Mean median mode range

Probability and statistics

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For $4, 7, 7, 9, 13$, find the mode and the range.

2 marks
Find the median of $3, 5, 8, 10$.

2 marks
Find the mean of $6, 8, 11, 15$.

3 marks
Five numbers have mean $8$. Four of them are $5, 7, 9$ and $10$. Find the fifth.

3 marks
For $2, 2, 3, 5, 8, 11$, find the mode, median, mean and range.

4 marks
Four numbers have mean $7$. A fifth number, $12$, is added. Find the new mean.

4 marks
Eight people have median age $24$, range $15$ and mean age $26$. The youngest is $18$. Find the oldest age and the total of the ages. The oldest person leaves. Find the new mean.

5 marks
The marks are $12, 15, 15, 18, 20, 24$. Find the mode, median and mean. What single mark should be added so the new mean is $16$? If instead the $24$ is removed, what is the new mean?

6 marks

Mark scheme

Answer. Mode $7$, range $9$
- $7$ appears twice.
- Range $= 13 - 4 = 9$.
Answer. $6.5$
- There are two middle values.
- $\dfrac{5 + 8}{2} = 6.5$
Answer. $10$
- The total is $40$.
- Mean $= \dfrac{40}{4} = 10$.
Answer. $9$
- The total must be $40$.
- $40 - 31 = 9$
Answer. Mode $2$, median $4$, mean $\dfrac{31}{6}$, range $9$
- The mode is $2$ and the range is $9$.
- The median is $\dfrac{3 + 5}{2} = 4$.
- The total is $31$, so the mean is $\dfrac{31}{6}$.
Answer. $8$
- The original total is $28$.
- New mean $= \dfrac{40}{5} = 8$.
Answer. Oldest $33$, total $208$, new mean $25$
- Oldest $= 18 + 15 = 33$.
- Total $= 8 \times 26 = 208$.
- New total $= 175$, so the new mean is $\dfrac{175}{7} = 25$.
Answer. Mode $15$, median $16.5$, mean $\dfrac{52}{3}$. Add $8$. After removing $24$ the mean is $16$.
- The mode is $15$ and the median is $\dfrac{15 + 18}{2} = 16.5$.
- The total is $104$, so the mean is $\dfrac{104}{6} = \dfrac{52}{3}$.
- Seven numbers with mean $16$ total $112$, so the new mark is $8$.
- Removing $24$ leaves a total of $80$ and a mean of $16$.