Averages (Mean, Median, Mode and Range): Question 3

Worked solution

Key idea

The mean links to the total of the scores by

$\text{mean} = \frac{\text{total of scores}}{\text{number of scores}} \quad\Longrightarrow\quad \text{total} = \text{mean} \times \text{number of scores}.$

To find a value that is added, compare the totals before and after, not the means directly.

Step 1: Total of the original 8 scores

$\text{total}_8 = 15 \times 8 = 120 \text{ marks}.$

Step 2: Total of all 9 scores

After the ninth student is included there are $9$ scores with a mean of $16$:

$\text{total}_9 = 16 \times 9 = 144 \text{ marks}.$

Step 3: The ninth score

The ninth score is the extra amount needed to raise the total from $120$ to $144$:

$\text{ninth score} = \text{total}_9 - \text{total}_8 = 144 - 120 = 24 \text{ marks}.$

Check

Adding $24$ to the original total: $120 + 24 = 144$, and $144 \div 9 = 16$, which matches the new mean. ✓

Why the other answers are wrong

• $16$: this is just the new mean, but a single new value does not have to equal the mean it produces.
• $1$: this is only the increase in the mean $(16-15)$, not the change in the total.
• $8$: this comes from $8\times(16-15)$, which wrongly multiplies the old count by the rise; the correct comparison uses both totals.

$\boxed{\text{ninth score} = 24 \text{ marks}}$