Worked Solutions
Statistics & Probability — Worked Solutions (Year 7 Maths)
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Worked examples for Year 7 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 — Mean, median, mode and range
Question
A student records the number of books read by seven classmates: $4, 7, 3, 7, 5, 7, 9$. Find the mean, median, mode and range of this data set.
Solution
Sort the data first: $3, 4, 5, 7, 7, 7, 9$. That makes every measure easier.
Mean: add them and divide by 7. $3 + 4 + 5 + 7 + 7 + 7 + 9 = 42$, so $42 \div 7 = 6$.
Median: the middle value of seven sorted numbers is the 4th, which is $7$.
Mode: the most common value is $7$ (it appears three times).
Range: largest minus smallest, $9 - 3 = 6$.
So mean $= 6$, median $= 7$, mode $= 7$, range $= 6$. Sort before you start — guessing the median from an unsorted list is a classic slip.
Let's put the numbers in order first, because the median and range both depend on order: $3, 4, 5, 7, 7, 7, 9$.
The mean is the average — we share the total equally. Adding gives $3 + 4 + 5 + 7 + 7 + 7 + 9 = 42$, and there are 7 values, so $42 \div 7 = 6$.
The median is the middle value. With 7 numbers the middle is the 4th one along, which is $7$.
The mode is the value that appears most often. Here $7$ shows up three times, more than any other, so the mode is $7$.
The range measures how spread out the data is: largest minus smallest, $9 - 3 = 6$.
So mean $= 6$, median $= 7$, mode $= 7$ and range $= 6$. Sorting first is what lets us pick out the middle and the spread with confidence.
Sorted: $3, 4, 5, 7, 7, 7, 9$.
- Mean: $42 \div 7 = 6$
- Median: 4th value $= 7$
- Mode: most frequent $= 7$
- Range: $9 - 3 = 6$
Mean $6$, median $7$, mode $7$, range $6$.
Where the marks go
- 1 mark: Correct mean: total $42 \div 7 = 6$
- 1 mark: Correct median ($7$) from the sorted data
- 1 mark: Correct mode ($7$) and range ($9 - 3 = 6$)
Key idea
Sort the data first, then the mean is the total divided by how many, the median is the middle value, the mode is the most frequent, and the range is largest minus smallest.
Example 2 — Probability of a single event
Question
A bag contains 5 red, 3 blue and 2 green marbles. One marble is drawn at random. Find the probability that it is not blue, giving your answer as a fraction in simplest form.
Solution
Count the total and the favourable outcomes.
Total marbles: $5 + 3 + 2 = 10$.
"Not blue" means red or green: $5 + 2 = 7$.
Probability: $\dfrac{7}{10}$, which is already in simplest form.
So $P(\text{not blue}) = \dfrac{7}{10}$. Watch the wording — "not blue" counts everything except the blue ones.
Probability compares the outcomes we want to all the outcomes possible, written as $\dfrac{\text{favourable}}{\text{total}}$.
First the total number of marbles: $5 + 3 + 2 = 10$.
We want "not blue", which means any marble that isn't blue — the reds and greens. That's $5 + 2 = 7$ marbles.
So the probability is $\dfrac{7}{10}$. Since 7 and 10 share no common factor, it's already in simplest form.
So $P(\text{not blue}) = \dfrac{7}{10}$. The key was reading "not blue" as everything that remains once we set the blue marbles aside.
$P = \dfrac{\text{favourable}}{\text{total}}$.
- Total: $5 + 3 + 2 = 10$
- Not blue: $5 + 2 = 7$
- $P(\text{not blue}) = \dfrac{7}{10}$ (already simplest)
Answer: $\dfrac{7}{10}$.
Where the marks go
- 1 mark: Correct total ($10$) and count of 'not blue' outcomes ($7$)
- 1 mark: Correct probability in simplest form: $\dfrac{7}{10}$
Key idea
Probability is favourable outcomes over total outcomes; "not blue" counts every marble except the blue ones.