Statistical Analysis — Worked Solutions (HSC Maths Advanced)
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Worked examples for HSC Maths Advanced statistical analysis. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Advanced — Statistical Analysis. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
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.
These examples cover a discrete probability distribution and the normal distribution using $z$-scores and the empirical (68–95–99.7) rule.
Example 1 — Discrete probability distribution
Question
A discrete random variable $X$ has the probability distribution below, where $k$ is a constant.
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X=x)$ | $0.1$ | $0.3$ | $k$ | $0.2$ |
Find the value of $k$, then calculate the expected value $E(X)$.
Solution
The probabilities must sum to 1, so that fixes $k$ immediately.
$0.1 + 0.3 + k + 0.2 = 1 \Rightarrow 0.6 + k = 1 \Rightarrow k = 0.4$.
Expected value is $E(X) = \sum x\,P(X=x)$:
$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7$.
So $k = 0.4$ and $E(X) = 1.7$. Don't multiply by $x = 0$ and lose track — that term is zero, but write it so the marker sees the full sum.
Every probability distribution has to account for all outcomes, so the four probabilities add to exactly 1. That single fact gives us $k$.
$0.1 + 0.3 + k + 0.2 = 1$. The known values total $0.6$, so $k = 1 - 0.6 = 0.4$.
The expected value is the long-run average of $X$ — each value weighted by how likely it is. We compute $E(X) = \sum x\,P(X=x)$:
$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2)$.
Working term by term: $0 + 0.3 + 0.8 + 0.6 = 1.7$.
So $k = 0.4$ and $E(X) = 1.7$. Notice $1.7$ isn't a value $X$ can actually take — that's fine, the expected value is an average, not a single outcome.
Probabilities sum to 1:
- $0.1 + 0.3 + k + 0.2 = 1$
- $k = 1 - 0.6 = 0.4$
$E(X) = \sum x\,P(X=x)$:
- $0(0.1) + 1(0.3) + 2(0.4) + 3(0.2)$
- $= 0 + 0.3 + 0.8 + 0.6 = 1.7$
$k = 0.4$, $E(X) = 1.7$.
Where the marks go
- 1 mark: States that the probabilities sum to 1
- 1 mark: Correct value $k = 0.4$
- 1 mark: Sets up $E(X) = \sum x\,P(X=x)$ correctly
- 1 mark: Correct expected value $E(X) = 1.7$
Key idea
Probabilities in a distribution sum to 1; the expected value is $E(X) = \sum x\,P(X=x)$.
Example 2 — The normal distribution and z-scores
Question
The heights of a large group of students are normally distributed with a mean of $\mu = 165$ cm and a standard deviation of $\sigma = 8$ cm.
Calculate the $z$-score of a student who is $181$ cm tall, and use the empirical (68–95–99.7) rule to find the percentage of students taller than $181$ cm.
Solution
Standardise first: $z = \dfrac{x - \mu}{\sigma} = \dfrac{181 - 165}{8} = \dfrac{16}{8} = 2$.
A $z$-score of $2$ means $181$ cm is exactly two standard deviations above the mean. By the empirical rule, $95\%$ of data lies within $z = \pm 2$, so $5\%$ lies outside, split equally between the two tails.
The upper tail (taller than $181$) is half of that: $2.5\%$.
So $z = 2$ and $2.5\%$ of students are taller than $181$ cm. The "half of the leftover" step is where marks are lost — symmetry of the bell curve is doing the work.
A $z$-score tells us how many standard deviations a value sits from the mean, using $z = \frac{x - \mu}{\sigma}$. Here $z = \frac{181 - 165}{8} = \frac{16}{8} = 2$, so this student is 2 standard deviations above average.
Now the empirical rule: about $95\%$ of a normal distribution lies within 2 standard deviations of the mean — that is, between $z = -2$ and $z = 2$. That leaves $100\% - 95\% = 5\%$ sitting in the two tails combined.
Because the bell curve is symmetric, those $5\%$ split evenly: $2.5\%$ in the lower tail and $2.5\%$ in the upper tail. "Taller than 181 cm" is the upper tail, so it's $2.5\%$.
So $z = 2$ and $2.5\%$ of students are taller than $181$ cm. Drawing the bell curve and shading the tail makes the symmetry obvious and stops you halving the wrong number.
$z$-score:
- $z = \frac{x - \mu}{\sigma} = \frac{181 - 165}{8} = 2$
Empirical rule at $z = \pm 2$:
- $95\%$ within $z = \pm 2$
- $5\%$ in the two tails
- Upper tail: $5\% \div 2 = 2.5\%$
$z = 2$; $2.5\%$ are taller than $181$ cm.
Where the marks go
- 1 mark: Correct standardisation formula $z = \frac{x-\mu}{\sigma}$
- 1 mark: Correct $z$-score $z = 2$
- 1 mark: Correct percentage $2.5\%$ using the empirical rule and symmetry
Key idea
Standardise with $z = \frac{x-\mu}{\sigma}$, then use the 68–95–99.7 rule and the curve's symmetry to read off tail percentages.
Frequently asked questions
Step-by-step solutions to exam-style questions on Statistical Analysis in HSC Maths Advanced, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Advanced syllabus for Statistical Analysis, using the methods and notation expected in the exam.