WACE Mathematics Methods — Continuous random variables and the normal distribution Questions & Answers

Q: How many SCSA Mathematics Methods questions cover Continuous random variables and the normal distribution?

AusGrader has 158 SCSA Mathematics Methods questions on Continuous random variables and the normal distribution, all with instant AI grading and detailed marking feedback.

Q2

2021

QCAA

Paper 1

1 mark

Q2

1 mark

The table shows the time a technician has spent servicing photocopiers.

Time (in minutes)	Frequency
$0 \leq t < 5$	10
$5 \leq t < 10$	20
$10 \leq t < 15$	30
$15 \leq t < 20$	40
$20 \leq t < 25$	100

What is the probability that a given service required at least 10 minutes but less than 20 minutes?

A

0.15

B

0.35

C

0.70

D

0.85

Reveal Answer

A

0.15

This value represents the probability for the interval $10 \leq t < 15$ only ( $30/200 = 0.15$ ). You must also include the interval $15 \leq t < 20$ .

B

0.35

Correct Answer

The total number of services is $10+20+30+40+100=200$ . The number of services between 10 and 20 minutes is $30+40=70$ . The probability is $\frac{70}{200} = 0.35$ .

C

0.70

This option likely confuses the frequency count (70) with the probability, or assumes a total frequency of 100. You must divide the specific frequency (70) by the total frequency (200).

D

0.85

This is the probability that the service took at least 10 minutes ( $t \geq 10$ ), calculated as $\frac{30+40+100}{200} = 0.85$ . The question excludes times of 20 minutes or more.

Q8

2023

VCAA

Paper 1

6 marks

Q8

Suppose that the queuing time, $T$ (in minutes), at a customer service desk has a probability density function given by

$f(t) = \begin{cases} kt(16 - t^2) & 0 \leq t \leq 4 \\ 0 & \text{elsewhere} \end{cases}$

for some $k \in R$ .

Q8a

1 mark

Show that $k = \frac{1}{64}$ .

Reveal Answer

\begin{align*} \int_0^4 kt(16 - t^2) dt &= 1\\ k \left[ 8t^2 - \frac{t^4}{4} \right]_0^4 &= 1\\ k \left( 8 \times 16 - \frac{16 \times 16}{4} \right) &= 1\\ 64k &= 1 \end{align*}

$\therefore k = \frac{1}{64}$

Marking Criteria

Descriptor	Marks
Correctly forms an integral equation equal to 1, antidifferentiates, and solves to show $k = \frac{1}{64}$	1

Q8b

2 marks

Find $\text{E}(T)$ .

Reveal Answer

$E(T) = \frac{1}{64} \int_0^4 \left(16t^2 - t^4\right) dt$

$= \frac{1}{64} \left[ \frac{16t^3}{3} - \frac{t^5}{5} \right]_0^4$

$= \frac{1}{64} \left( \frac{1024}{3} - \frac{1024}{5} - 0 \right)$

$= \frac{1}{64} \times \frac{2048}{15}$

$= \frac{64}{30} = \frac{32}{15} = 2\frac{2}{15}$

Marking Criteria

Descriptor	Marks
Sets up the correct integral for the expected value, e.g., $\text{E}(T) = \frac{1}{64} \int_0^4 \left(16t^2 - t^4\right) dt$	1
Evaluates the integral correctly to find the final answer of $\frac{32}{15}$ or $2\frac{2}{15}$	1

Q8c

3 marks

What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?

Reveal Answer

\begin{align*} \text{Pr}(2 < T < 4 | T > 1) &= \frac{\text{Pr}(2 < T < 4)}{\text{Pr}(T > 1)}\\ &= \frac{\frac{1}{64} \int_2^4 (16t - t^3) dt}{\frac{1}{64} \int_1^4 (16t - t^3) dt} = \frac{\int_2^4 (16t - t^3) dt}{\int_1^4 (16t - t^3) dt}\\ &= \frac{\left[ 8t^2 - \frac{t^4}{4} \right]_2^4}{\left[ 8t^2 - \frac{t^4}{4} \right]_1^4}\\ &= \frac{(64 - (32 - 4))}{\left( 64 - \left( 8 - \frac{1}{4} \right) \right)}\\ &= \frac{36}{\left( \frac{225}{4} \right)}\\ &= \frac{144}{225} = \frac{16}{25} = 0.64 \end{align*}

Marking Criteria

Descriptor	Marks
Correctly formulates the conditional probability expression, such as $\frac{\text{Pr}(2 < T < 4)}{\text{Pr}(T > 1)}$	1
Correctly substitutes the integrals into the numerator and denominator as a single fraction	1
Correctly evaluates the expression to find the final probability of $\frac{16}{25}$ or $0.64$	1

Q11

2023

SCSA

Paper 2

13 marks

Q11

Mrs Euler is having her car serviced at BIMDAS Mechanics. She drops her vehicle off at 8 am and is told that her car will be ready for collection at some time between 1 pm and 5 pm that day.

Let the random variable $B$ denote the time after noon (12 pm) at which a vehicle is ready for collection at BIMDAS Mechanics. The probability density function for $B$ is shown in the graph below.

The probability of a vehicle being ready for collection between 2 pm and 3 pm is 0.1.

Q11d

Mr Euler is also having his car serviced, but by Addition Autos. He drops his vehicle off at 8 am and is told that his car will be ready for collection at some time between 1 pm and 5 pm that day.

Let the random variable $A$ denote the time after noon (12 pm) that a vehicle is ready for collection at Addition Autos. The cumulative distribution function for $A$ is given by

$P(A \le a) = \begin{cases} 0, & a < 1 \\ \frac{10a - a^2 - 9}{16}, & 1 \le a \le 5 \\ 1, & a > 5 \end{cases}$

Q11a

2 marks

Determine the value of $k$ .

Reveal Answer

The area under the curve must be equal to 1. Hence
$4 \times 0.1 + 0.5 \times 2 \times (k - 0.1) = 1$
$\Rightarrow k + 0.3 = 1$
$\Rightarrow k = 0.7$

Marking Criteria

Descriptor	Marks
states that the area under the curve must equal 1	1
obtains correct value of $k$	1

Q11b

2 marks

An incomplete expression for the probability density function of $B$ is given below. Fill in the boxes to complete the missing parts of the expression.

$f(b) = \begin{cases} 0.1, & \text{[box]} \\ \text{[box]}, & 3 \le b \le 5 \\ 0, & \text{otherwise} \end{cases}$

Reveal Answer

The probability density function for $B$ is given by
$f(b) = \begin{cases} 0.1, & \bold{1 \le b < 3} \\ \bold{0.3b - 0.8}, & 3 \le b \le 5 \\ 0, & \text{otherwise} \end{cases}$

Marking Criteria

Descriptor	Marks
correctly completes the interval	1
correctly completes the linear function	1

Q11c

3 marks

Determine the expected time that Mrs Euler's vehicle will be ready for collection at BIMDAS Mechanics.

Reveal Answer

$E(B) = \int_{1}^{3} 0.1b \ db + \int_{3}^{5} b(0.3b - 0.8) \ db$
$= 3.8$

Therefore, the expected pickup time is 3:48 pm.

Marking Criteria

Descriptor	Marks
states a correct integral expression for the expected value of $B$	1
determines the correct expected value of $B$	1
states the expected value as a time	1

Q11d (i)

1 mark

Determine the probability that Mr Euler's vehicle will be ready to collect

by 3 pm.

Reveal Answer

$P(A \le 3) = \frac{10(3) - 3^2 - 9}{16}$
$= 0.75$

Marking Criteria

Descriptor	Marks
calculates correct probability	1

Q11d (ii)

2 marks

between 3 pm and 4 pm.

Reveal Answer

$P(3 \le A \le 4) = P(A \le 4) - P(A \le 3)$
$= \frac{10(4) - 4^2 - 9}{16} - \frac{10(3) - 3^2 - 9}{16}$
$= \frac{15}{16} - \frac{12}{16}$
$= \frac{3}{16}$
$= 0.1875$

Marking Criteria

Descriptor	Marks
expresses the probability as the difference $P(A \le 4) - P(A \le 3)$	1
calculates correct probability	1

Q11e

3 marks

Determine the expected time at which Mr Euler's vehicle will be ready for collection at Addition Autos.

Reveal Answer

The probability density function is given by
$p(a) = \frac{d}{da}\left( \frac{10a - a^2 - 9}{16} \right)$
$= \frac{5}{8} - \frac{a}{8}$

for $1 \le a \le 5$ (0 otherwise). Hence the expected value is given by
$E(A) = \int_{1}^{5} a\left( \frac{5}{8} - \frac{a}{8} \right) da$
$= \frac{7}{3} \left( = 2\frac{1}{3} \right)$

Therefore, the expected pickup time is 2:20 pm.

Marking Criteria

Descriptor	Marks
determines correct expression for the probability density function for $1 \le a \le 5$	1
determines the correct expected value for $A$	1
states the expected value as a time	1

Q19

2024

QCAA

Paper 2

6 marks

Q19

6 marks

The normal distribution probability density function is

$p(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$

, with the parameters mean, $\mu$ , and standard deviation, $\sigma$ .

The speeds of electric scooter (e-scooter) riders on a particular section of a bike path are approximately normally distributed with a mean of 18 km/h. It is known that $p(10) = 0.0135$ .

The speed limit for e-scooters on this section of bike path is 23 km/h.

A speed camera is set up and records the speeds of 75 e-scooter riders. Every rider travelling faster than the speed limit is given a ＄143 fine. Before setting up the speed camera, the following suggestion was made.

The total of the fines expected to be issued will be more than ＄1500.

Evaluate the reasonableness of this suggestion.

Reveal Answer

Given that $p(10)=0.0135$

$0.0135=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{10-18}{\sigma}\right)^2}$

$0.0135=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{-8}{\sigma}\right)^2}$

Using a GDC:

Two solutions are obtained for the std deviation

$\sigma=4.0002$ and $28.4020$

Reject 28.4020 as it is not a possible standard deviation
because for example three standard deviations less than
the mean would produce a negative speed or three above
would result in an impossible speed on an e-scooter
( $>100\ \text{km/h}$ ).

Use a GDC to determine $P(X\ge 23)$ with $N\sim(18,4.00)$ ,

$P(X\ge 23)=0.10566$

The number of riders is:

$0.10566\times 75=7.9245$

The total fines obtained:

$7.9245\times 143=1133.20$

The expected total fines is about ＄1133, which is less
than the suggested ＄1500, so the suggestion is not
reasonable.

Marking Criteria

Descriptor	Marks
Correctly substitutes the known information, $x=10$ and $p(10)=0.0135$ , into the given normal distribution formula	1
Determines a possible value for the standard deviation	1
Determines the proportion of riders above 23 km/h	1
Determines the number of riders above 23 km/h	1
Determines the expected total fines	1
Provides appropriate statement of reasonableness	1

Q10

2021

QCAA

Paper 1

1 mark

Q10

1 mark

Handspans of teenagers are approximately normally distributed, with a mean of 15 cm and a standard deviation of 2 cm.

Which of the following groups is expected to be the largest?

A

teenagers with handspans that are between 7 cm and 11 cm

B

teenagers with handspans that are between 11 cm and 15 cm

C

teenagers with handspans that are between 13 cm and 17 cm

D

teenagers with handspans that are between 17 cm and 21 cm

Reveal Answer

A

teenagers with handspans that are between 7 cm and 11 cm

This range falls between 2 and 4 standard deviations below the mean ( $z = -4$ to $z = -2$ ). This represents the far left tail of the curve, which contains a negligible percentage of the population.

B

teenagers with handspans that are between 11 cm and 15 cm

This range covers the area from 2 standard deviations below the mean up to the mean ( $z = -2$ to $z = 0$ ). While substantial, it captures less area than an interval of the same width centered directly on the peak of the distribution.

C

teenagers with handspans that are between 13 cm and 17 cm

Correct Answer

This range corresponds to exactly one standard deviation below and above the mean ( $15 \pm 2$ cm). Because the normal distribution is symmetric and peaks at the mean, the interval centered on the mean contains the largest proportion of data (approximately 68%).

D

teenagers with handspans that are between 17 cm and 21 cm

This range falls between 1 and 3 standard deviations above the mean ( $z = 1$ to $z = 3$ ). This represents the tapering right tail of the distribution, which contains significantly fewer teenagers than the central range.

Q14

2024

VCAA

Paper 2

1 mark

Q14

1 mark

Let $h$ be the probability density function for a continuous random variable $X$ , where

h(x) = \begin{cases} \frac{x}{6} + k & -3 \le x < 0 \\ -\frac{x}{2} + k & 0 \le x \le 1 \\ 0 & \text{elsewhere} \end{cases}

and $k$ is a positive real number.

The value of $\Pr(X < 0.5)$ is

A

$\frac{1}{2}$

B

$\frac{15}{16}$

C

$\frac{3}{16}$

D

$\frac{49}{48}$

Reveal Answer

A

$\frac{1}{2}$

This is the value of the constant $k$ , not the probability $\Pr(X < 0.5)$ . The constant $k = \frac{1}{2}$ is found by setting the total area under the PDF to 1.

B

$\frac{15}{16}$

Correct Answer

First, find $k = \frac{1}{2}$ by setting the integral of $h(x)$ from $-3$ to $1$ equal to $1$ . Then, calculate $\Pr(X < 0.5)$ by evaluating $\int_{-3}^{0.5} h(x) dx = \int_{-3}^{0} (\frac{x}{6} + \frac{1}{2}) dx + \int_{0}^{0.5} (-\frac{x}{2} + \frac{1}{2}) dx = \frac{3}{4} + \frac{3}{16} = \frac{15}{16}$ .

C

$\frac{3}{16}$

This is only the probability $\Pr(0 \le X < 0.5)$ . It incorrectly omits the probability $\Pr(-3 \le X < 0) = \frac{3}{4}$ which must be included since the lower bound of the domain is $-3$ .

D

$\frac{49}{48}$

This is an incorrect calculation. A probability value cannot exceed 1, so $\frac{49}{48}$ is impossible for any valid probability density function.

Q14

2022

VCAA

Paper 2

1 mark

Q14

1 mark

A continuous random variable, $X$ , has a probability density function given by

f(x) = \begin{cases} \frac{2}{9} x e^{-\frac{1}{9}x^2} & x \geq 0 \\ 0 & x < 0 \end{cases}

The expected value of $X$ , correct to three decimal places, is

A

$1.000$

B

$2.659$

C

$3.730$

D

$6.341$

E

$9.000$

Reveal Answer

A

$1.000$

This is the total area under the probability density function curve, $\int_0^\infty f(x) dx = 1$ , which verifies it is a valid PDF but does not give the expected value.

B

$2.659$

Correct Answer

The expected value is calculated using $E(X) = \int_0^\infty x f(x) dx = \int_0^\infty \frac{2}{9} x^2 e^{-\frac{1}{9}x^2} dx$ . Evaluating this integral yields $\frac{3\sqrt{\pi}}{2} \approx 2.659$ .

C

$3.730$

This is an incorrect evaluation of the expected value. The correct formula requires evaluating the integral $\int_0^\infty x f(x) dx$ .

D

$6.341$

This value results from an incorrect integration or applying the wrong formula for the expected value of a continuous random variable.

E

$9.000$

This is the second moment, $E(X^2) = \int_0^\infty x^2 f(x) dx = 9$ , rather than the expected value $E(X)$ .

Q11

2020

VCAA

Paper 2

1 mark

Q11

1 mark

The lengths of plastic pipes that are cut by a particular machine are a normally distributed random variable, $X$ , with a mean of 250 mm.

$Z$ is the standard normal random variable.

If $\Pr(X < 259) = 1 - \Pr(Z > 1.5)$ , then the standard deviation of the lengths of plastic pipes, in millimetres, is

A

1.5

B

3

C

6

D

9

E

12

Reveal Answer

A

1.5

Incorrect. This value represents the z-score ( $z = 1.5$ ) from the standard normal distribution, not the standard deviation of $X$ .

B

3

Incorrect. This would be the standard deviation if the z-score was 3, but the given z-score is 1.5.

C

6

Correct Answer

Correct. Since $1 - \Pr(Z > 1.5) = \Pr(Z < 1.5)$ , the z-score for $X = 259$ is $1.5$ . Using the standardization formula $z = \frac{x - \mu}{\sigma}$ , we have $1.5 = \frac{259 - 250}{\sigma}$ , which gives $\sigma = 6$ .

D

9

Incorrect. This is the difference between the given value and the mean ( $259 - 250 = 9$ ), which is the numerator of the z-score formula, not the standard deviation.

E

12

Incorrect. This value might result from an arithmetic error, such as incorrectly multiplying the difference from the mean by a factor instead of dividing by the z-score.

Q16

2020

QCAA

Paper 2

4 marks

Q16

4 marks

Bottles of soft drink should contain a volume with a mean of 591 mL, but some variation is expected.

Any bottle at or below the 20th percentile of the volume distribution is rejected. A percentile is a measure in statistics that shows the values below which a given percentage of observations occur.

Thirty-five per cent of the bottles contain 593 mL or more of soft drink.

Assuming the volumes are normally distributed, determine the smallest volume (in mL) that will be accepted.

Reveal Answer

Given $\mu = 591$
Using GDC
z-score associated with 65th percentile
$= 0.38532$

$0.38532 = \frac{593 - 591}{\sigma}$

$\sigma = 5.1905$

Using GDC
z-score associated with 20% rejection region $= -0.841621$

To determine the smallest volume that will be accepted ( $x$ )
$-0.841621 = \frac{x - 591}{5.1905}$

$x = 587$ mL

Marking Criteria

Descriptor	Marks
Correctly determines the z-score associated with the 65th percentile	1
Determines $\sigma$	1
Correctly determines the z-score associated with the 20% rejection region	1
Determines the smallest volume	1

Q18

2022

QCAA

Paper 1

4 marks

Q18

4 marks

A percentile is a measure in statistics showing the value below which a given percentage of observations occur.

The continuous random variable $X$ has the probability density function
$f(x) = \begin{cases} 2x-2, & 1 \le x \le 2 \\ 0 , & \text{otherwise} \end{cases}$

Determine the 36th percentile of $X$ .

Reveal Answer

$\int_1^a 2x - 2 dx = 0.36$
$x^2 - 2x \Big|_1^a = 0.36$
$(a^2 - 2a) - (1 - 2) = 0.36$
$a^2 - 2a + 1 = 0.36$
$a^2 - 2a + 0.64 = 0$
$\therefore a = \frac{2\pm\sqrt{4-4\times1\times0.64}}{2}$
$\therefore a = \frac{2 \pm \sqrt{1.44}}{2}$
$\therefore a = \frac{2 \pm 1.2}{2}$
$\therefore a = 1.6$ or $0.4$
Given $1 \le x \le 2$
$\therefore a = 1.6$

Marking Criteria

Descriptor	Marks
correctly determines the definite integral	1
determines the quadratic equation	1
determines values of a	1
evaluates the reasonableness of solutions	1

Q2

2024

QCAA

Paper 2

1 mark

Q2

1 mark

Calculate the expected value of a continuous random variable $X$ with the probability density function

$p(x) = \begin{cases} \frac{1}{4}x^2, & 0 \le x \le \sqrt[3]{12} \\ 0, & \text{otherwise} \end{cases}$

A

1.72

B

1.15

C

1.00

D

0.11

Reveal Answer

A

1.72

Correct Answer

The expected value is calculated using the integral $E[X] = \int_{0}^{\sqrt[3]{12}} x \cdot \frac{1}{4}x^2 \, dx$ . Solving this yields $\frac{3}{4}\sqrt[3]{12} \approx 1.72$ .

B

1.15

This value is approximately the midpoint of the range $[0, \sqrt[3]{12}]$ , which would be the mean if the distribution were uniform. However, since $p(x) \propto x^2$ , the distribution is skewed toward higher values.

C

1.00

This represents the total area under the probability density function (total probability), which is always 1, rather than the expected value.

D

0.11

This value is far too low; since the probability density increases with $x$ , the expected value must be closer to the upper bound of the interval than the lower bound.

Q10

2023

QCAA

Paper 2

1 mark

Q10

1 mark

A student is trying to determine which subject they performed best in compared to other students. Results from recent tests in four subjects (A to D) are shown. Assume student results in each subject are normally distributed.

In which subject did the student perform best compared to other students?

	Class mean	Class standard deviation	Student's result
A	62	22	77
B	55	25	74
C	61	15	70
D	73	20	82

A

Row A

B

Row B

C

Row C

D

Row D

Reveal Answer

A

Row A

The z-score for this subject is $z = \frac{77 - 62}{22} \approx 0.68$ . While this is a positive result, it is lower than the z-score for Subject B.

B

Row B

Correct Answer

To compare performance across different distributions, we calculate the z-score using $z = \frac{x - \mu}{\sigma}$ . For this subject, $z = \frac{74 - 55}{25} = 0.76$ , which is the highest value among the options, indicating the best relative performance.

C

Row C

The z-score for this subject is $z = \frac{70 - 61}{15} = 0.60$ . This indicates a lower relative standing compared to Subject B.

D

Row D

The z-score for this subject is $z = \frac{82 - 73}{20} = 0.45$ . Even though the raw score (82) is the highest, the relative performance is actually the lowest of the four subjects.

Q14

2020

VCAA

Paper 2

1 mark

Q14

1 mark

The random variable $X$ is normally distributed.

The mean of $X$ is twice the standard deviation of $X$ .

If $\Pr(X > 5.2) = 0.9$ , then the standard deviation of $X$ is closest to

A

7.238

B

14.476

C

3.327

D

1.585

E

3.169

Reveal Answer

A

7.238

Correct Answer

Correct. Since $\Pr(X > 5.2) = 0.9$ , the corresponding z-score is approximately $-1.2816$ . Using the standardization formula $z = \frac{x - \mu}{\sigma}$ with $\mu = 2\sigma$ , we get $-1.2816 = \frac{5.2 - 2\sigma}{\sigma}$ , which solves to $\sigma \approx 7.238$ .

B

14.476

Incorrect. This value represents the mean of the distribution ( $\mu = 2\sigma \approx 14.476$ ), not the standard deviation.

C

3.327

Incorrect. This is a distractor that likely results from an algebraic error or incorrectly setting up the relationship between the mean and standard deviation.

D

1.585

Incorrect. This value is obtained by incorrectly using a positive z-score ( $1.2816$ ), which corresponds to $\Pr(X < 5.2) = 0.9$ rather than $\Pr(X > 5.2) = 0.9$ .

E

3.169

Incorrect. This represents the mean ( $\mu = 2\sigma$ ) calculated under the incorrect assumption that $\Pr(X < 5.2) = 0.9$ .

Q14

2025

VCAA

Paper 2

1 mark

Q14

1 mark

Let $f$ be the probability density function for a continuous random variable $X$ , where

f(x) = \begin{cases} k\sin(x) & 0 \le x < \frac{\pi}{4} \\ k\cos(x) & \frac{\pi}{4} \le x \le \frac{\pi}{2} \\ 0 & \text{otherwise} \end{cases}

and $k$ is a positive real number.

The value of $k$ is

A

$\frac{1}{\sqrt{2}}$

B

$\frac{1}{2 - \sqrt{2}}$

C

$\sqrt{2} + 2$

D

$2 - \sqrt{2}$

Reveal Answer

A

$\frac{1}{\sqrt{2}}$

Incorrect. This value does not make the total area under the probability density function equal to 1, likely resulting from an error in evaluating the trigonometric integrals.

B

$\frac{1}{2 - \sqrt{2}}$

Correct Answer

Correct. For $f(x)$ to be a valid probability density function, its integral over all $x$ must equal 1. Evaluating $\int_0^{\pi/4} k\sin(x)dx + \int_{\pi/4}^{\pi/2} k\cos(x)dx = 1$ yields $k(2 - \sqrt{2}) = 1$ , which gives $k = \frac{1}{2 - \sqrt{2}}$ .

C

$\sqrt{2} + 2$

Incorrect. This might result from incorrectly rationalizing the denominator or making an arithmetic error when solving $k(2 - \sqrt{2}) = 1$ .

D

$2 - \sqrt{2}$

Incorrect. This is the value of the integral when $k=1$ . Since the total area must be 1, $k$ must be the reciprocal of this value.

Q14

2020

QCAA

Paper 2

6 marks

Q14

Let $X$ denote the time in minutes between the arrival of trains at a station. The cumulative distribution function of $X$ is defined by

$F(x) = \begin{cases} 2 - \frac{10}{x}, & 5 \le x \le 10 \\ 0, & \text{otherwise} \end{cases}$

Q14a

3 marks

Determine the probability density function of $X$ .

Reveal Answer

$\text{pdf}(f(x)) = \frac{d}{dx}\left(\frac{-10}{x}\right)$

$f(x) = \begin{cases} \frac{10}{x^2}, & 5 \le x \le 10 \\ 0, & \text{otherwise} \end{cases}$

Marking Criteria

Descriptor	Marks
correctly identifies the use of the derivative of the cumulative distribution function	1
correctly determines the derivative of the cumulative distribution function (pdf)	1
uses appropriate convention to communicate the probability density function as a piecewise function with given domains	1

Q14b

1 mark

Determine the probability that $5 < X < 7$ .

Reveal Answer

$F(7) - F(5) = \frac{-10}{7} - \frac{-10}{5}$
$= \frac{4}{7}$

Marking Criteria

Descriptor	Marks
correctly determines the probability that there are 5 to 7 minutes between train arrivals	1

Q14c

2 marks

Determine the mean time between the arrival of trains at the station.

Reveal Answer

Mean $= \int_5^{10} \frac{10}{x} dx$
Using GDC
Mean $= 6.93$ minutes

Marking Criteria

Descriptor	Marks
provides a statement identifying the use of expected value for a continuous random variable	1
determines mean time between arrivals	1

SCSA Mathematics Methods Continuous random variables and the normal distribution

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SCSA Mathematics Methods Continuous random variables and the normal distribution

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Sample Answer

Sample Answer

Sample Answer

Sample Answer

Frequently Asked Questions

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