table.
| Goals | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| Frequency | $5$ | $8$ | $4$ | $3$ |
▶ Show solution
$\sum f = 5 + 8 + 4 + 3 = 20$
$\sum fx = 0(5) + 1(8) + 2(4) + 3(3) = 0 + 8 + 8 + 9 = 25$
Mean $= \dfrac{25}{20} = 1.25$ goals
Estimate the mean from this grouped table, and state the modal class.
| Time $t$ (min) | Frequency |
|---|---|
| $0 \leq t \lt 10$ | $5$ |
| $10 \leq t \lt 20$ | $12$ |
| $20 \leq t \lt 30$ | $9$ |
| $30 \leq t \lt 40$ | $4$ |
▶ Show solution
Midpoints: $5$, $15$, $25$, $35$.
$\sum fx = 5(5) + 12(15) + 9(25) + 4(35) = 25 + 180 + 225 + 140 = 570$
$\sum f = 30$
Estimated mean $= \dfrac{570}{30} = 19$ minutes
Modal class $= 10 \leq t \lt 20$
Find the interquartile range of: $5, 8, 10, 13, 16, 19, 22, 26, 30$.
▶ Show solution
$n = 9$, so $Q_1$ is at position $\dfrac{10}{4} = 2.5$: midpoint of $8$ and $10$, so $Q_1 = 9$.
$Q_3$ at position $\dfrac{30}{4} = 7.5$: midpoint of $22$ and $26$, so $Q_3 = 24$.
IQR $= 24 - 9 = 15$
A data set has $Q_1 = 14$ and $Q_3 = 26$. Determine whether $48$ is an outlier.
▶ Show solution
IQR $= 26 - 14 = 12$
$1.5 \times 12 = 18$
Upper limit $= 26 + 18 = 44$
$48 \gt 44$, so yes, $48$ is an outlier.
A box plot has minimum $15$, $Q_1 = 22$, median $30$, $Q_3 = 34$, maximum $40$. Find the range and IQR, and describe the skew.
▶ Show solution
Range $= 40 - 15 = 25$
IQR $= 34 - 22 = 12$
Median to $Q_1$ is $30 - 22 = 8$; median to $Q_3$ is $34 - 30 = 4$.
The left half of the box is longer, and the lower whisker ($7$) is longer than the upper ($6$), so the data are negatively skewed — stretched towards the lower values.
Explain why the median is a better average than the mean for house prices in a town.
▶ Show solution
House prices usually include a small number of very expensive properties. These act as outliers and pull the mean upwards, so the mean is higher than most people actually pay.
The median is the middle price and is unaffected by those few extreme values, so it better represents a typical house price.
Two classes sat a test. Class P: median $58$, IQR $22$. Class Q: median $54$, IQR $9$. Write a full comparison.
▶ Show solution
Average: Class P has the higher median ($58$ vs $54$), so on average Class P scored better on the test.
Spread: Class Q has a much smaller IQR ($9$ vs $22$), so Class Q's marks are far more consistent, while Class P's are widely spread.
Overall: Class P did better on average, but Class Q performed much more uniformly — Class P probably contains both very strong and very weak students.
The mean of $10$ numbers is $16$. A further number is added and the mean becomes $17$.
(a) Find the new number. (b) The original $10$ numbers had a range of $12$ and a minimum of $9$. State whether the new number changes the range, explaining your answer. (c) The smallest value, $9$, is then removed from the set of $11$. Find the new mean of the remaining $10$ numbers.
▶ Show solution
(a) Original total $= 16 \times 10 = 160$.
New total $= 17 \times 11 = 187$.
New number $= 187 - 160 = \mathbf{27}$
(b) The original minimum is $9$ and the range is $12$, so the original maximum is $9 + 12 = 21$.
The new number, $27$, is greater than $21$, so it becomes the new maximum.
New range $= 27 - 9 = 18$, so yes, the range increases by $6$.
(c) The set of $11$ numbers totals $187$. Removing the value $9$ leaves $10$ numbers.
New total $= 187 - 9 = 178$
New mean $= \dfrac{178}{10} = \mathbf{17.8}$