How many QCAA Mathematical Methods questions cover Sampling and proportions?

AusGrader has 47 QCAA Mathematical Methods questions on Sampling and proportions, all with instant AI grading and detailed marking feedback.

QCAA Mathematical Methods — Sampling and proportions Questions & Answers

Q3

2025

QCAA

Paper 1

1 mark

Q3

1 mark

A council wants to survey residents about a new dog park. Which sampling method would best minimise bias in the survey?

A

Questioning every third resident entering a supermarket near an existing dog park.

B

Collecting responses from residents who clicked a survey link on the website.

C

Asking residents visiting a dog park on a randomly selected day.

D

Selecting residents using a random number generator.

Reveal Answer

A

Questioning every third resident entering a supermarket near an existing dog park.

Surveying near an existing dog park introduces location bias, as people in that area might have stronger opinions about dog parks than the general population.

B

Collecting responses from residents who clicked a survey link on the website.

This relies on voluntary response sampling, which introduces self-selection bias because only residents with strong opinions are likely to participate.

C

Asking residents visiting a dog park on a randomly selected day.

Surveying at a dog park introduces selection bias by overrepresenting dog owners and excluding residents who do not currently use dog parks.

D

Selecting residents using a random number generator.

Correct Answer

Using a random number generator creates a simple random sample, giving every resident an equal chance of being chosen and effectively minimizing bias.

Q16

2024

QCAA

Paper 2

4 marks

Q16

4 marks

At council meetings in a particular town, new proposals are only discussed if more than 80% of the community are in favour of the proposal.
To discover community opinion on a new bus route proposal, the council conducted several surveys, each with a sample size of 120. The distribution of the sample proportions from the surveys had a standard deviation of 0.04.
Make a justified decision as to whether the new bus route proposal would be discussed at a council meeting.

Reveal Answer

Analytical procedure:

$\sigma=\sqrt{\dfrac{p(1-p)}{n}}$ (formula book)

$0.04=\sqrt{\dfrac{p(1-p)}{120}}$

$0.0016=\dfrac{p(1-p)}{120}$

$0.192=p(1-p)$

$0.192=p-p^2$

$p^2-p+0.192=0$

Use GDC equation solver or graph $y=0.04$ and the
square root function to find intersections:

$p=0.25921$

$=26\%$

or

$p=0.7408$

$=74\%$

Both population proportions (26% and 74%) are less than
80%.

Therefore, the bus route proposal would not be discussed
at a council meeting.

Marking Criteria

Descriptor	Marks
Correctly substitutes the given information into the standard deviation formula for sample proportion	1
Determines a possible value for the population proportion	1
Determines a second possible population proportion value	1
Makes a justified decision regarding the proposal	1

Q10

2022

QCAA

Paper 1

1 mark

Q10

1 mark

A survey plans to draw conclusions based on a random sample of 1% of Queensland's adult population.
To be regarded as a random sample, every

A

adult in the population will be placed in an alphabetical list and every 100th person will be selected for the sample.

B

adult in the population can choose to participate until the sample size has been reached.

C

subgroup within the population will be represented in a similar proportion in the sample.

D

adult in the population will have an equal chance of being selected for the sample.

Reveal Answer

A

adult in the population will be placed in an alphabetical list and every 100th person will be selected for the sample.

This describes systematic sampling (selecting every $n$ th person), which is a specific technique distinct from simple random sampling.

B

adult in the population can choose to participate until the sample size has been reached.

This describes voluntary response sampling, which is a non-probability method that often leads to bias rather than a random sample.

C

subgroup within the population will be represented in a similar proportion in the sample.

This describes a representative sample or stratified sampling; while random samples aim to be representative, they are defined by the selection probability, not the final composition.

D

adult in the population will have an equal chance of being selected for the sample.

Correct Answer

The fundamental definition of a simple random sample is that every member of the population has an equal probability of being selected.

Q13

2020

QCAA

Paper 2

6 marks

Q13

An online retailer claims that 90% of all orders are shipped within 12 hours of being received. On a particular day, 121 orders were received and 102 orders were shipped within 12 hours.

Q13a

1 mark

State the sample proportion of orders shipped within 12 hours.

Reveal Answer

$\hat{p} = \frac{102}{121} = 0.84$

Marking Criteria

Descriptor	Marks
Correctly determines the sample proportion	1

Q13b

3 marks

The distribution of the sample proportion of all orders that are shipped within 12 hours of being received on any day is approximately normal.

Assuming the online retailer's claim is true, find the probability that, in a random sample of 121, less than 85% of all orders are shipped within 12 hours.

Reveal Answer

Using normal distribution
$\mu = 0.9$
$\sigma = \sqrt{\frac{0.9 \times 0.1}{121}}$

Using GDC
$P(\hat{p} < 0.85)$
$= 0.0334$

Marking Criteria

Descriptor	Marks
Correctly identifies the mean of the population	1
Determines expression for standard deviation of the normal distribution	1
Determines probability of producing sample proportion or less	1

Q13c

2 marks

Use the result from 13b) to evaluate the reasonableness of the online retailer's claim.

Reveal Answer

There is 3.34% chance of shipping less than 85% of orders when the population parameter is 0.90.
Observing a sample proportion of 0.843 (or even lower) would have occurred by chance less than 3.34% of the time if the retailer's claim is true. Therefore, we suspect that the retailer's claim is dubious.

Marking Criteria

Descriptor	Marks
Identifies meaning of the probability of producing the sample proportion	1
Evaluates reasonableness of the retailer's claim	1

Q7

2023

QCAA

Paper 2

1 mark

Q7

1 mark

The distribution of a certain sample proportion has a mean of 0.70 and a standard deviation of 0.02.
Determine the sample size.

A

525

B

750

C

1750

D

2500

Reveal Answer

A

525

Correct Answer

Using the formula for the standard deviation of a sample proportion, $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ , we solve $0.02 = \sqrt{\frac{0.70(0.30)}{n}}$ to find $n = \frac{0.21}{0.0004} = 525$ .

B

750

This sample size is incorrect. If $n=750$ , the standard deviation would be $\sqrt{\frac{0.21}{750}} \approx 0.017$ , which does not match the given value of $0.02$ .

C

1750

This sample size is too large. Using $n=1750$ in the standard deviation formula yields approximately $0.011$ , rather than the required $0.02$ .

D

2500

This sample size is incorrect. A sample size of $2500$ would result in a much smaller standard deviation of $\sqrt{\frac{0.21}{2500}} \approx 0.009$ .

Q12

2021

VCAA

Paper 2

1 mark

Q12

1 mark

For a certain species of bird, the proportion of birds with a crest is known to be $\frac{3}{5}$ .

Let $\hat{P}$ be the random variable representing the proportion of birds with a crest in samples of size $n$ for this specific bird.

The smallest sample size for which the standard deviation of $\hat{P}$ is less than 0.08 is

A

7

B

27

C

37

D

38

E

43

Reveal Answer

A

7

Setting up the inequality $\sqrt{\frac{0.6(0.4)}{n}} < 0.08$ yields $n > 37.5$ . A sample size of 7 is much too small and results in a standard deviation of about 0.185.

B

27

Setting up the inequality $\sqrt{\frac{0.6(0.4)}{n}} < 0.08$ yields $n > 37.5$ . A sample size of 27 results in a standard deviation of about 0.094, which is greater than 0.08.

C

37

Setting up the inequality $\sqrt{\frac{0.6(0.4)}{n}} < 0.08$ yields $n > 37.5$ . A sample size of 37 results in a standard deviation of about 0.0805, which is slightly greater than the required 0.08.

D

38

Correct Answer

The standard deviation of a sample proportion is $\sqrt{\frac{p(1-p)}{n}}$ . Solving $\sqrt{\frac{0.6(0.4)}{n}} < 0.08$ yields $n > 37.5$ , making 38 the smallest integer sample size that satisfies the condition.

E

43

Although a sample size of 43 results in a standard deviation less than 0.08, it is not the smallest possible sample size to do so since $n=38$ also satisfies the condition.

Q4

2022

QCAA

Paper 2

1 mark

Q4

1 mark

The distribution for a sample proportion $\hat{p}$ has a mean of 0.15 and a standard deviation of 0.0345.

The sample size is

A

10

B

14

C

107

D

116

Reveal Answer

A

10

This sample size is too small. Using the formula for the standard deviation of a sample proportion, a sample size of 10 would result in a standard deviation of approximately $\sqrt{\frac{0.15(0.85)}{10}} \approx 0.113$ .

B

14

This is incorrect. A sample size of 14 would yield a standard deviation of approximately $\sqrt{\frac{0.15(0.85)}{14}} \approx 0.095$ , which does not match the given value.

C

107

Correct Answer

Using the formula for the standard deviation of a sample proportion $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ , we can solve for $n$ : $n = \frac{0.15(1-0.15)}{0.0345^2} \approx 107.12$ , which rounds to 107.

D

116

This value is incorrect. Solving the equation $0.0345 = \sqrt{\frac{0.15(0.85)}{n}}$ for $n$ results in approximately 107, not 116.

Q16

2025

SCSA

Paper 2

12 marks

Q16

Unexplained respiratory symptoms reported by athletes are sometimes incorrectly thought to be exercise-induced asthma. A researcher wants to investigate the proportion of Australian athletes with unexplained respiratory symptoms who do have exercise-induced asthma. Using a nationwide repository of medical records, the researcher collects a random sample of 71 athletes referred by their doctor for unexplained respiratory symptoms.

Q16a

1 mark

Identify and explain a possible source of bias in the sampling method.

Reveal Answer

Exclusion bias: Athletes with unexplained respiratory symptoms who do not seek medical intervention, or who do not identify themselves as athletes to their doctor, will not be represented in the sample.

Marking Criteria

Descriptor	Marks
correctly explains a valid source of bias	1

Q16b

2 marks

Ignore the potential bias in the sampling method in the remaining parts of the question.

Suppose that 25 athletes from the sample were found to have exercise-induced asthma.

Calculate a 95% confidence interval for the true proportion of athletes with unexplained respiratory symptoms who do have exercise-induced asthma.

Reveal Answer

The sample proportion
$\hat{p} = \frac{25}{71}$
Hence the 95% confidence interval is given by
$95\% \text{ CI} = \left( \frac{25}{71} - 1.96\sqrt{\frac{\frac{25}{71}\left(1-\frac{25}{71}\right)}{71}}, \ \frac{25}{71} + 1.96\sqrt{\frac{\frac{25}{71}\left(1-\frac{25}{71}\right)}{71}} \right)$
$= (0.2410, 0.4632)$

Marking Criteria

Descriptor	Marks
calculates the correct sample proportion	1
calculates the correct 95% confidence interval	1

Q16c

1 mark

Determine the margin of error of the 95% confidence interval from part (b).

Reveal Answer

The margin of error is given by
$E = \frac{0.4632 - 0.2410}{2} = 0.1111$

Marking Criteria

Descriptor	Marks
correctly determines the margin of error	1

Q16d (i)

1 mark

All else remaining unchanged, what would you expect to happen to the margin of error if the sample size was increased.

Reveal Answer

The margin of error would decrease if the sample size is increased.

Marking Criteria

Descriptor	Marks
states that the margin of error would decrease	1

Q16d (ii)

1 mark

All else remaining unchanged, what would you expect to happen to the margin of error if the confidence level was increased.

Reveal Answer

The margin of error would increase if the confidence level is increased.

Marking Criteria

Descriptor	Marks
states that the margin of error would increase	1

Q16d (iii)

2 marks

All else remaining unchanged, what would you expect to happen to the margin of error if the sample proportion of athletes with exercise-induced asthma decreased. Justify your answer.

Reveal Answer

Since the sample proportion is less than 0.5 (0.5 is the sample proportion yielding the largest margin of error), a decrease in sample proportion would move the margin of error away from being maximum. Hence the margin of error would decrease.

Marking Criteria

Descriptor	Marks
states that the margin of error would decrease	1
provides a valid justification based on the sample proportion moving away from 0.5	1

Q16e

2 marks

Determine the minimum sample size required to guarantee a margin of error for the 95% confidence interval of at most 0.04.

Reveal Answer

We require
$0.04 \ge 1.96\sqrt{\frac{0.5(1-0.5)}{n}}$
$\Rightarrow \left(\frac{0.04}{1.96}\right)^2 \ge \frac{0.25}{n}$
$\Rightarrow n \ge 0.25\left(\frac{1.96}{0.04}\right)^2$
$= 600.25$
Since $n$ must be an integer the minimum value of the sample size is $n = 601$ .

Marking Criteria

Descriptor	Marks
correctly substitutes values into the margin of error equation to obtain an equation for $n$	1
correctly solves for $n$ , rounding up to an integer value	1

Q16f

2 marks

A separate large-scale study from the United States of America claims that 20% of American athletes with unexplained respiratory symptoms do have exercise-induced asthma.

Based on the 95% confidence interval calculated in part (b) on page 20, is the proportion of Australian athletes with unexplained respiratory symptoms who do have exercise-induced asthma different from the American proportion? Justify your answer.

Reveal Answer

Yes. The claimed American proportion is outside the confidence interval for Australian athletes, so it is possible to conclude that the proportions are different at the 95% level of confidence.

Marking Criteria

Descriptor	Marks
states that the proportion is different	1
provides a justification based on the American proportion being outside the Australian confidence interval	1

Q4

2021

VCAA

Paper 2

14 marks

Q4

A teacher coaches their school's table tennis team.

The teacher has an adjustable ball machine that they use to help the players practise.

The speed, measured in metres per second, of the balls shot by the ball machine is a normally distributed random variable $W$ .

The teacher sets the ball machine with a mean speed of 10 metres per second and a standard deviation of 0.8 metres per second.

Q4c

The teacher adjusts the height setting for the ball machine. The machine now shoots balls high above the table tennis table.

Unfortunately, with the new height setting, 8% of balls do not land on the table.

Let $\hat{P}$ be the random variable representing the sample proportion of balls that do not land on the table in random samples of 25 balls.

Q4e

The teacher can also adjust the spin setting on the ball machine.

The spin, measured in revolutions per second, is a continuous random variable $X$ with the probability density function

f(x) = \begin{cases} \frac{x}{500} & 0 \leq x < 20 \\ \frac{50 - x}{750} & 20 \leq x \leq 50 \\ 0 & \text{elsewhere} \end{cases}

Q4a

1 mark

Determine $\Pr(W \geq 11)$ , correct to three decimal places.

Reveal Answer

$0.106$

Marking Criteria

Descriptor	Marks
Correctly determines the probability as $0.106$	1

Q4b

1 mark

Find the value of $k$ , in metres per second, which 80% of ball speeds are below. Give your answer in metres per second, correct to one decimal place.

Reveal Answer

$10.7$

Marking Criteria

Descriptor	Marks
Correctly finds the value of $k$ as $10.7$	1

Q4c

2 marks

Find the mean and the standard deviation of $\hat{P}$ .

Reveal Answer

$\text{E}(\hat{P}) = 0.08 = \frac{2}{25}$ , $\text{sd}(\hat{P}) = \frac{\sqrt{46}}{125}$

Marking Criteria

Descriptor	Marks
Correctly calculates the mean: $\text{E}(\hat{P}) = 0.08$ or $\frac{2}{25}$	1
Correctly calculates the exact standard deviation: $\text{sd}(\hat{P}) = \frac{\sqrt{46}}{125}$	1

Q4d

2 marks

Use the binomial distribution to find $\Pr(\hat{P} > 0.1)$ , correct to three decimal places.

Reveal Answer

$X \sim \text{Bi}(25, 0.08)$ , $\text{Pr}(X > 2.5)$ , $\text{Pr}(X \ge 3) = 0.323$

Marking Criteria

Descriptor	Marks
Identifies the correct binomial distribution and probability statement, e.g., $X \sim \text{Bi}(25, 0.08)$ and $\text{Pr}(X \ge 3)$	1
Correctly evaluates the probability as $0.323$	1

Q4e

1 mark

Find the maximum possible spin applied by the ball machine, in revolutions per second.

Reveal Answer

$50$

Marking Criteria

Descriptor	Marks
Correctly identifies the maximum possible spin as $50$	1

Q4f

2 marks

Find the median spin, in revolutions per second, correct to one decimal place.

Reveal Answer

$\int_0^m f(x) dx = \frac{1}{2}$ , $m = 22.6$

Marking Criteria

Descriptor	Marks
Sets up the correct equation using the integral from $0$ to $m$ , e.g., $\int_0^m f(x) dx = \frac{1}{2}$	1
Calculates the correct median spin, $m = 22.6$	1

Q4g

3 marks

Find the standard deviation of the spin, in revolutions per second, correct to one decimal place.

Reveal Answer

$\sigma = \sqrt{\int_0^{50} (x^2 f(x)) dx - \left(\int_0^{50} x f(x) dx\right)^2} = 10.3$

Marking Criteria

Descriptor	Marks
Sets up the correct integral expression for the variance or standard deviation	1
Correctly evaluates the variance or intermediate integrals	1
Correctly calculates the standard deviation as $10.3$	1

Q4h

2 marks

The teacher adjusts the spin setting so that the median spin becomes 30 revolutions per second. This will transform the original probability density function $f$ to a new probability density function $g$ , where $g(x) = a f\left(\frac{x}{b}\right)$ .

Find the values of $a$ and $b$ for which the new median spin is 30 revolutions per second, giving your answer correct to two decimal places.

Reveal Answer

$\int_0^{30} \left(af\left(\frac{x}{b}\right)\right) dx = \frac{1}{2}$ , $\int_0^{50b} \left(af\left(\frac{x}{b}\right)\right) dx = 1$ , $a = 0.75$ and $b = 1.33$

Marking Criteria

Descriptor	Marks
Sets up the correct equations using integrals to find $a$ and $b$ , e.g., $\int_0^{30} \left(af\left(\frac{x}{b}\right)\right) dx = \frac{1}{2}$ and $\int_0^{50b} \left(af\left(\frac{x}{b}\right)\right) dx = 1$	1
Calculates the correct values for $a$ and $b$ , $a = 0.75$ and $b = 1.33$	1

Q14

2022

QCAA

Paper 2

8 marks

Q14

Ravi randomly sampled 200 different pet owners in Brisbane and found that 50 celebrate their pet's birthday.

Q14a

2 marks

Determine an approximate 95% confidence interval for the proportion of Brisbane pet owners who celebrate their pet's birthday.

Reveal Answer

Using GDC to determine confidence interval associated with $n = 200, \hat{p} = 0.25, z = 1.96$

$(0.19, 0.31)$

Marking Criteria

Descriptor	Marks
correctly identifies all of the information required to establish the confidence interval	1
correctly determines the confidence interval	1

Q14b

2 marks

Two of Ravi's friends also randomly sampled Brisbane pet owners. The results are shown in the table.

Friend's name	Number sampled	Number who celebrate their pet's birthday
Khadija	100	26
Tim	150	34

Khadija suggested a more precise estimate for the proportion of Brisbane pet owners who celebrate their pet's birthday could be obtained by combining their results.

Using all available data, determine an approximate 95% confidence interval for the proportion of Brisbane pet owners who celebrate their pet's birthday.

Reveal Answer

Combining results
$n = 450, \hat{p} = \frac{11}{45}$

Using GDC
$(0.2047, 0.2842)$

Marking Criteria

Descriptor	Marks
correctly determines $n$ and $\hat{p}$ for the combined sample	1
determines confidence interval	1

Q14c

2 marks

Use the results from Questions 14a) and 14b) to evaluate the reasonableness of Khadija's suggestion.

Reveal Answer

By combining the results, the sample size is increased and the confidence interval width is reduced.
The new sample statistic provides a better estimate for the population parameter.

Marking Criteria

Descriptor	Marks
identifies changed width of confidence interval	1
evaluates the reasonableness of Khadija’s suggestion	1

Q14d

2 marks

The proportion of all Brisbane pet owners who celebrate their pet's birthday is 0.24.

Using the normal approximation, determine the probability that in a randomly selected sample of size 200, more than 30% of pet owners celebrate their pet's birthday.

Reveal Answer

Using approximation to the normal distribution
Mean $= 0.24$
Standard deviation $= \sqrt{\frac{0.24 \times 0.76}{200}} = 0.0302$
Using GDC
$P(\hat{p} > 0.30) = 0.0235$

Marking Criteria

Descriptor	Marks
correctly determines the mean and standard deviation of the normal distribution	1
determines the probability	1

Q10

2025

SCSA

Paper 2

12 marks

Q10

A cognitive ability test is developed for Australian students aged 15 years. Reported scores are normally distributed with a mean of 50 and a standard deviation of 10. Let the random variable $X$ denote the score of a randomly selected 15-year-old Australian student.

Q10a

2 marks

Calculate the percentage of Australian 15-year-old students you expect to obtain a score of at least 64 on the test.

Reveal Answer

We are given that $X \sim \text{N}(50, 10^2)$ . Hence

$\text{P}(X \ge 64) = 0.0808$

and so, we expect 8.08% of Australian 15-year-old students to obtain a score of at least 64.

Marking Criteria

Descriptor	Marks
correctly calculates the probability of a student obtaining a score of at least 64	1
converts probability to a percentage	1

Q10b

2 marks

Calculate the minimum score a student needs to achieve to be in the top 1% of Australian 15-year-old students.

Reveal Answer

Solving for $\text{P}(X \ge x) = 0.01$ yields $x = 73.26\dots$ . Hence the minimum score required for a student to be in the top 1% is 73.27 (or 74 if reporting only integer scores).

Marking Criteria

Descriptor	Marks
writes out correct probability statement	1
correctly solves for the correct minimum score	1

Q10c

1 mark

Students who obtain scores in the range of 43 to 57 are classified as 'average'.

Calculate the probability that a randomly selected student is classified as 'average'.

Reveal Answer

$\text{P}(43 \le X \le 57) = 0.5161$

Marking Criteria

Descriptor	Marks
correctly calculates probability	1

Q10d

2 marks

A sample of 50 students is to be randomly selected.

Use the approximate normality of the distribution of sample proportions to approximate the probability that the sample proportion of students classified as 'average' is no more than 0.46.

Reveal Answer

The sample proportion $\hat{p}$ of students classified as 'average' has a mean of

$\mu = p = 0.5161$

and standard deviation

$\sigma = \sqrt{\frac{0.5161(1 - 0.5161)}{50}} = 0.0707$

Hence, we use the approximate distribution $\hat{p} \sim \text{N}(0.5161, 0.0707^2)$ . The probability that $\hat{p}$ is no more than 0.46 given by

$\text{P}(\hat{p} < 0.46) = 0.2137$

Marking Criteria

Descriptor	Marks
states the correct mean and standard deviation of the distribution of sample proportions	1
calculates correct probability	1

Q10e

3 marks

It has been decided to transform the test scores using the equation

$Y = aX + b$

such that $\text{P}(Y \le 82) = 0.1151$ and $\text{P}(Y \ge 130) = 0.0228$ .

Determine the mean and standard deviation of $Y$ , rounding your answers to integer values.

Reveal Answer

Let $Z \sim \text{N}(0, 1)$ . Given that

$\text{P}(Y \le 82) = \text{P}(Z \le -1.2) = 0.1151$

and

$\text{P}(Y \ge 130) = \text{P}(Z \ge 2) = 0.0228$

It follows that

$-1.2 = \frac{82 - \mu}{\sigma}$ and $2 = \frac{130 - \mu}{\sigma}$

where $\mu$ and $\sigma$ are the mean and standard deviation of $Y$ respectively. Solving these two equations yields

$\mu = 100$

$\sigma = 15$

Marking Criteria

Descriptor	Marks
determines correct $Z$ scores for $Y = 82$ and $Y = 130$	1
writes correct equations for the mean and standard deviation	1
correctly solves for the mean and standard deviation, expressing answers as integers	1

Q10f

2 marks

Hence, determine the values of $a$ and $b$ used to transform the original scores.

Reveal Answer

Given that $Y = aX + b$ it follows from the means and standard deviations of $X$ and $Y$ that

$100 = 50a + b$

$15 = 10a$

Hence $a = 1.5$ and $b = 25$ .

Marking Criteria

Descriptor	Marks
states correct equations to solve for $a$ and $b$	1
correctly solves for $a$ and $b$	1

Q3

2022

VCAA

Paper 2

14 marks

Q3

Mika is flipping a coin. The unbiased coin has a probability of $\frac{1}{2}$ of landing on heads and $\frac{1}{2}$ of landing on tails.

Let $X$ be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

Q3b

The height reached by each of Mika's coin flips is given by a continuous random variable, $H$ , with the probability density function

f(h) = \begin{cases} ah^2 + bh + c & 1.5 \leq h \leq 3 \\ 0 & \text{elsewhere} \end{cases}

where $h$ is the vertical height reached by the coin flip, in metres, between the coin and the floor, and $a$ , $b$ and $c$ are real constants.

Q3c

Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of $p$ of landing on heads and $(1 - p)$ of landing on tails, where $p$ is a constant value between 0 and 1.

Bella flips her coin 25 times in order to estimate $p$ .

Let $\hat{P}$ be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.

Q3a (i)

1 mark

Find $\Pr(X = 5)$ .

Reveal Answer

$\frac{1}{32} = 0.03125$

Marking Criteria

Descriptor	Marks
Calculates the correct exact probability: $\frac{1}{32}$ or $0.03125$	1

Q3a (ii)

1 mark

Find $\Pr(X \geq 2)$ .

Reveal Answer

$\frac{13}{16} = 0.8125$

Marking Criteria

Descriptor	Marks
Calculates the correct exact probability: $\frac{13}{16}$ or $0.8125$	1

Q3a (iii)

2 marks

Find $\Pr(X \geq 2 \mid X < 5)$ , correct to three decimal places.

Reveal Answer

$\text{Pr}(X \ge 2 | X < 5) = \frac{\text{Pr}(2 \le X \le 4)}{\text{Pr}(X \le 4)} = \frac{0.78125}{0.96875} = 0.806$

Marking Criteria

Working

Descriptor	Marks
Sets up the correct conditional probability expression or substitution, e.g., $\frac{\text{Pr}(2 \le X \le 4)}{\text{Pr}(X \le 4)}$ or $\frac{0.78125}{0.96875}$	1

Answer

Descriptor	Marks
Calculates the correct answer to three decimal places: $0.806$	1

Q3a (iv)

2 marks

Find the expected value and the standard deviation for $X$ .

Reveal Answer

$\text{E}(X) = \frac{5}{2} = 2.5$ , $\text{sd}(X) = \frac{\sqrt{5}}{2}$

Marking Criteria

Expected Value

Descriptor	Marks
Calculates the correct expected value: $\text{E}(X) = 2.5$ or $\frac{5}{2}$	1

Standard Deviation

Descriptor	Marks
Calculates the correct exact standard deviation: $\text{sd}(X) = \frac{\sqrt{5}}{2}$	1

Q3b (i)

1 mark

State the value of the definite integral $\int_{1.5}^3 f(h) dh$ .

Reveal Answer

$1$

Marking Criteria

Descriptor	Marks
States the correct value of the definite integral: $1$	1

Q3b (ii)

3 marks

Given that $\Pr(H \leq 2) = 0.35$ and $\Pr(H \geq 2.5) = 0.25$ , find the values of $a$ , $b$ and $c$ .

Reveal Answer

$a = -0.8 = -\frac{4}{5}$ , $b = 3.4 = \frac{17}{5}$ , $c = -2.78\dot{3} = -\frac{167}{60}$

Marking Criteria

Working

Descriptor	Marks
Sets up the correct system of definite integrals: $\int_{1.5}^3 f(h) dh = 1$ , $\int_{1.5}^2 f(h) dh = 0.35$ , and $\int_{2.5}^3 f(h) dh = 0.25$	1
Demonstrates a correct method to evaluate the integrals and solve the system of equations for the constants	1

Answer

Descriptor	Marks
Calculates all three correct exact values: $a = -0.8$ (or $-\frac{4}{5}$ ), $b = 3.4$ (or $\frac{17}{5}$ ), and $c = -\frac{167}{60}$	1

Q3b (iii)

1 mark

The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, $D$ , with probability density function $g$ .

The function $g$ is a transformation of the function $f$ given by $g(d) = f(rd + s)$ ,
where $d$ is the minimum distance between the coin and the ceiling, and $r$ and $s$ are real constants.

Find the values of $r$ and $s$ .

Reveal Answer

$r = -1, s = 3$

Marking Criteria

Descriptor	Marks
Finds both correct values: $r = -1$ and $s = 3$	1

Q3c (i)

1 mark

Is the random variable $\hat{P}$ discrete or continuous? Justify your answer.

Reveal Answer

Discrete, countable

Marking Criteria

Descriptor	Marks
Identifies the random variable as discrete and provides a valid justification (e.g., the number of heads is countable)	1

Q3c (ii)

1 mark

If $\hat{p} = 0.4$ , find an approximate 95% confidence interval for $p$ , correct to three decimal places.

Reveal Answer

$(0.208, 0.592)$

Marking Criteria

Descriptor	Marks
Calculates the correct 95% confidence interval, correct to three decimal places: $(0.208, 0.592)$	1

Q3c (iii)

1 mark

Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.

If $\hat{p} = 0.4$ , how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?

Reveal Answer

$100$

Marking Criteria

Descriptor	Marks
Calculates the correct number of coin flips required: $100$	1

Q12

2020

SCSA

Paper 2

21 marks

Q12

It is estimated that 20% of small businesses fail in the first year. A business advisory group takes a random sample of 500 new businesses that started in January 2018. An analyst employed by the group suggests the use of the binomial distribution is appropriate in this case.

Q12e

The business advisory group believes that the proportion of new businesses that fail within a year can be reduced by providing financial advice. They took another random sample of 500 businesses that started in January 2019 and provided them with regular financial advice. In this random sample, at the end of the year 80 businesses had failed.

Q12a

2 marks

What is the probability that at most 120 of the businesses fail in the first year?

Reveal Answer

Let the random variable $X$ denote the number of new businesses that fail out of the 500. Then $X \sim Bin(500, 0.2)$ . $P(X \le 120) = 0.9877$

Marking Criteria

Descriptor	Marks
states the parameters of the Binomial distribution of an appropriate random variable	1
calculates the probability	1

Q12b

3 marks

What is the approximate distribution of the sample proportion of small businesses that fail by the end of the year in this sample? Justify your answer.

Reveal Answer

Sample proportion $\hat{p} = N\left(0.2, \frac{0.2 \times 0.8}{500}\right)$ ,
That is, $\hat{p} = N(0.2, 0.00032)$ , as the sample size is large

Marking Criteria

Descriptor	Marks
states the distribution is normal as the sample size is large	1
gives the value of the mean	1
gives the value of the variance	1

Q12c

2 marks

What is the probability that the sample proportion of businesses that fail by the end of the year is less than 0.18?

Reveal Answer

$P(\hat{p} < 0.18) = 0.1318$

Marking Criteria

Descriptor	Marks
uses the correct value of mean and standard deviation	1
obtains the correct probability	1

Q12d

2 marks

By January 2019, 90 of the 500 new businesses had failed. Calculate a 95% confidence interval for the proportion of new businesses that fail in the first year.

Reveal Answer

Sample proportion $\hat{p} = \frac{90}{500} = 0.18$
95% confidence interval $\left( 0.18 - 1.96 \times \sqrt{\frac{0.18 \times 0.82}{500}}, \ 0.18 + 1.96 \times \sqrt{\frac{0.18 \times 0.82}{500}} \right)$
$= (0.1463, 0.2136)$

Marking Criteria

Descriptor	Marks
calculates the lower bound of the interval correctly	1
calculates the upper bound of the interval correctly	1

Q12e

2 marks

Calculate the sample proportion and its margin of error at the 95% confidence level.

Reveal Answer

Sample proportion $\hat{p} = \frac{80}{500} = 0.16$
$E = 1.96 \times \sqrt{\frac{0.16 \times 0.84}{500}} = 0.0321$

Marking Criteria

Descriptor	Marks
calculates the sample proportion correctly	1
calculates $E$ correctly	1

Q12f

4 marks

Calculate a 95% confidence interval for the proportion of businesses that failed. What do you conclude regarding the value of the financial advice provided to the new businesses?

Reveal Answer

Sample proportion $\hat{p} = \frac{80}{500} = 0.16$
95% confidence interval $\left( 0.16 - 1.96 \times \sqrt{\frac{0.16 \times 0.84}{500}}, \ 0.16 + 1.96 \times \sqrt{\frac{0.16 \times 0.84}{500}} \right)$
$= (0.1279, 0.1921)$
Comparing this confidence interval with the previous one $(0.1463, 0.2136)$ , we can see that they overlap.
Therefore, it does not appear that the financial advice has reduced the proportion of businesses that fail in the first year.

Marking Criteria

Descriptor	Marks
calculates the lower bound of the interval correctly	1
calculates the upper bound of the interval correctly	1
conclusion refers to the confidence intervals overlapping	1
states the correct conclusion	1

Q12g

2 marks

If the sample size was reduced, what would be the effect on the confidence interval? Justify your answer.

Reveal Answer

The width of the confidence interval would be increased, as the margin of error of the sample proportion will increase, thus increasing the error.

Marking Criteria

Descriptor	Marks
claims the width of the interval would increase	1
refers to the increase in margin of error or increase in error	1

Q12h

4 marks

State two assumptions that the analyst made in recommending the use of the binomial model in this case and discuss whether they are valid.

Reveal Answer

We assume that the businesses fail independently of each other. This is unlikely to be valid because:
(a) two similar businesses may both fail or both survive
(b) if two similar businesses in an area then one may dominate and the other fails.
We assume that the probability of a business failing is the same for all businesses. This is unlikely to be valid, as businesses of different types are expected to have different probabilities of failing.
The probability of failure is fixed (for the year). This is unlikely to be true, and the probability will depend on changing conditions over time.

Marking Criteria

Descriptor	Marks
states one assumption	1
states with reasons that assumption is unlikely to be valid	1
states second assumption	1
states with reasons that assumption is unlikely to be valid	1

Q14

2020

SCSA

Paper 2

10 marks

Q14

A suburban council hires a consultant to estimate the proportion of residents of the suburb who use its library.

Q14a

3 marks

The consultant decides to estimate a 95% confidence interval for the proportion to within an error of 0.01. What minimum sample size should be selected?

Reveal Answer

$n > \left( \frac{1.96\sqrt{0.5 \times 0.5}}{0.01} \right)^2 = 9604$

Marking Criteria

Descriptor	Marks
uses the correct z-value	1
uses 0.5 in the expression for standard error	1
determines the sample size (as an integer)	1

Q14b

3 marks

If resource limitations dictate that the maximum sample size that can be managed is 500, what is the maximum margin of error in estimating a 99% confidence interval?

Reveal Answer

$\varepsilon = 2.5758 \times \sqrt{\frac{0.5 \times 0.5}{500}} = 0.058$
that is, within 5.8%

Marking Criteria

Descriptor	Marks
uses the correct z-value	1
uses 0.5 in the expression for standard error	1
calculates the error	1

Q14c

4 marks

The consultant decides to select the sample by standing on the roadside outside the library at lunchtime and asking a random sample of the passers-by whether they use the library.

Identify and explain two possible sources of bias with this sampling scheme.

Reveal Answer

The sample is at a fixed time, so only people around at that time will be sampled.
The location is fixed, so:
(i) only people at that location will be sampled or
(ii) not everyone from the suburb will pass by that area, so this is not a random sample of the residents.

Marking Criteria

Descriptor	Marks
identifies one possible source of bias	1
explains why it is a possible source of bias	1
identifies another possible source of bias	1
explains why it is a possible source of bias	1

Q10

2024

SCSA

Paper 2

13 marks

Q10

A book called 'Why I Love Mathematics' is having its first print run. This is scheduled to last for one week, using four printing presses. The publisher claims that, historically, 10 books have printing errors for every two hundred that are printed.

A sample of 200 books is to be chosen to determine how many contained errors. The proposed sampling procedure is to select the first 200 books printed over a 6-hour window using the newest printing press.

Q10a (i)

2 marks

Identify and explain one possible source of bias in the proposed sampling procedure.

Reveal Answer

Answers could include:

the sample only includes books from the newest printing press, which may perform differently to the other three presses
the sample was gathered over a single narrow time period, so books printed later in the week are not included (this may involve different operators and the performance of the presses might change during their use over the week).

Marking Criteria

Descriptor	Marks
identifies a source of bias	1
provides a correct explanation for that source	1

Q10a (ii)

2 marks

Identify two changes to the sampling procedure that would reduce bias.

Reveal Answer

The sample could be randomly selected:

from books printed across the entire duration of the print run
from all printing presses.

Marking Criteria

Descriptor	Marks
suggests a change to improve randomisation across the entire print run	1
suggests a second change to improve randomisation across the entire print run	1

Q10b

2 marks

Assume that the sample was gathered using an improved procedure, and that the publisher's claim is correct.

Use the approximate normality of the distribution of sample proportions to determine the probability that the sample proportion of books with errors is less than 0.04.

Reveal Answer

The distribution of $\hat{p}$ can be approximated as

$\hat{p} \sim \text{N}(0.05, 0.0002375)$

Hence,

$\text{P}(\hat{p} < 0.04) = 0.2582$

Marking Criteria

Descriptor	Marks
calculates the correct distribution parameters (mean and standard deviation/variance)	1
calculates the correct probability	1

Q10c

1 mark

In a different random sample of 600 books, it is found that the proportion of books containing an error is 0.1, with a margin of error of 0.024.

Determine a 95% confidence interval for the proportion of books that will have printing errors.

Reveal Answer

A 95% confidence interval is given by

$95\% \text{ CI} = (0.1 - 0.024, 0.1 + 0.024) = (0.0760, 0.1240)$

Marking Criteria

Descriptor	Marks
correctly calculates confidence interval	1

Q10d

2 marks

On the basis of the confidence interval determined in part (c), is the proportion of books with printing errors different from what was claimed by the publisher?

Reveal Answer

The proportion of books with errors claimed by the publisher was $\frac{10}{200} = 0.05$ . This proportion is not within the confidence interval and so there is sufficient evidence to conclude that the claim of the publisher is incorrect at the 95% confidence level.

Marking Criteria

Descriptor	Marks
states that the claimed proportion is not within the confidence interval	1
states that there is sufficient evidence at the above confidence level to conclude that the claimed proportion is incorrect	1

Q10e

2 marks

Suggest two changes that could be made in order to decrease the margin of error of the confidence interval.

Reveal Answer

The margin of error could be decreased by

increasing the size of the sample
decreasing the confidence level.

Marking Criteria

Descriptor	Marks
states one possible change	1
states a second possible change	1

Q10f

2 marks

Determine the minimum sample size that would be necessary to guarantee that the margin of error of the resulting 95% confidence interval was at most 0.02.

Reveal Answer

For the worst-case scenario set $\hat{p} = 0.5$ . Solving for a margin of error equal to 0.02 gives

$0.02 = 1.96\sqrt{\frac{0.5(1 - 0.5)}{n}}$

$\Rightarrow n = 2401$

A sample size of at least 2401 would ensure the margin of error is at most 0.02.

Marking Criteria

Descriptor	Marks
sets $\hat{p} = 0.5$ and provides a correct expression for the sample size	1
correctly solves for the minimum sample size.	1

QCAA Mathematical Methods Sampling and proportions

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QCAA Mathematical Methods Sampling and proportions

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