Statistical Analysis — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC Maths Extension 1 the binomial distribution. 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 Extension 1 — 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.
Example 1 — A binomial probability
Question
A fair six-sided die is rolled $5$ times. Find the probability of obtaining exactly two sixes. Give your answer as an exact fraction.
Solution
Binomial with $n = 5$, $p = \tfrac16$ (a six), $q = \tfrac56$. Want exactly $2$ successes:
$$P(X = 2) = \binom{5}{2}\left(\frac16\right)^2\left(\frac56\right)^3.$$
$\binom{5}{2} = 10$, $\left(\tfrac16\right)^2 = \tfrac{1}{36}$, $\left(\tfrac56\right)^3 = \tfrac{125}{216}$.
$$P(X=2) = 10\cdot\frac{1}{36}\cdot\frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}.$$
State $n$, $p$ and the number of successes up front — that's the setup mark.
Each roll is an independent trial that either gives a six (success, $p = \tfrac16$) or not (failure, $q = \tfrac56$), with a fixed number of rolls — that's exactly a binomial situation. The formula is $P(X = k) = \dbinom{n}{k}p^k q^{n-k}$.
Here $n = 5$ and we want $k = 2$ sixes:
$$P(X = 2) = \binom{5}{2}\left(\frac16\right)^2\left(\frac56\right)^3.$$
Work through the pieces: $\binom{5}{2} = 10$ counts which two of the five rolls are sixes; $\left(\tfrac16\right)^2 = \tfrac{1}{36}$ is the chance those two are sixes; $\left(\tfrac56\right)^3 = \tfrac{125}{216}$ is the chance the other three are not.
$$10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}.$$
The $\binom{5}{2}$ matters because the two sixes could land on any pair of rolls — we count all those ways.
- Binomial: $n = 5$, $p = \frac16$, $q = \frac56$, $k = 2$
- $P(X=2) = \binom{5}{2}\left(\frac16\right)^2\left(\frac56\right)^3$
- $= 10\cdot\frac{1}{36}\cdot\frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}$
Where the marks go
- 1 mark: Identifies binomial with $n = 5$, $p = \frac16$
- 1 mark: Correct expression $\binom{5}{2}(\frac16)^2(\frac56)^3$
- 1 mark: Exact answer $\frac{625}{3888}$
Key idea
For "exactly $k$ successes" use $P(X=k) = \binom{n}{k}p^k q^{n-k}$, with the binomial coefficient counting which trials succeed.
Example 2 — Mean and variance of a binomial
Question
A multiple-choice quiz has $80$ questions, each with $4$ options. A student guesses every answer. Let $X$ be the number of correct answers. Find the mean and standard deviation of $X$.
Solution
Binomial: $n = 80$, $p = \tfrac14$, $q = \tfrac34$.
Mean: $E(X) = np = 80 \cdot \tfrac14 = 20$.
Variance: $\operatorname{Var}(X) = npq = 80 \cdot \tfrac14 \cdot \tfrac34 = 15$.
Standard deviation: $\sqrt{15} \approx 3.87$.
Use $np$ for the mean and $npq$ for the variance — then square-root for the SD. Don't quote the variance when asked for SD.
Guessing 80 independent questions, each correct with probability $p = \tfrac14$, is a binomial experiment with $n = 80$. For a binomial there are tidy formulas so we don't sum anything by hand.
The mean (expected number correct) is $E(X) = np$: on average a guesser gets $80 \cdot \tfrac14 = 20$ right. That makes sense — a quarter of 80.
The variance is $\operatorname{Var}(X) = npq$, where $q = 1 - p = \tfrac34$: $80 \cdot \tfrac14 \cdot \tfrac34 = 15$.
The question asks for the standard deviation, which is the square root of the variance: $\sqrt{15} \approx 3.87$.
We square-root because standard deviation is measured in the same units as $X$ (questions), whereas variance is in squared units.
- $n = 80$, $p = \frac14$, $q = \frac34$
- Mean $= np = 20$
- Variance $= npq = 80\cdot\frac14\cdot\frac34 = 15$
- SD $= \sqrt{15} \approx 3.87$
Where the marks go
- 1 mark: Identifies binomial with $n = 80$, $p = \frac14$
- 1 mark: Mean $E(X) = np = 20$
- 1 mark: Variance $npq = 15$
- 1 mark: Standard deviation $\sqrt{15} \approx 3.87$
Key idea
For a binomial, mean $= np$ and variance $= npq$; the standard deviation is $\sqrt{npq}$.
Frequently asked questions
Step-by-step solutions to exam-style questions on Statistical Analysis in HSC Maths Extension 1, 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 Extension 1 syllabus for Statistical Analysis, using the methods and notation expected in the exam.