Worked Solutions
Statistical Analysis — Worked Solutions (HSC Maths Standard 2)
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Worked examples for HSC Maths Standard 2 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.
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 — Bivariate data and least-squares regression
Question
A study records the number of hours per week $x$ that students spend on a revision app and their score $y$ (out of 100) on a class quiz. The least-squares regression line is
$$\hat{y} = 6.2x + 41,$$
and Pearson's correlation coefficient is $r = 0.84$.
(a) Interpret the gradient of the regression line in context.
(b) Predict the quiz score for a student who uses the app for 5 hours per week.
(c) Comment on the strength and direction of the correlation.
Solution
(a) The gradient is 6.2. For each extra hour per week on the app, the predicted quiz score rises by 6.2 marks.
(b) Substitute $x = 5$: $\hat{y} = 6.2(5) + 41 = 31 + 41 = 72$. Predicted score 72 out of 100.
(c) $r = 0.84$ is positive and close to 1, so a strong positive linear correlation — more app time goes with higher scores.
State the gradient with units and context; "6.2" alone won't earn the interpretation mark.
(a) The gradient tells us how the predicted score changes for each one-unit increase in $x$. Here $x$ is hours per week, so for every extra hour on the app the model predicts a rise of 6.2 marks in the quiz score.
(b) To predict, we just substitute the hours into the line: $\hat{y} = 6.2 \times 5 + 41 = 31 + 41 = 72$, so about 72 out of 100. This is reliable because 5 hours sits within the range of the data we'd expect.
(c) The correlation coefficient $r = 0.84$ is positive (the points trend upward) and close to 1, which signals a strong relationship. So we describe it as a strong positive linear correlation — but remember, this is association, not proof that the app causes higher scores.
(a) Gradient $= 6.2$: each extra hour/week → predicted score up by 6.2 marks.
(b) $x = 5$:
- $\hat{y} = 6.2(5) + 41 = 72$ marks
(c) $r = 0.84$:
- Positive, close to 1
- Strong positive linear correlation
Where the marks go
- 1 mark: Correct contextual interpretation of the gradient (6.2 marks per hour)
- 1 mark: Correct substitution $x = 5$ into the line
- 1 mark: Correct predicted score of 72
- 1 mark: Describes correlation as strong and positive
Key idea
The regression gradient is the predicted change in $y$ per unit of $x$; substitute to predict, and read $r$ for the strength (near $\pm 1$) and direction (sign) of the linear relationship.
Example 2 — The normal distribution
Question
The masses of apples from an orchard are normally distributed with a mean of 150 g and a standard deviation of 12 g.
(a) Find the $z$-score of an apple weighing 174 g.
(b) Using the empirical (68–95–99.7) rule, what percentage of apples weigh between 126 g and 174 g?
(c) What percentage of apples weigh more than 174 g?
Solution
(a) $z = \dfrac{x - \mu}{\sigma} = \dfrac{174 - 150}{12} = \dfrac{24}{12} = 2$. So 174 g is 2 standard deviations above the mean.
(b) 126 g is $z = \dfrac{126 - 150}{12} = -2$. The interval is $\mu \pm 2\sigma$, which by the empirical rule holds 95% of apples.
(c) Outside $\pm 2\sigma$ is $5\%$, split evenly between the two tails, so above 174 g is $\tfrac{5}{2} = $ 2.5%.
Convert to $z$ first, then everything is just the 68–95–99.7 percentages.
(a) A $z$-score measures how many standard deviations a value sits from the mean: $z = \dfrac{x - \mu}{\sigma} = \dfrac{174 - 150}{12} = 2$. So this apple is exactly 2 standard deviations above average.
(b) Notice 126 g gives $z = \dfrac{126 - 150}{12} = -2$, so we're looking at everything between $z = -2$ and $z = +2$ — the $\mu \pm 2\sigma$ band. The empirical rule says about 95% of data lies within two standard deviations of the mean.
(c) The remaining $100\% - 95\% = 5\%$ sits in the two tails. Because the normal curve is symmetric, that 5% splits equally, leaving 2.5% above 174 g.
(a) $z = \dfrac{174 - 150}{12} = 2$.
(b) $126$ g $\to z = -2$; interval is $\mu \pm 2\sigma$:
- Empirical rule → 95%
(c) Tails $= 100\% - 95\% = 5\%$, symmetric:
- Above 174 g $= 2.5\%$
Where the marks go
- 1 mark: Correct $z$-score of 2
- 1 mark: Identifies 95% within $\mu \pm 2\sigma$
- 1 mark: Correct 2.5% above 174 g using symmetry
Key idea
Standardise with $z = \dfrac{x - \mu}{\sigma}$, then apply the 68–95–99.7 rule and the symmetry of the normal curve to find percentages.