Worked Solutions
Statistics & Probability — Worked Solutions (Year 8 Maths)
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Worked examples for Year 8 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 — Interpreting data displays
Question
A class of $20$ students recorded the number of books they read over the holidays. The results were:
$$2,\ 3,\ 1,\ 0,\ 4,\ 2,\ 3,\ 5,\ 2,\ 1,\ 3,\ 2,\ 0,\ 4,\ 2,\ 3,\ 1,\ 2,\ 6,\ 4$$
(a) Find the mean number of books read.
(b) Find the median number of books read.
Solution
(a) The mean is the total divided by how many values there are. Add them: the sum is $50$. There are $20$ students, so mean $= 50 \div 20 = 2.5$ books.
(b) For the median, order the data and find the middle. With $20$ values (an even count) the median is the average of the $10$th and $11$th values. Sorted, both of these are $2$, so the median $= (2 + 2) \div 2 = 2$ books.
For an even number of values, always average the two middle ones — don't just pick one. And order the list before you go hunting for the middle.
(a) The mean is the "fair share" — if every student had read the same number, how many would that be? We add all the values to get the total, which comes to $50$, then share it among the $20$ students: $50 \div 20 = 2.5$ books.
(b) The median is the middle value once the data is in order. Here there are $20$ values, an even number, so there isn't a single middle — instead we sit between the $10$th and $11$th values. After sorting, both of those are $2$, so the median is $(2 + 2) \div 2 = 2$ books.
It helps to remember the mean can be a decimal (a fair share doesn't have to be whole), while the median is found by position — which is why we order the list first.
$n = 20$.
- Sum $= 50$
- (a) Mean $= 50 \div 20 = 2.5$ books
- Even $n$: median = average of $10$th and $11$th sorted values
- Both $= 2$
- (b) Median $= (2 + 2) \div 2 = 2$ books
Answers: mean $2.5$; median $2$.
Where the marks go
- 1 mark: Correct total of $50$
- 1 mark: Correct mean of $2.5$ books
- 1 mark: Correct median of $2$ books (averaging the two middle values)
Key idea
The mean is the total divided by the count; the median is the middle of the ordered data, averaging the two middle values when the count is even.
Example 2 — Probability of compound events
Question
A fair coin is tossed and a fair six-sided die is rolled.
(a) List the sample space and state how many equally likely outcomes there are.
(b) Find the probability of getting a head and an even number.
Solution
(a) Each coin result (H or T) pairs with each die result (1–6), so the outcomes are H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6. That's $2 \times 6 = 12$ equally likely outcomes.
(b) Favourable outcomes are a head with an even number: H2, H4, H6 — that's $3$. So $P(\text{head and even}) = \dfrac{3}{12} = \dfrac{1}{4}$.
Count the total with multiplication ($2 \times 6$), then count only the outcomes that fit both conditions. Always simplify the final fraction.
(a) Because the coin and the die are independent, every coin outcome can go with every die outcome. Listing them gives H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 — and since there are $2$ coin results and $6$ die results, there are $2 \times 6 = 12$ outcomes, all equally likely.
(b) We want both things to happen: a head and an even number. The even numbers are $2, 4, 6$, so the matching outcomes are H2, H4, H6 — three of them. Probability is favourable outcomes over total outcomes: $P = \dfrac{3}{12} = \dfrac{1}{4}$.
The reason we multiply $2 \times 6$ for the total is that the events combine — each of the $2$ coin results branches into $6$ die results — which is the heart of how compound events work.
Coin × die.
- (a) Sample space: H1–H6, T1–T6 ⇒ $2 \times 6 = 12$ outcomes
- Favourable (head and even): H2, H4, H6 ⇒ $3$
- (b) $P = \dfrac{3}{12} = \dfrac{1}{4}$
Answers: $12$ outcomes; $P = \dfrac{1}{4}$.
Where the marks go
- 1 mark: Lists the sample space correctly
- 1 mark: States $12$ equally likely outcomes
- 1 mark: Correct probability $\dfrac{1}{4}$
Key idea
For two independent events the total number of outcomes is the product of each event's outcomes; probability is favourable outcomes over total outcomes.