How many VCAA Specialist Mathematics questions cover Data analysis, probability and statistics?

AusGrader has 128 VCAA Specialist Mathematics questions on Data analysis, probability and statistics, all with instant AI grading and detailed marking feedback.

VCE Specialist Mathematics — Data analysis, probability and statistics Questions & Answers

Q20

2024

VCAA

Paper 2

1 mark

Q20

1 mark

The masses of avocados in a crop may be assumed to be normally distributed, with a mean of $200$ grams and a standard deviation of $7.5$ grams.

After an avocado of mass $M$ grams is peeled and the stone is removed, the mass of edible flesh $F$ grams is given by $F = 0.70M$ . Four avocados are randomly selected from the crop.

What is the probability, correct to four decimal places, that a total of more than $570$ grams of edible flesh is obtained?

A

$0.0868$

B

$0.1705$

C

$0.2128$

D

$0.3170$

Reveal Answer

A

$0.0868$

This answer is incorrect and likely results from an error in calculating the standard deviation of the combined mass of the four avocados.

B

$0.1705$

Correct Answer

The total edible mass $T$ of 4 avocados has a mean of $4 \times (0.70 \times 200) = 560$ g and a variance of $4 \times (0.70 \times 7.5)^2 = 110.25$ . Calculating $P(T > 570)$ gives $P(Z > \frac{570 - 560}{\sqrt{110.25}}) = P(Z > 0.9524) \approx 0.1705$ .

C

$0.2128$

This is incorrect. It stems from misapplying the properties of variance when combining independent normally distributed variables.

D

$0.3170$

This is the probability that 4 times the mass of a single avocado is greater than $570$ grams. It incorrectly uses $Var(4F) = 16Var(F)$ instead of the correct sum of independent variances $Var(F_1+F_2+F_3+F_4) = 4Var(F)$ .

Q4

2025

QCAA

Paper 1

1 mark

Q4

1 mark

$X$ is a random variable with mean $\mu$ and standard deviation $\sigma$ .

From random samples of $X$ values, each of size $n$ , the sample mean is calculated. This sampling and calculation is repeated a large number of times.

The mean of the distribution of the sample means would be approximately

A

$\frac{\overline{x}}{n}$

B

$\frac{\mu}{\sqrt{n}}$

C

$\overline{x}$

D

$\mu$

Reveal Answer

A

$\frac{\overline{x}}{n}$

This formula does not represent the mean of the sampling distribution. The mean of the sample means is equal to the population mean and is not divided by the sample size $n$ .

B

$\frac{\mu}{\sqrt{n}}$

This incorrectly combines the population mean $\mu$ with the denominator for the standard error. The standard deviation of the sample means is $\frac{\sigma}{\sqrt{n}}$ , but the mean remains $\mu$ .

C

$\overline{x}$

$\overline{x}$ represents the mean of a single specific sample. The question asks for the mean of the distribution of all possible sample means, which is a population parameter.

D

$\mu$

Correct Answer

According to the properties of sampling distributions, the expected value (or mean) of the distribution of sample means is always exactly equal to the population mean $\mu$ .

Q1

2021

QCAA

Paper 2

1 mark

Q1

1 mark

The time taken to complete orders at a pizza store is normally distributed with a mean time ( $\mu$ ) of 10 minutes.
The owner of the pizza store records the time taken to complete orders for a random sample of 20 pizzas each day over a 30-day period. From this data, an approximate 90% confidence interval for $\mu$ is calculated at the end of each day.
How many of these confidence intervals would be expected to contain $\mu$ ?

A

3

B

18

C

27

D

30

Reveal Answer

A

3

This represents $10\%$ of the 30 days ( $0.10 \times 30 = 3$ ). This is the expected number of intervals that would \textit{fail} to contain the mean, not the number that would contain it.

B

18

This represents only $60\%$ of the 30 days ( $0.60 \times 30 = 18$ ). Given a $90\%$ confidence level, the expected number of successful intervals should be higher.

C

27

Correct Answer

By definition, a $90\%$ confidence interval is expected to contain the true population parameter $90\%$ of the time in repeated sampling. Therefore, the expected number is $0.90 \times 30 = 27$ .

D

30

This assumes that every single interval will contain the mean ( $100\%$ ). While possible, the expected value is determined by the specific confidence level of $90\%$ , not $100\%$ .

Q6

2024

VCAA

Paper 2

9 marks

Q6

A machine fills bottles with olive oil. The volume of olive oil dispensed into each bottle may be assumed to be normally distributed with mean $\mu$ millilitres (mL) and standard deviation $\sigma = 4.2$ mL. When the machine is working properly $\mu = 1000$ .

The volume dispensed is monitored regularly by taking a random sample of nine bottles and finding the mean volume dispensed.

The machine will be paused and adjusted if the mean volume of olive oil in the nine bottles is significantly less than $1000$ mL at the $5\%$ level of significance.

When checked, a random sample of nine bottles gave a mean volume of $997.5$ mL.
A one-sided statistical test is to be performed.

Q6e

A new machine is purchased, and it is observed that the volume dispensed by the new machine in $50$ randomly chosen bottles provided a sample mean of $1005$ mL and a sample standard deviation of $4$ mL.

Q6a

1 mark

Write down suitable null and alternative hypotheses $H_0$ and $H_1$ for the test.

Reveal Answer

$H_0: \mu = 1000, \ H_1: \mu < 1000$

Marking Criteria

Descriptor	Marks
Correctly states the null and alternative hypotheses: $H_0: \mu = 1000$ and $H_1: \mu < 1000$	1

Q6b (i)

1 mark

Find the $p$ value for this test correct to three decimal places.

Reveal Answer

Sample mean has standard deviation $\frac{\sigma}{\sqrt{n}} = \frac{4.2}{3} = 1.4$ , so $p\text{-value} = \text{Pr}\left(Z < \frac{997.5 - 1000}{1.4}\right)$

$p = 0.037$ .

Marking Criteria

Descriptor	Marks
Calculates the correct $p$ -value of $0.037$	1

Q6b (ii)

1 mark

Using the $p$ value found in part b.i, state with a reason whether the machine should be paused.

Reveal Answer

$p < 0.05$ , so pause the machine

Marking Criteria

Descriptor	Marks
Correctly states that the machine should be paused and provides a valid reason (e.g., $p < 0.05$ )	1

Q6c

2 marks

Assuming that the mean volume dispensed by the machine each time is in fact $997$ mL and not $1000$ mL, find the probability of a type II error for the test using nine bottles at the $5\%$ level of significance. Assume that the population standard deviation is $4.2$ mL, and give your answer correct to two decimal places.

Reveal Answer

Letting $\bar{X}$ be the sample mean, the null hypothesis will be rejected when $\bar{X} < c$ where $\text{Pr}\left(Z < \frac{c - 1000}{1.4}\right) = 0.05$ , i.e. $\frac{c - 1000}{1.4} = -1.6449$ , so $c = 997.697$ .

Thus the probability of type II error is $\text{Pr}(\bar{X} > 997.697 | \mu = 997) = \text{Pr}\left(Z > \frac{0.697}{1.4}\right) \approx 0.31$ .

Marking Criteria

Descriptor	Marks
Finds the correct critical value ( $c = 997.697$ ) or sets up the correct probability expression for a Type II error	1
Calculates the correct probability of a Type II error: $0.31$	1

Q6d

1 mark

Let $\overline{X}$ denote the sample mean of a random sample of nine bottles. As a quality-control measure, the machine will be paused if $\overline{X} < a$ or if $\overline{X} > b$ , where $\Pr(\overline{X} < a) = 0.01$ and $\Pr(\overline{X} > b) = 0.01$ .

Assume $\mu = 1000$ mL and $\sigma = 4.2$ mL.

Find the values of $a$ and $b$ correct to one decimal place.

Reveal Answer

The sample mean has mean 1000 and standard deviation $\frac{\sigma}{\sqrt{n}} = \frac{42}{\sqrt{9}} = 1.4$

Solve $\text{Pr}\left(Z < \frac{a - 1000}{1.4}\right) = 0.01$ and $\text{Pr}\left(Z > \frac{b - 1000}{1.4}\right) = 0.01$ numerically.

$a = 996.7, \ b = 1003.3$

Marking Criteria

Descriptor	Marks
Calculates both correct values: $a = 996.7$ and $b = 1003.3$	1

Q6e

1 mark

Find a $95\%$ confidence interval for the population mean volume dispensed by the new machine, giving values correct to one decimal place. You may assume a population standard deviation of $4$ mL.

Reveal Answer

95% confidence interval $= \left(\bar{x} - 1.96\frac{4}{\sqrt{50}}, \bar{x} - 1.96\frac{4}{\sqrt{50}}\right)$

The answer is $(1003.9, 1006.1)$ .

Marking Criteria

Descriptor	Marks
Calculates the correct $95\%$ confidence interval: $(1003.9, 1006.1)$	1

Q6f

1 mark

Forty samples, each consisting of $50$ randomly chosen bottles, are taken, and a $95\%$ confidence interval is calculated for each sample.

In how many of these confidence intervals would the population mean volume dispensed by the machine be expected to lie?

Reveal Answer

38

Marking Criteria

Descriptor	Marks
Calculates the correct expected number of confidence intervals: $38$	1

Q6g

1 mark

What minimum size sample should be used so that, with $95\%$ confidence, the sample mean is within $1$ mL of the population mean volume dispensed by the new machine?

Assume a population standard deviation of $4$ mL.

Reveal Answer

The confidence interval extends $1.96\frac{4}{\sqrt{n}}$ each side of the mean, so solve $1.96\frac{4}{\sqrt{n}} < 1$ .

This yields $n > (1.96 \times 4)^2 \approx 61.5$ . As $n$ is an integer, then $n \ge 62$ .

Answer is 62.

Marking Criteria

Descriptor	Marks
Calculates the correct minimum sample size: $62$	1

Q14

2024

QCAA

Paper 2

5 marks

Q14

The height of Year 12 students at a school is normally distributed, with a mean height of 168.6 cm and standard deviation of 12.7 cm.
The heights of a random sample of 20 of these students are recorded.

Q14a

1 mark

Explain why it can be assumed that the sample means for random samples of the heights of students from this school are normally distributed.

Reveal Answer

The distribution of sample means is normally distributed as the population from which a random sample is taken is normally distributed.

Marking Criteria

Descriptor	Marks
correctly explains the assumption based on the normality of the population distribution	1

Q14b

2 marks

Determine the probability that the mean height of this sample will be greater than 170 cm.

Reveal Answer

$\mu_{\bar{X}} = 168.6 \text{ cm}$

$\sigma_{\bar{X}} = \frac{12.7}{\sqrt{20}} \approx 2.84 \text{ cm}$

Using GDC

$P(\bar{X} > 170) \approx 0.31$

Marking Criteria

Descriptor	Marks
correctly determines the value for $\sigma_{\bar{X}}$	1
determines probability	1

Q14c

1 mark

There is a 75% probability that the mean height of this sample will lie within $\pm h$ cm of the population mean.

Determine $P(\bar{X} \ge 168.6 + h)$ .

Reveal Answer

$P(\bar{X} \geq 168.6 + h) = 0.125$

Marking Criteria

Descriptor	Marks
correctly determines the probability	1

Q14d

1 mark

There is a 75% probability that the mean height of this sample will lie within $\pm h$ cm of the population mean.

Use your result from Question 14c) to determine the value of $h$ .

Reveal Answer

Using GDC

$168.6 + h \approx 171.867$

$h \approx 3.27 \text{ cm}$

Marking Criteria

Descriptor	Marks
determines $h$	1

Q10

2022

QCAA

Paper 2

1 mark

Q10

1 mark

In a town, the mean number of residents per household is 3.79 people with a standard deviation of 1.47 people.
Using a random sample of 45 households from the town, determine the probability that the mean number of residents per household will be more than 4.

A

0.17

B

0.33

C

0.83

D

0.96

Reveal Answer

A

0.17

Correct Answer

First, calculate the z-score: $z = \frac{4 - 3.79}{1.47/\sqrt{45}} \approx 0.96$ . The probability $P(Z > 0.96)$ is $1 - 0.8315 \approx 0.17$ .

B

0.33

This value does not result from the standard normal distribution calculation using the Central Limit Theorem parameters provided.

C

0.83

This represents the probability that the mean is less than 4 ( $P(Z < 0.96) \approx 0.83$ ). You must subtract this from 1 to find the probability of being more than 4.

D

0.96

This value is the calculated z-score ( $z \approx 0.96$ ), not the probability associated with that z-score.

Q2

2024

QCAA

Paper 2

1 mark

Q2

1 mark

Rounded to two decimal places, the z-value used in the calculation of an approximate 95% confidence interval for $\mu$ is

A

0.95

B

1.64

C

1.96

D

2.58

Reveal Answer

A

0.95

This value represents the confidence level itself (0.95), not the critical z-score derived from the standard normal distribution.

B

1.64

This z-value (approximately 1.645) is typically used for a 90% confidence interval, corresponding to a tail area of 0.05.

C

1.96

Correct Answer

For a 95% confidence interval, the significance level is $\alpha = 0.05$ . The critical value $z_{\alpha/2}$ leaves $0.025$ in the upper tail, which corresponds to $1.96$ .

D

2.58

This z-value is typically used for a 99% confidence interval, corresponding to a tail area of 0.005.

Q6

2025

VCAA

Paper 2

10 marks

Q6

The volume of water, $V \text{ mL}$ , consumed by a student during a school day may be assumed to be normally distributed with a mean of $1000 \text{ mL}$ and a standard deviation of $80 \text{ mL}$ .

Q6b

The canteen at a particular school stocks two brands of water in bottles, Wasser and Apa.

The manufacturer of Wasser bottled water knows that the volume of water dispensed into bottles may be assumed to be normally distributed with a standard deviation of $5 \text{ mL}$ . Engineers at the company take a random sample of $30 \text{ bottles}$ and measure the volume of water in each bottle. The sample mean is found to be $750 \text{ mL}$ .

Q6e

The volume of water dispensed into Apa water bottles may be assumed to be normally distributed with a mean of $750 \text{ mL}$ and a standard deviation of $5 \text{ mL}$ . After a service, a random sample of $50 \text{ bottles}$ gave a sample mean of $748 \text{ mL}$ . The company now claims that the mean volume of water dispensed is less than the stated mean of $750 \text{ mL}$ .

A one-tailed statistical test at the $1\%$ level of significance is proposed.

Q6a (i)

1 mark

Write down the mean and standard deviation of the sampling distribution for the average volume of water consumed by randomly selected samples of $25 \text{ students}$ .

Give your answers in millilitres.

Reveal Answer

$\mu = 1000, \sigma = 16$

Marking Criteria

Descriptor	Marks
States the correct mean and standard deviation, $\mu = 1000$ and $\sigma = 16$ .	1

Q6a (ii)

1 mark

What is the probability, correct to four decimal places, that the average volume of water consumed by a random sample of $25 \text{ students}$ on a particular school day is more than $970 \text{ mL}$ ?

Reveal Answer

0.9696

Marking Criteria

Descriptor	Marks
States the correct probability, $0.9696$ .	1

Q6b

1 mark

Find a $95\%$ confidence interval for the mean volume of water dispensed into each Wasser bottle.

Give your values in millilitres, correct to one decimal place.

Reveal Answer

$(748.2, 751.8)$

Marking Criteria

Descriptor	Marks
States the correct confidence interval, $(748.2, 751.8)$ .	1

Q6c

1 mark

The engineers decide to take $300 \text{ random samples}$ , each containing $30 \text{ bottles}$ , and calculate the respective $95\%$ confidence intervals. All samples are independent.

In how many of these confidence intervals would the engineers expect the value of the true mean volume dispensed to be included?

Reveal Answer

285

Marking Criteria

Descriptor	Marks
States the correct expected number of confidence intervals, $285$ .	1

Q6d

1 mark

What is the minimum size of the sample required to ensure that the difference between the sample mean and the mean volume dispensed is no more than $1 \text{ mL}$ at the $95\%$ confidence level?

Reveal Answer

97

Found solving the inequality
$1 \ge 1.96 \times \frac{5}{\sqrt{n}}$
$n \ge 96.04$

Marking Criteria

Descriptor	Marks
States the correct minimum sample size, $97$ .	1

Q6e

1 mark

Write down the null and alternative hypotheses that will be used in testing the company's claim.

Reveal Answer

$H_0: \mu = 750$
$H_1: \mu < 750$

Marking Criteria

Descriptor	Marks
States the correct null and alternative hypotheses, $H_0: \mu = 750$ and $H_1: \mu < 750$ .	1

Q6f (i)

1 mark

Determine the $p$ value for this test.

Give your answer correct to four decimal places.

Reveal Answer

0.0023

Marking Criteria

Descriptor	Marks
States the correct $p$ value, $0.0023$ .	1

Q6f (ii)

1 mark

Is the company's claim correct?

Explain your conclusion in terms of the $p$ value.

Reveal Answer

Yes, the company's claim is correct as $0.0023 < 0.01$ (the significance level).

Marking Criteria

Descriptor	Marks
Concludes that the claim is correct and provides a valid explanation comparing the $p$ value to the significance level (e.g., $0.0023 < 0.01$ ).	1

Q6g

1 mark

At the $1\%$ level of significance for a sample size of $50 \text{ bottles}$ , find the critical value of the sample mean, below which a sample mean value would support the conclusion that the mean volume of water dispensed is now less than $750 \text{ mL}$ .

Give your answer correct to three decimal places.

Reveal Answer

748.355

$\text{Pr}(\bar{X} < c) = 0.01$

Marking Criteria

Descriptor	Marks
States the correct critical value, $748.355$ .	1

Q6h

1 mark

Assume that, after the service, the true mean volume of water in the Apa bottles was found to be $747.5 \text{ mL}$ and that the population standard deviation, $\sigma$ , is $5 \text{ mL}$ .

At the $1\%$ level of significance, for a sample size of $50$ , find the probability that the company will conclude that the service has not reduced the mean volume of water in an Apa bottle.

Give your answer correct to three decimal places.

Reveal Answer

0.113

Marking Criteria

Descriptor	Marks
States the correct probability, $0.113$ .	1

Q18

2020

SCSA

Paper 2

11 marks

Q18

The mass of chocolate that is placed into each biscuit produced by the BikkiesAreUs company has been observed to be normally distributed with mean $\mu = 7.5$ grams and standard deviation $\sigma = 1.5$ grams.

Q18a

4 marks

Determine the probability, correct to 0.01, that the total amount of chocolate used for 50 biscuits is less than 365 grams.

Reveal Answer

\begin{align*} \text{Let } \overline{M} &= \text{sample mean for the mass of chocolate per biscuit for 50 biscuits (g)}\\ &= N\left(7.5, \sigma_{\overline{M}}^2\right) \text{ where } \sigma_{\overline{M}} = \frac{1.5}{\sqrt{50}} = 0.21213... \end{align*}

$\text{For a total of 365 g, the sample mean } \overline{M} = \frac{365}{50} = 7.3 \text{ grams per biscuit}$

$\text{Require } P\left(\overline{M} < 7.3\right) = 0.1729... = 0.17$

Marking Criteria

Descriptor	Marks
states that the sample mean is a normal random variable	1
states the correct parameters for the normal random variable	1
calculates the sample mean correctly for the total 365 grams	1
determines the correct probability (to 0.01)	1

Q18b

3 marks

If the probability that the mean amount of chocolate used per biscuit differs from $\mu$ by less than 0.2 grams is 98%, determine $n$ , the number of biscuits that need to be sampled.

Reveal Answer

$\sigma_{\overline{M}} = \frac{1.5}{\sqrt{n}}$

$\text{We require } P(-k < z < k) = 0.98 \text{ this gives } k = 2.326$

$\text{Hence } 2.326\left(\frac{1.5}{\sqrt{n}}\right) < 0.2 \implies \text{Solving gives } n > 304.32$

i.e. we require at least 305 biscuits to have the sample mean differ by less than 0.2 grams

Marking Criteria

Descriptor	Marks
uses the standard $z$ score that represents 98% confidence	1
forms the correct inequality/equation to solve for $n$	1
states the correct minimum integer value for $n$	1

Q18c

4 marks

A competitor company called YouBeautChokkies produces similar biscuits. A sample of 144 biscuits was taken and it was found that the standard deviation of the mass of chocolate used in each biscuit was 1.8 grams and the total amount of chocolate used in the sample of 144 biscuits was 1.09 kg.

Charlie Chokka, a representative from the YouBeautChokkies company, stated that "we are using significantly more chocolate for each biscuit than BikkiesAreUs. If you want that real chocolate taste, then buy from us!"

Perform the necessary calculations to comment on Charlie's claim.

Reveal Answer

Let $\mu_Y$ be the population mean for the mass of chocolate per biscuit for the YBC company (grams).

For the YBC total of 1090 grams, this gives $\overline{M} = 7.56944...$ grams.

$\text{The distribution for } \overline{M} \sim N\left(7.56944, \sigma_{\overline{M}}^2\right) \text{ where } \sigma_{\overline{M}} = \frac{1.8}{\sqrt{144}} = 0.15$

$\text{Confidence Interval for } \mu_Y \text{ 95\% level :}$

\begin{align*} 7.56944 - 1.96\left(\sigma_{\overline{M}}\right) &< \mu_Y < 7.56944 + 1.96\left(\sigma_{\overline{M}}\right)\\ \text{i.e. } 7.2754 &< \mu_Y < 7.8634 \end{align*}

$\text{Confidence Interval for } \mu_Y \text{ 99\% level :}$

\begin{align*} 7.56944 - 2.576\left(\sigma_{\overline{M}}\right) &< \mu_Y < 7.56944 + 2.576\left(\sigma_{\overline{M}}\right)\\ 7.1830 &< \mu_Y < 7.9558 \end{align*}

$\text{The BAU population mean } \mu = 7.5$ is WITHIN the confidence interval using $\overline{M} = 7.56944 \text{ and } \sigma = 1.8$ . i.e. the claim is NOT vindicated.

i.e. the YBC company are NOT using significantly more chocolate per biscuit than compared to BAU.

Marking Criteria

Descriptor	Marks
determines the expected variation using $n = 144$	1
determines an appropriate confidence interval for the YouBeautChokkies population mean	1
states that the BikkiesAreUs population mean 7.5 is within the confidence interval	1
concludes correctly by writing a comment about the claim	1

Q19

2021

QCAA

Paper 2

7 marks

Q19

7 marks

Consider the following information.

	mean	variance
Continuous random variable $X$	$E(X) = \mu = \int_{-\infty}^{\infty} x p(x)dx$	$Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 p(x)dx$

The waiting time (minutes) until workers at a certain call centre receive their $n$ th phone call, where $n \in Z^+$ , is a random variable $T$ with probability density function

$f(t) = \begin{cases} \frac{k^n t^{n-1}}{(n-1)!} e^{-\frac{t}{3}}, & t \ge 0 \\ 0 & , \text{otherwise} \end{cases}$

where $k$ is a positive constant.

The waiting time until workers receive their 5th call is collected from a random sample of 80 workers.
Determine the probability that the mean waiting time from this sample is more than 16 minutes.

Reveal Answer

Using the property of a PDF
$\int_{-\infty}^{\infty} p(x) dx = 1$
Using $n=5$ in the given PDF
$\int_0^\infty \frac{k^5 t^4}{4!} e^{-\frac{t}{3}} dt = 1$

Solving the equation: $k = \frac{1}{3}$
Mean of distribution for waiting time until 5th call, $\mu$
$E(X) = \int_{-\infty}^{\infty} x p(x) dx$
$\mu = \int_0^\infty t \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = \int_0^\infty \frac{\left(\frac{1}{3}\right)^5 t^5}{4!} e^{-\frac{t}{3}} dt$
$= 15 \text{ minutes}$

Variance of distribution for 5th call
$Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 p(x) dx$
$= \int_0^\infty (t-15)^2 \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = 45 \text{ minutes}^2$
$\therefore \sigma = \sqrt{45} \text{ minutes}$

Consider the distribution of the sample mean of the waiting time until the 5th phone call is received, $\bar{T}$ .
As the sample size is large, the distribution of $\bar{T}$ can be considered normal.
$\mu_{\bar{T}} = 15$ and $\sigma_{\bar{T}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{45}}{\sqrt{80}} = 0.75$

Using normal cdf on GDC: $P(\bar{T} > 16) \approx 0.09$

Marking Criteria

Descriptor	Marks
Correctly determines equation in terms of k	1
Solves equation to determine k	1
Determines population mean	1
Determine population variance	1
Justifies that the distribution of T can be considered normal	1
Determines mean and standard deviation of the sample mean	1
Determines required probability	1

Q11

2024

QCAA

Paper 2

4 marks

Q11

A company claims that the mean battery life of their latest model of smartphone is 9.5 hours.
To test this claim, the battery lives of a random sample of 40 of the smartphones were measured.
A sample mean of 9.31 hours and a standard deviation of 0.52 hours were calculated from this data.

Q11a

1 mark

Determine an approximate 95% confidence interval for $\mu$ . Give your answer to at least two decimal places.

Reveal Answer

Given $n=40, \bar{x}=9.31$ and $s=0.52$

Using GDC

$CI(95\%) = (9.15, 9.47)$ hours

Marking Criteria

Descriptor	Marks
correctly calculates 95% confidence interval to at least two decimal places	1

Q11b

1 mark

Determine an approximate 99% confidence interval for $\mu$ . Give your answer to at least two decimal places.

Reveal Answer

Using GDC

$CI(99\%) = (9.10, 9.52)$ hours

Marking Criteria

Descriptor	Marks
correctly calculates 99% confidence interval to at least two decimal places	1

Q11c

2 marks

A manager comments that either confidence interval could be used to support the company’s claim.
Use your results from Questions 11a) and 11b) to evaluate the reasonableness of the manager’s comment. Justify your decision using mathematical reasoning.

Reveal Answer

The 95% confidence interval does not include the claimed mean battery life of 9.5 hours, although the 99% CI does.

So the comment is not reasonable.

Marking Criteria

Descriptor	Marks
justifies decision using mathematical reasoning	1
provides appropriate statement of reasonableness	1

Q19

2024

VCAA

Paper 2

1 mark

Q19

1 mark

When conducting a hypothesis test, a type II error occurs when

A

a null hypothesis is not rejected when the alternative hypothesis is true.

B

a null hypothesis is rejected when it is true.

C

a null hypothesis is rejected when the alternative hypothesis is true.

D

a null hypothesis is not rejected when it is doubtful.

Reveal Answer

A

a null hypothesis is not rejected when the alternative hypothesis is true.

Correct Answer

A Type II error, also known as a false negative, occurs specifically when you fail to reject the null hypothesis even though the alternative hypothesis is actually true.

B

a null hypothesis is rejected when it is true.

Rejecting a true null hypothesis is the definition of a Type I error, often called a false positive, rather than a Type II error.

C

a null hypothesis is rejected when the alternative hypothesis is true.

Rejecting the null hypothesis when the alternative hypothesis is true represents a correct statistical decision, not an error.

D

a null hypothesis is not rejected when it is doubtful.

In hypothesis testing, errors are defined by the actual objective truth of the hypotheses, not subjective terms like being doubtful.

Q6

2023

VCAA

Paper 1

4 marks

Q6

Josie travels from home to work in the city. She drives a car to a train station, waits, and then rides on a train to the city. The time, $X_c$ minutes, taken to drive to the station is normally distributed with a mean of 20 minutes ( $\mu_c = 20$ ) and standard deviation of 6 minutes ( $\sigma_c = 6$ ). The waiting time, $X_w$ minutes, for a train is normally distributed with a mean of 8 minutes ( $\mu_w = 8$ ) and standard deviation of $\sqrt{3}$ minutes ( $\sigma_w = \sqrt{3}$ ). The time, $X_t$ minutes, taken to ride on a train to the city is also normally distributed with a mean of 12 minutes ( $\mu_t = 12$ ) and standard deviation of 5 minutes ( $\sigma_t = 5$ ). The three times are independent of each other.

Q6a

2 marks

Find the mean and standard deviation of the total time, in minutes, it takes for Josie to travel from home to the city.

Reveal Answer

$X_c \sim \text{N}(20, 6^2)$ , $X_w \sim \text{N}\left(8, (\sqrt{3})^2\right)$ and $X_t \sim \text{N}(12, 5^2)$
So
$\text{E}(X_{\text{total}}) = \text{E}(X_c) + \text{E}(X_w) + \text{E}(X_t) = 20 + 8 + 12 = 40$

and

$\text{Var}(X_{\text{total}}) = \text{Var}(X_c) + \text{Var}(X_w) + \text{Var}(X_t) = 36 + 3 + 25 = 64$

$\Rightarrow \text{sd}(X_{\text{total}}) = \sqrt{64} = 8$

Marking Criteria

Descriptor	Marks
Calculates the correct mean of the total time ( $40$ )	1
Calculates the correct standard deviation of the total time ( $8$ )	1

Q6b

2 marks

Josie's waiting time for a train on each work day is independent of her waiting time for a train on any other work day. The probability that, for 12 randomly chosen work days, Josie's average waiting time is between 7 minutes 45 seconds and 8 minutes 30 seconds is equivalent to $\text{Pr}(a < Z < b)$ , where $Z \sim \text{N}(0, 1)$ and $a$ and $b$ are real numbers.

Find the values of $a$ and $b$ .

Reveal Answer

\overline{X}_w \sim \text{N}\left(8, \left(\frac{\sqrt{3}}{\sqrt{12}}\right)^2\right) = \text{N}\left(8, \left(\frac{1}{2}\right)^2\right)

Then

\begin{align*} \text{Pr}(7.75 < \overline{X}_w < 8.5) &= \text{Pr}\left(\frac{7.75-8}{\frac{1}{2}} < Z < \frac{8.5-8}{\frac{1}{2}}\right) \\ &= \text{Pr}\left(-\frac{\frac{1}{4}}{\frac{1}{2}} < Z < \frac{\frac{1}{2}}{\frac{1}{2}}\right) \\ &= \text{Pr}\left(-\frac{1}{2} < Z < 1\right) \end{align*}

Therefore $a = -\frac{1}{2}$ and $b = 1$ .

Marking Criteria

Descriptor	Marks
Determines the correct standard deviation of the sample mean ( $\frac{1}{2}$ ) or sets up the correct standardisation equation	1
States the correct values for $a$ and $b$ ( $a = -\frac{1}{2}$ and $b = 1$ )	1

Q15

2023

QCAA

Paper 2

7 marks

Q15

The travel time for students attending a certain university is assumed to be normally distributed, with a population mean of 25.2 minutes and standard deviation of 4.7 minutes.

Travel times are collected from a random sample of 120 of these students and used to calculate a sample mean, $\bar{X}_1$ , in minutes.

Q15a

2 marks

Determine $P(\bar{X}_1 \leq 25)$ .

Reveal Answer

Given $\mu_{\bar{x}} = 25.2$
$\sigma_{\bar{x}_1} = \frac{\sigma}{\sqrt{n}} = \frac{4.7}{\sqrt{120}}$
$= 0.429 \text{ minutes}$

Using GDC
$P(\bar{X}_1 \le 25) = 0.32$

Marking Criteria

Descriptor	Marks
correctly calculates $\sigma_{\bar{x}}$ for the first sample	1
calculates required probability	1

Q15b

1 mark

Given $P(\bar{X}_1 > k) = 0.9$ , determine the value of $k$ .

Reveal Answer

$P(\bar{X}_1 > k) = 0.9$
Using GDC
$k = 24.65$ minutes

Marking Criteria

Descriptor	Marks
calculates $k$	1

Q15c

4 marks

Travel times are collected from a second random sample of the university's students and used to calculate a second sample mean, $\bar{X}_2$ , in minutes.

Given $P(\bar{X}_2 \leq 25) \approx 0.4$ , determine the number of students in the second sample.

Reveal Answer

$P(z \le z_1) \approx 0.4 \Rightarrow z_1 = -0.253$
$z = \frac{\bar{X}_2 - \mu}{\frac{\sigma}{\sqrt{n}}}$
$-0.253 = \frac{25 - 25.2}{\frac{4.7}{\sqrt{n}}}$

Using GDC
$n \approx 35.3$

The sample size is 35.

Marking Criteria

Descriptor	Marks
correctly calculates the z-value based on given probability	1
determines an equation in terms of the sample size (n)	1
determines an approximate value of n	1
evaluates the reasonableness of the solution by rounding n to an integer value	1

Q9

2020

QCAA

Paper 1

1 mark

Q9

1 mark

The scores on a test are assumed to be normally distributed.
Researchers use the results from a random sample of scores to calculate a confidence interval for the population mean. However, a shorter confidence interval width is required so the researchers decide to use a second sample for their calculations.
Assuming that the standard deviations for both samples are the same, the researchers can ensure that a shorter confidence interval width is produced by

A

decreasing the sample size and decreasing the confidence level.

B

decreasing the sample size and increasing the confidence level.

C

increasing the sample size and decreasing the confidence level.

D

increasing the sample size and increasing the confidence level.

Reveal Answer

A

decreasing the sample size and decreasing the confidence level.

Decreasing the sample size increases the standard error ( $\frac{\sigma}{\sqrt{n}}$ ), which widens the interval and counteracts the narrowing effect of a lower confidence level.

B

decreasing the sample size and increasing the confidence level.

Both decreasing the sample size and increasing the confidence level contribute to a wider confidence interval, not a shorter one.

C

increasing the sample size and decreasing the confidence level.

Correct Answer

A confidence interval width is determined by $2 \times z^* \times \frac{\sigma}{\sqrt{n}}$ . Increasing the sample size ( $n$ ) reduces the standard error, and decreasing the confidence level reduces the critical value ( $z^*$ ), both of which shorten the interval.

D

increasing the sample size and increasing the confidence level.

Increasing the confidence level requires a larger critical value, which widens the interval and opposes the narrowing effect of the increased sample size.

VCAA Specialist Mathematics Data analysis, probability and statistics

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