Worked Solutions
Statistics & Probability — Worked Solutions (Year 9 Maths)
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Worked examples for Year 9 Maths Statistics & Probability. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Comparing data sets and spread
Question
A class records the number of books read over a holiday: $2, 4, 4, 5, 7, 8, 12$. Find the median, the range and the interquartile range (IQR), and explain which of the range or IQR better describes the spread of this data.
Solution
The data is already in order, and there are 7 values, so the median is the 4th value: $5$.
Range $=$ highest $-$ lowest $= 12 - 2 = 10$.
For the IQR, split the data around the median. Lower half $2, 4, 4$ gives $Q_1 = 4$; upper half $7, 8, 12$ gives $Q_3 = 8$. So $\text{IQR} = Q_3 - Q_1 = 8 - 4 = 4$.
The $12$ is an outlier that stretches the range. The IQR ignores the extremes, so it better describes the typical spread. State the reason — the marker wants the why, not just "IQR".
First check the data is in order — it is — and count the values: there are 7. With an odd count the median is simply the middle one, the 4th value, which is $5$.
The range is the full stretch from smallest to largest: $12 - 2 = 10$.
The IQR looks at just the middle 50%, so we find the quartiles. The median splits the list, and we don't include it. The lower three values $2, 4, 4$ have a middle of $4$, so $Q_1 = 4$; the upper three $7, 8, 12$ have a middle of $8$, so $Q_3 = 8$. The IQR is the gap between them: $8 - 4 = 4$.
Now, which describes the spread better? That single $12$ is much larger than the rest, and it inflates the range to $10$. The IQR strips away the extreme values, so its value of $4$ reflects how spread out most of the class is. That's why the IQR is more reliable when there's an outlier.
Ordered, $n = 7$.
- Median = 4th value = $5$
- Range $= 12 - 2 = 10$
- $Q_1 = 4$ (median of $2, 4, 4$), $Q_3 = 8$ (median of $7, 8, 12$)
- IQR $= 8 - 4 = 4$
The $12$ is an outlier inflating the range; IQR ignores extremes, so IQR better describes the spread.
Where the marks go
- 1 mark: Correct median $5$
- 1 mark: Correct range $10$
- 1 mark: Correct IQR $4$ from $Q_1 = 4$ and $Q_3 = 8$
- 1 mark: Justifies IQR as the better measure because the outlier inflates the range
Key idea
The range uses only the extremes, so a single outlier distorts it; the IQR uses the middle 50% and is more reliable when outliers are present.
Example 2 — Relative frequency and probability
Question
A spinner is spun $200$ times and lands on red $46$ times. Find the relative frequency of landing on red, and use it to estimate how many times the spinner would land on red in $500$ spins.
Solution
Relative frequency is just the fraction of spins that gave the result: $\dfrac{46}{200} = 0.23$.
Use that as the experimental probability to predict $500$ spins: $0.23 \times 500 = 115$.
So expect about $115$ reds. Keep the relative frequency as a decimal so the next step is a clean multiplication.
Relative frequency means "how often did it actually happen out of all the tries", so we divide the reds by the total spins: $\dfrac{46}{200}$. That simplifies to $0.23$, or $23\%$.
Because we don't know the spinner's exact design, we use this experimental result as our best estimate of the probability. To predict $500$ spins, we apply that same rate: $0.23 \times 500 = 115$.
So we'd expect around $115$ reds. The idea here is that relative frequency from a large number of trials becomes a trustworthy estimate of probability — and probability times the number of trials gives an expected count.
Relative frequency $= \frac{46}{200} = 0.23$.
Expected count over 500 spins:
- $0.23 \times 500 = 115$
Estimate: about $115$ reds.
Where the marks go
- 1 mark: Correct relative frequency $\frac{46}{200} = 0.23$
- 1 mark: Multiplies relative frequency by $500$
- 1 mark: Correct estimate of $115$ reds
Key idea
Relative frequency (successes ÷ trials) estimates the probability; multiplying it by a new number of trials gives the expected number of outcomes.