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

AusGrader has 151 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

2021

VCAA

Paper 2

1 mark

Q20

1 mark

An office has two coffee machines that operate independently of each other. The time taken for each machine to produce a cup of coffee is normally distributed with a mean of 30 seconds and a standard deviation of 5 seconds. On a particular morning, a cup of coffee is produced from each machine.

The probability that the time taken by each coffee machine to produce one cup of coffee will differ by less than 3 seconds is closest to

A

0.164

B

0.236

C

0.329

D

0.451

E

0.671

Reveal Answer

A

0.164

This is the probability that the first machine takes up to 3 seconds longer than the second machine, $P(0 < X_1 - X_2 < 3)$ , rather than the absolute difference.

B

0.236

This incorrect value results from adding the standard deviations ( $5 + 5 = 10$ ) instead of the variances when finding the distribution of the difference.

C

0.329

Correct Answer

The difference in times $D = X_1 - X_2$ is normally distributed with mean $0$ and variance $5^2 + 5^2 = 50$ . The probability $P(-3 < D < 3)$ is approximately $0.329$ .

D

0.451

This assumes the standard deviation of the difference is $5$ , failing to account for the combined variance of both independent machines.

E

0.671

This is the probability that the times differ by more than 3 seconds, which is the complement of the correct answer ( $1 - 0.329 = 0.671$ ).

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 $100\%$ .

Q3

2021

VCAA

Paper 1

5 marks

Q3

A company produces a particular type of light globe called Shiny. The company claims that the lifetime of these globes is normally distributed with a mean of 200 weeks and it is known that the standard deviation of the lifetime of Shiny globes is 10 weeks. Customers have complained, saying Shiny globes were lasting less than the claimed 200 weeks. It was decided to investigate the complaints. A random sample of 36 Shiny globes was tested and it was found that the mean lifetime of the sample was 195 weeks.

Use $\Pr(-1.96 < Z < 1.96) = 0.95$ and $\Pr(-3 < Z < 3) = 0.9973$ to answer the following questions.

Q3a

1 mark

Write down the null and alternative hypotheses for the one-tailed test that was conducted to investigate the complaints.

Reveal Answer

$H_0 : \mu = 200$
$H_1 : \mu < 200$

Marking Criteria

Descriptor	Marks
States the correct null and alternative hypotheses ( $H_0 : \mu = 200$ and $H_1 : \mu < 200$ )	1

Q3b (i)

2 marks

Determine the $p$ value, correct to three decimal places, for the test.

Reveal Answer

$\text{Pr}\left(\bar{X} < 195 | \mu = 200\right)$
$= \text{Pr}\left(Z < \frac{195 - 200}{\frac{5}{3}}\right)$
$= \text{Pr}(Z < -3)$
$= 0.001$

Marking Criteria

Working

Descriptor	Marks
Calculates the correct standardised $Z$ value or sets up the correct probability expression (e.g., $\text{Pr}(Z < -3)$ )	1

Answer

Descriptor	Marks
Evaluates the correct $p$ value, correct to three decimal places ( $0.001$ )	1

Q3b (ii)

1 mark

What should the company be told if the test was carried out at the 1% level of significance?

Reveal Answer

Reject the null hypothesis.

Marking Criteria

Descriptor	Marks
States the correct conclusion to reject the null hypothesis	1

Q3c

1 mark

The company decided to produce a new type of light globe called Globeplus.

Find an approximate 95% confidence interval for the mean lifetime of the new globes if a random sample of 25 Globeplus globes is tested and the sample mean is found to be 250 weeks. Assume that the standard deviation of the population is 10 weeks. Give your answer correct to two decimal places.

Reveal Answer

$\left(250 - 1.96 \times \frac{10}{5}, 250 + 1.96 \times \frac{10}{5}\right) = (246.08, 253.92)$

Marking Criteria

Descriptor	Marks
Calculates the correct 95% confidence interval ( $(246.08, 253.92)$ )	1

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 $Var(F_1+F_2+F_3+F_4) = 4Var(F)$ .

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

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

Q3

2022

VCAA

Paper 1

2 marks

Q3

The time taken by a coffee machine to dispense a cup of coffee varies normally with a mean of 10 seconds and a standard deviation of 1.5 seconds.

Q3a

2 marks

Find the probability that more than 34 seconds is needed to dispense a total of four cups of coffee. Give your answer correct to two decimal places.

Reveal Answer

The time to dispense four cups of coffee is a normally distributed random variable with a mean of 40 seconds and a standard deviation of 3 seconds.

If $Z \sim \text{N}(0,1)$ , then the probability that the wait time is greater than 34 is

$\text{Pr}\left(Z > \frac{34-40}{3}\right) = \text{Pr}(Z > -2) \approx 0.975$

Rounding to two decimal places gives 0.98.

Marking Criteria

Descriptor	Marks
Determines the correct mean (40 seconds) and standard deviation (3 seconds) for the total time of four cups, or sets up the correct standardisation $\text{Pr}(Z > -2)$ .	1
Calculates the correct probability, rounded to two decimal places (0.98).	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

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

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

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

Q18

2021

VCAA

Paper 2

1 mark

Q18

1 mark

A scientist investigates the distribution of the masses of fish in a particular river. A 95% confidence interval for the mean mass of a fish, in grams, calculated from a random sample of 100 fish is (70.2, 75.8).

The sample mean divided by the population standard deviation is closest to

A

1.3

B

2.6

C

5.1

D

10.2

E

13.0

Reveal Answer

A

1.3

This is incorrect. The sample mean is 73 and the population standard deviation is approximately 14.29, which does not yield a ratio of 1.3.

B

2.6

This is incorrect. This value is half of the correct ratio, which might result from incorrectly using the full interval width (5.6) instead of the margin of error (2.8) to calculate the standard deviation.

C

5.1

Correct Answer

This is correct. The sample mean is the midpoint of the interval, $\bar{x} = 73$ . The margin of error is 2.8, so $2.8 = 1.96 \times \frac{\sigma}{\sqrt{100}}$ , giving $\sigma \approx 14.29$ . The ratio is $\frac{73}{14.29} \approx 5.1$ .

D

10.2

This is incorrect. This is double the correct ratio, likely resulting from forgetting to divide the interval width by 2 when calculating the margin of error, which would incorrectly halve the calculated standard deviation.

E

13.0

This is incorrect. This value does not represent the ratio of the sample mean (73) to the population standard deviation ( $\approx 14.29$ ).

VCAA Specialist Mathematics Data analysis, probability and statistics

Working

Answer

Frequently Asked Questions

Ready to practise VCAA Specialist Mathematics?

VCAA Specialist Mathematics Data analysis, probability and statistics

Sample Answer

Sample Answer

Working

Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Frequently Asked Questions

Ready to practise VCAA Specialist Mathematics?