Calculate the percentage of the population of Ottawa that speak English but not French.

Calculate the number of people in Ottawa that speak both English and French.

Write down your answer to part (b) in the form $a\times {10}^k$ where and k $\in \mathbb{Z}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$97-36$     (M1)

Note:     Award (M1) for subtracting 36 from 97.

OR

(M1)

Note:     Award (M1) for 61 and 36 seen in the correct places in the Venn diagram.

$=61(\mathrm{\%})$     (A1)     (C2)

Note:     Accept 61.0 (%).

[2 marks]

$\frac{36}{100}\times 985000$     (M1)

Note:     Award (M1) for multiplying 0.36 (or equivalent) by $985000$.

$=355000(354600)$     (A1)     (C2)

[2 marks]

$3.55\times {10}^5(3.546\times {10}^5)$     (A1)(ft)(A1)(ft)     (C2)

Note:     Award (A1)(ft) for 3.55 (3.546) must match part (b), and (A1)(ft) $\times {10}^5$.

Award (A0)(A0) for answers of the type: $35.5\times {10}^4$. Follow through from part (b).

[2 marks]

Show that $\cos \theta =\frac{3}{4}$.

Given that $\tan \theta >0$, find $\tan \theta$.

Let $y=\frac{1}{\cos x}$, for . The graph of $y$between $x=\theta$ and $x=\frac{\pi }{4}$ is rotated 360° about the $x$-axis. Find the volume of the solid formed.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of summing to 1     (M1)

eg$$$\sum p=1$

correct equation     A1

eg$$$\cos \theta +2\cos 2\theta =1$

correct equation in $\cos \theta$     A1

eg$$$\cos \theta +2(2{\cos }^2\theta -1)=1,4{\cos }^2\theta +\cos \theta -3=0$

evidence of valid approach to solve quadratic     (M1)

eg$$factorizing equation set equal to $0,\frac{-1\pm \sqrt{1-4\times 4\times (-3)}}{8}$

correct working, clearly leading to required answer     A1

eg$$$(4\cos \theta -3)(\cos \theta +1),\frac{-1\pm 7}{8}$

correct reason for rejecting $\cos \theta \ne -1$     R1

eg$$$\cos \theta$ is a probability (value must lie between 0 and 1), $\cos \theta >0$

Note:     Award R0 for $\cos \theta \ne -1$ without a reason.

$\cos \theta =\frac{3}{4}$    AG  N0

valid approach     (M1)

eg$$sketch of right triangle with sides 3 and 4, ${\sin }^{2}x+{\cos }^{2}x=1$

correct working     

(A1)

eg$$missing side $=\sqrt{7},\frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}}$

$\tan \theta =\frac{\sqrt{7}}{3}$     A1     N2

[3 marks]

attempt to substitute either limits or the function into formula involving $f^2$     (M1)

eg$$$\pi {\int }_{\theta }^{\frac{\pi }{4}}{f}^2,\int {\left(\frac{1}{\cos x}\right)}^2$

correct substitution of both limits and function     (A1)

eg$$$\pi {\int }_{\theta }^{\frac{\pi }{4}}{\left(\frac{1}{\cos x}\right)}^2\text{d}x$

correct integration     (A1)

eg$$$\tan x$

substituting their limits into their integrated function and subtracting     (M1)

eg$$$\tan \frac{\pi }{4}-\tan \theta$

Note:     Award M0 if they substitute into original or differentiated function.

$\tan \frac{\pi }{4}=1$    (A1)

eg$$$1-\tan \theta$

$V=\pi -\frac{\pi \sqrt{7}}{3}$     A1     N3

[6 marks]

Find the height of Flower null.

Using this information, write down an equation in $p$ and $q$.

Write down a second equation in $p$ and $q$.

Using your answers to parts (b) and (c), find the height of Flower $\text{A}$.

Using your answers to parts (b) and (c), find the height of Flower $\text{D}$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.

$24-8$  OR  $24-\left(32-24\right)$  OR  $24=\frac{32+h}{2}$        (M1)

Note: Award (M1) for subtracting $8$ from the median, or equivalent.

$16 \left(\text{cm}\right)$       (A1)  (C2)

[2 marks]

$q-p=50$  (or equivalent)      (A1)  (C1)

[1 mark]

$\frac{p+16+32+q}{4}=27$   OR   $p+q=60$  (or equvalent)      (A1)(ft)  (C1)

Note: Follow through from part (a).

[1 mark]

$5 \left(\text{cm}\right)$       (A1)(ft)  (C1)

Note: Follow through from parts (b) and (c).

[1 mark]

$55 \left(\text{cm}\right)$       (A1)(ft)  (C1)

Note: Follow through from parts (b) and (c).

[1 mark]

Find the value of $t$.

Find the value of $v$.

Find the probability that a randomly selected student from the class plays tennis or volleyball, but not both.

valid approach to find $t$       (M1)

eg  $t+3=19, 19-3$

$t=16$ (may be seen on Venn diagram)        A1 N2

[2 marks]

valid approach to find $v$       (M1)

eg  $t+3+v+6=30, 30-19-6$

$v=5$ (may be seen on Venn diagram)        A1 N2

[2 marks]

valid approach       (M1)

eg  $16+5$, $21$ students, $1-\frac{3+6}{30}$, 

$\frac{21}{30} \left(=\frac{7}{10}\right)$        A1 N2

[2 marks]

Find the value of $a$.

Write down the value of the median distance in kilometres (km).

Find the value of $b$.

Find $m$.

The first $150$ athletes that completed the race won a prize.

Given that an athlete took between $22$ and $m$ minutes to complete the $5 \text{km}$ race, calculate the probability that they won a prize.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach     (M1)

eg    $Q_3-Q_1 , Q_3-1.1 , 4.5-a=1.1$

$a=3.4$      A1   N2

[2 marks]

$\frac{32}{5} \left(=6.4\right)$ (km)       A1   N1

[1 mark]

METHOD 1 (standard deviation first)

valid approach        (M1)

eg    $\text{standard deviation}=\sqrt{\text{variance}} , \sqrt{\frac{16}{9}}$

standard deviation$=\frac{4}{3}$ (km)       (A1)

valid approach to convert their standard deviation        (M1)

eg     $\frac{4}{3}\times \frac{5}{8} , \sqrt{\frac{16}{9}}=\frac{8}{5}M$

$\frac{20}{24}$ (miles)  $\left(=\frac{5}{6}\right)$      A1   N3

Note: If no working shown, award M1A1M0A0 for the value $\frac{4}{3}$.
If working shown, and candidate’s final answer is $\frac{4}{3}$, award M1A1M0A0.

METHOD 2 (variance first)

valid approach to convert variance        (M1)

eg   ${\left(\frac{5}{8}\right)}^2 , \frac{64}{25} , \frac{16}{9}\times {\left(\frac{5}{8}\right)}^2$

variance $=\frac{25}{36}$       (A1)

valid approach        (M1)

eg    $\text{standard deviation}=\sqrt{\text{variance}} , \sqrt{\frac{25}{36}} , \sqrt{\frac{16}{9}\times {\left(\frac{5}{8}\right)}^2}$

$\frac{20}{24}$ (miles)  $\left(=\frac{5}{6}\right)$      A1   N3

[4 marks]

correct frequency for $22$ minutes       (A1)

eg    $20$

adding their frequency (do not accept $22+400$)       (M1)

eg    $20+400 , 420$ athletes

$m=30$ (minutes)         A1   N3

[3 marks]

$27$ (minutes)       (A1)

correct working      (A1)

eg    $130$ athletes between $22$ and $27$ minutes, $\text{P}\left(22<t<27\right)=\frac{150-20}{600} , \frac{13}{60}$

evidence of conditional probability or reduced sample space      (M1)

eg    $\text{P}\left(A \overline{) B}\right) , \text{P}\left(t<27 \overline{) 22<t<30}\right) , \frac{\text{P}\left(22<t<27\right)}{\text{P}\left(22<t<m\right)} , \frac{150}{400}$

correct working      (A1)

eg    $\frac{{\displaystyle \frac{130}{600}}}{{\displaystyle \frac{400}{600}}} , \frac{150-20}{400}$

$\frac{130}{400} \left(\frac{13}{40}=\frac{78000}{240000}=\frac{390}{1200}=0.325\right)$      A1     N5

Note: If no other working is shown, award A0A0M1A0A0 for answer of $\frac{150}{400}$.
Award N0 for answer of $\frac{3}{8}$ with no other working shown.

[5 marks]

Find the value of $p$.

Hence, find the value of $\text{E}\left(X\right)$.

State the range of possible values of $r$.

Hence, find the range of possible values of $q$.

Hence, find the range of possible values for $\text{E}\left(Y\right)$.

Agnes and Barbara play a game using these dice. Agnes rolls die $\text{A}$ once and Barbara rolls die $\text{B}$ once. The probability that Agnes’ score is less than Barbara’s score is $\frac{1}{2}$.

Find the value of $\text{E}\left(Y\right)$.

recognising probabilities sum to $1$          (M1)

$p+p+p+\frac{1}{2}p=1$

$p=\frac{2}{7}$           A1

[2 marks]

valid attempt to find $\text{E}\left(X\right)$         (M1)

$1\times p+2\times p+3\times p+4\times \frac{1}{2}p\left(=8p\right)$

$\text{E}\left(X\right)=\frac{16}{7}$           A1

[2 marks]

$0\le r\le 1$       A1

[1 mark]

attempt to find a value of $q$         (M1)

$0\le 1-3q\le 1$  OR  $r=0\Rightarrow q=\frac{1}{3}$  OR  $r=1\Rightarrow q=0$

$0\le q\le \frac{1}{3}$       A1

[2 marks]

$\text{E}\left(Y\right)=1\times q+2\times q+3\times q+4\times r$ ($=2+2r$  OR  $4-6q$)         (A1)

one correct boundary value       A1

$1\times \frac{1}{3}+2\times \frac{1}{3}+3\times \frac{1}{3}+4\times 0 \left(=2\right)$ OR

$1\times 0+2\times 0+3\times 0+4\times 1 \left(=4\right)$ OR

$2+2\left(0\right) \left(=2\right)$ OR

$2+2\left(1\right) \left(=4\right)$ OR

$4-6\left(0\right) \left(=4\right)$ OR $4-6\left(\frac{1}{3}\right) \left(=2\right)$

$2\le \text{E}\left(Y\right)\le 4$       A1

[3 marks]

METHOD 1

evidence of choosing at least four correct outcomes from

$1\&2, 1\&3, 1\&4, 2\&3, 2\&4, 3\&4$         (M1)

$\frac{6}{7}q+\frac{6}{7}r$  OR  $3pq+3pr$  OR  $pq+pq+p\left(1-3q\right)+pq+p\left(1-3q\right)+p\left(1-3q\right)$         (A1)

solving for either $q$ or $r$        M1

$\frac{6}{7}\left(q+1-3q\right)=\frac{1}{2}$  OR  $\frac{6}{7}\left(\frac{1-r}{3}+r\right)=\frac{1}{2}$  OR  $3pq+3p\left(1-3q\right)=\frac{1}{2}$

        OR  $3p\left(\frac{1-r}{3}\right)+3 pr=\frac{1}{2}$

EITHER two correct values

$q=\frac{5}{24}$  and  $r=\frac{3}{8}$      A1A1

OR one correct value

$q=\frac{5}{24}$  OR  $r=\frac{3}{8}$      A1

substituting their value for $q$ or $r$      A1

$4-6\left(\frac{5}{24}\right)$  OR  $2+2\left(\frac{3}{8}\right)$

THEN

$\text{E}\left(Y\right)=\frac{11}{4}$      A1

METHOD 2 (solving for $\text{E}\left(Y\right)$)

evidence of choosing at least four correct outcomes from

$1\&2, 1\&3, 1\&4, 2\&3, 2\&4, 3\&4$         (M1)

$\frac{6}{7}q+\frac{6}{7}r$  OR  $3pq+3pr$  OR  $pq+pq+p\left(1-3q\right)+pq+p\left(1-3q\right)+p\left(1-3q\right)$         (A1)

rearranging to make $q$ the subject      M1

$q=\frac{4-\text{E}\left(Y\right)}{6}$

$3pq+3p\left(1-3q\right)=\frac{1}{2}$      M1

$\frac{6}{7}\times \left(\frac{4-\text{E}\left(Y\right)}{6}\right)+\frac{6}{7}\left(1-3\left(\frac{4-\text{E}\left(Y\right)}{6}\right)\right)=\frac{1}{2}$       A1

$\frac{2\left(\text{E}\left(Y\right)-1\right)}{7}=\frac{1}{2}$

$\text{E}\left(Y\right)=\frac{11}{4}$       A1

[6 marks]

Using the given information, complete the following Venn diagram.

Find the number of surveyed students who did not like any of the three flavours.

A student is chosen at random from the surveyed students.

Find the probability that this student likes kiwi fruit smoothies given that they like mango smoothies.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

    (A1)(A1) (C2)

Note: Award (A1) for 18, 12 and 1 in correct place on Venn diagram, (A1) for 3, 16 and 11 in correct place on Venn diagram.

[2 marks]

85 − (3 + 16 + 11 + 18 + 12 + 1 + 2)      (M1)

Note: Award (M1) for subtracting the sum of their values from 85.

22      (A1)(ft)  (C2)

Note: Follow through from their Venn diagram in part (a).
If any numbers that are being subtracted are negative award (M1)(A0).

[2 marks]

$\frac{14}{35}\left(\frac{2}{5}\text{,}0.4\text{,}40\text{\% }\right)$     (A1)(ft)(A1)(ft)  (C2)

Note: Award (A1) for correct numerator; (A1) for correct denominator. Follow through from their Venn diagram.

[2 marks]

State whether speed is a continuous or discrete variable.

Write down the median speed for these animals.

Write down the range of the animal speeds.

For these eight animals find the mean speed.

For these eight animals write down the standard deviation.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

continuous    (A1) (C1)

[1 mark]

75.5 (km h⁻¹)   (A1) (C1)

Note: Answer must be exact.

[1 mark]

294 (km h⁻¹)   (A1) (C1)

[1 mark]

$\frac{300+97+80+80+71+64+21+6}{8}$  OR  $\frac{719}{8}$       (M1)

Note: Award (M1) for correct sum divided by 8.

89.9  (89.875)(km h⁻¹)   (A1) (C2)

[2 marks]

84.6  (84.5597…)(km h⁻¹)   (A1) (C1)

Note: If the response to part (d)(i) is awarded zero marks, a correct response to part (d)(ii) is awarded (C2).

[1 mark]

State what 13 represents in the given diagram.

Write down the interquartile range for this data.

Write down the approximate number of snacks whose amount of sugar ranges from 18 to 20 grams.

The health inspector visits two school cafeterias. She inspects the same number of meals at each cafeteria. The data is shown in the following box-and-whisker diagrams.

Meals prepared in the school cafeterias are required to have less than 10 grams of sugar.

State, giving a reason, which school cafeteria has more meals that do not meet the requirement.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

median      (A1) (C1)  

[1 mark]

18 − 12      (A1) 

Note: Award (M1) for correct quartiles seen.

6 (g)      (A1) (C2)

[2 marks]

125      (A1) (C1)

[1 mark]

Cafeteria 2       (A1) (C1)

75 % > 50 % (do not meet the requirement)        (R1) (C1)

OR

25 % < 50 % (meet the requirement)       (R1) (C1)

Note: Do not award (A1)(R0). Award the (R1) for a correct comparison of percentages for both cafeterias, which may be in words. The percentage values or fractions must be seen. It is possible to award (A0)(R1).

[2 marks]

Show that $p=0.4$ and $q=0.2$.

Find $\text{P}(X>2)$.

Assuming that rolls of the die are independent, find the probability that Nicky wins the game.

Determine the value of $b$.

Find the value of $a$, providing evidence for your answer.

uses $\sum \text{P}\left(X=x\right)=1$ to form a linear equation in $p$ and $q$           (M1)

correct equation in terms of $p$ and $q$ from summing to $1$          A1

$p+0.3+q+0.1=1$  OR  $p+q=0.6$  (or equivalent)

uses $\text{E}\left(X\right)=2$ to form a linear equation in $p$ and $q$           (M1)

correct equation in terms of $p$ and $q$ from $\text{E}\left(X\right)=2$          A1

$p+0.6+3q+0.4=2$  OR  $p+3q=1$  (or equivalent)

Note: The marks for using $\sum \text{P}\left(X=x\right)=1$ and the marks for using $\text{E}\left(X\right)=2$ may be awarded independently of each other.

evidence of correctly solving these equations simultaneously          A1

for example, $2q=0.4\Rightarrow q=0.2$  or  $p+3\times \left(0.6-p\right)=1\Rightarrow p=0.4$

so $p=0.4$ and $q=0.2$          AG

[5 marks]

valid approach        (M1)

$\text{P}(X>2)=\text{P}(X=3)+\text{P}(X=4)$  OR  $\text{P}(X>2)=1-\text{P}(X=1)-\text{P}(X=2)$

$=0.3$          A1

[2 marks]

recognises at least one of the valid scores ($6, 7$, or $8$) required to win the game           (M1)

Note: Award M0 if candidate also considers scores other than $6, 7$, or $8$ (such as $5$).

let $T$ represent the score on the last two rolls

a score of $6$ is obtained by rolling $\left(2,4\right), \left(4,2\right)$ or $\left(3,3\right)$

$\text{P}\left(T=6\right)=2\left(0.3\right)\left(0.1\right)+{\left(0.2\right)}^2 \left(=0.1\right)$          A1

a score of $7$ is obtained by rolling $\left(3,4\right)$ or $\left(4,3\right)$

$\text{P}\left(T=7\right)=2\left(0.2\right)\left(0.1\right) \left(=0.04\right)$          A1

a score of $8$ is obtained by rolling $\left(4,4\right)$

$\text{P}\left(T=8\right)={\left(0.1\right)}^2 \left(=0.01\right)$          A1

Note: The above 3 A1 marks are independent of each other.

$\text{P}\left(\text{Nicky wins}\right)=0.1+0.04+0.01$

$=0.15$          A1

[5 marks]

$3+b=8$           (M1)

$b=5$          A1 

[2 marks]

METHOD 1

EITHER

$\text{P}\left(S=5\right)=\frac{4}{16}$

$\text{P}\left(S=a+2\right)=\frac{4}{16}$          A1 

$\Rightarrow a+2=5$

OR

$\text{P}\left(S=6\right)=\frac{3}{16}$

$\text{P}\left(S=a+3\right)=\frac{2}{16}$  and  $\text{P}\left(S=5+1\right)=\frac{1}{16}$          A1 

$\Rightarrow a+3=6$

OR

$\text{P}\left(S=4\right)=\frac{3}{16}$

$\text{P}\left(S=a+1\right)=\frac{2}{16}$  and  $\text{P}\left(S=1+3\right)=\frac{1}{16}$          A1 

$\Rightarrow a+1=4$

THEN

$\Rightarrow a=3$          A1 

Note: Award A0A0 for $a=3$ obtained without working/reasoning/justification.

METHOD 2

EITHER

correctly lists a relevant part of the sample space          A1 

for example, $\left\{S=4\right\}=\left\{\left(3,1\right),\left(1,a\right),\left(1,a\right)\right\}$  or  $\left\{S=5\right\}=\left\{\left(2,a\right),\left(2,a\right),\left(2,a\right),\left(2,a\right)\right\}$

or $\left\{S=6\right\}=\left\{\left(3,a\right),\left(3,a\right),\left(1,5\right)\right\}$

$a+3=6$

OR

eliminates possibilities (exhaustion) for $a<5$

convincingly shows that $a\ne 2,4$          A1 

$a\ne 4$, for example, $\text{P}\left(S=7\right)=\frac{2}{16}$ from $\left(2,5\right),\left(2,5\right)$ and so

$\left(3,a\right),\left(3,a\right)\Rightarrow a+3\ne 7$

THEN

$\Rightarrow a=3$          A1 

[2 marks]

A majority of candidates knew to set up equations using the sum of the probabilities in the distribution equal to 1 and/or the expected value equal to 2, however some candidates simply substituted the given values of $p$ and $q$ into their equations, which is considered working backwards and not doing what is required by the command term "show that". For the candidates who did set up both equations in terms of $p$ and $q$, nearly all were successful in solving the resulting system of equations. Many candidates answered part (b) correctly using the given values from part (a). In part (c), most candidates recognized a sum of 6 (or more) was required in the final two rolls, but very few were able to find all the different outcomes to make this happen, especially for sums that can happen in more than one way, such as $3+4$ and $4+3$. While some candidates were able to correctly answer parts (d) and (e), some did not attempt
these questions parts, and many did not justify their final answer in part (e).

Find n.

Write down the value of the new mean.

Find the value of the new variance.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

correct approach      (A1)

eg  $\frac{800}{n}=20$

40      A1 N2

[2 marks]

200   A1 N1

[1 mark]

METHOD 1

recognizing variance = σ ²      (M1)

eg  3² = 9

correct working to find new variance      (A1)

eg  σ ² × 10², 9 × 100

900      A1 N3

METHOD 2

new standard deviation is 30      (A1)

recognizing variance = σ ²      (M1)

eg 3² = 9, 30²

900      A1 N3

[3 marks]

Plot and label the point M$\left(\overline{m},\overline{p}\right)$ on the scatter diagram.

Draw the line of best fit, by eye, on the scatter diagram.

Using your line of best fit, estimate the physics test score for a student with a score of 20 in their mathematics test.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1) (C2)

Note: Award (A1) for mean point plotted and (A1) for labelled M.

[2 marks]

straight line through their mean point crossing the p-axis at 5±2     (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for a straight line through their mean point. Award (A1)(ft) for a correct p-intercept if line is extended.

[2 marks]

point on line where m = 20 identified and an attempt to identify y-coordinate     (M1)

10.5     (A1)(ft)    (C2)

Note: Follow through from their line in part (b).

[2 marks]

Complete the following table.

State whether length of trout is a continuous or discrete variable.

Write down the modal class.

Any trout with length 40 cm or less is returned to the lake.

Calculate the percentage of the fisherman’s catch that is returned to the lake.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

     (A2)     (C2)

Note:     Award (A2) for all correct entries, (A1) for 3 correct entries.

[2 marks]

continuous     (A1)     (C1)

[1 mark]

    (A1)     (C1)

Note:     Accept equivalent notation such as $(60,70]$ or $]60,70]$.

Award (A0) for “60-70” (incorrect notation).

[1 mark]

$\frac{4}{22}\times 100$     (M1)

Note:     Award (M1) for their 4 divided by their 22.

$=18.2(18.1818\dots )$     (A1)(ft)     (C2)

Note:     Follow through from their part (a). Do not accept 0.181818….

[2 marks]

Complete the Venn diagram for these students.

One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.

Determine whether the events $S$ and $M$ are independent.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

     (A1)(A1)     (C2)

Note:     Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.

[2 marks]

$\frac{30}{90}\left(\frac{1}{3},0.333333\dots ,33.3333\dots \mathrm{\%}\right)$     (A1)(ft)(A1)(ft)     (C2)

Note:     Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.

[2 marks]

$\text{P}(S)\times \text{P}(M)=\frac{3}{4}\times \frac{1}{3}=\frac{1}{4}$     (R1)

Note:     Award (R1) for multiplying their by $\frac{1}{3}$.

therefore the events are independent $\left(\text{as P}(S\cap M)=\frac{1}{4}\right)$     (A1)(ft)     (C2)

Note:     Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.

Do not award (R0)(A1)(ft).

Do not award final (A1) if $\text{P}(S)\times \text{P}(M)$ is not calculated. Follow through from part (a).

[2 marks]

Find P(−1.6 < $z$ < 2.4). Write your answer in terms of $a$ and $b$.

Given that $z$ > −1.6, find the probability that z < 2.4 . Write your answer in terms of $a$ and $b$.

Write down the standardized value for $x=1$.

It is also known that P($x$ > 2) = $b$.

Find $s$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing area under curve = 1        (M1)

eg   $a+x+b=1$,  $100-a-b$,  $1-a+b$

               A1  N2

[2 marks]

$\text{P}\left(z>-1.6\right)=1-a$ (seen anywhere)        (A1)

recognizing conditional probability        (M1)

eg   $\text{P}\left(A|B\right)$,  $\text{P}\left(B|A\right)$

correct working       (A1)

eg   ,   

      A1  N4

Note: Do not award the final A1 if correct answer is seen followed by incorrect simplification.

[4 marks]

$z=-1.6$ (may be seen in part (d))     A1  N1

Note: Depending on the candidate’s interpretation of the question, they may give $\frac{1-m}{s}$ as the answer to part (c). Such answers should be awarded the first (M1) in part (d), even when part (d) is left blank. If the candidate goes on to show $z=-1.6$ as part of their working in part (d), the A1 in part (c) may be awarded.

[1 mark]

attempt to standardize $x$ (do not accept $\frac{x-\mu }{\sigma }$)       (M1)

eg   $\frac{1-m}{s}$ (may be seen in part (c)), $\frac{m-2}{s}$,  $\frac{x-m}{\sigma }$

correct equation with each $z$-value      (A1)(A1)

eg   $-1.6=\frac{1-m}{s}$,  $2.4=\frac{2-m}{s}$,  $m+2.4s=2$

valid approach (to set up equation in one variable)       M1

eg   $2.4=\frac{2-\left(1.6s+1\right)}{s}$,   $\frac{1-m}{-1.6}=\frac{2-m}{2.4}$

correct working        (A1)

eg   $1.6s+1=2-2.4s$,  $4s=1$,  $m=\frac{7}{5}$

$s=\frac{1}{4}$       A1 N2

[6 marks]

Find the probability that the first ball chosen is labelled $\text{A}$.

Find the probability that the first ball chosen is labelled $\text{A}$ or labelled $\text{N}$.

Find the probability that the second ball chosen is labelled $\text{A}$, given that the first ball chosen was labelled $\text{N}$.

Find the probability that both balls chosen are labelled $\text{N}$.

$\frac{3}{9} \left(\frac{1}{3}, 0.333, 0.333333\dots , 33.3\%\right)$        (A1)   (C1)

[1 mark]

$\frac{5}{9} \left(0.556, 0.555555\dots , 55.6\%\right)$        (A1)   (C1)

[1 mark]

$\frac{3}{8} \left(0.375, 37.5\%\right)$        (A1)(A1)   (C2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{2}{9}\times \frac{1}{8}$        (M1)

Note: Award (M1) for a correct compound probability calculation seen.

$\frac{2}{72} \left(\frac{1}{36}, 0.0278, 0.0277777\dots , 2.78\%\right)$       (A1)  (C2)

[2 marks]

Write down the independent variable.

Write down the boiling temperature of the liquid.

Jim describes the correlation as very strong. Circle the value below which best represents the correlation coefficient.

$$0.9920.2510-0.251-0.992$$

Jim’s model is $d=-2.24t+105$, for $0⩽t⩽20$. Use his model to predict the decrease in temperature for any 2 minute interval.

$t$     A1     N1

[1 mark]

105     A1     N1

[1 mark]

$-0.992$     A2     N2

[2 marks]

valid approach     (M1)

eg$$$\frac{\text{d}d}{\text{d}t}=-2.24;2\times 2.24,2\times -2.24,d(2)=-2\times 2.24\times 105,$

finding $d(t_2)-d(t_1)$ where $t_2=t_1+2$

4.48 (degrees)     A1     N2

Notes:     Award no marks for answers that directly use the table to find the decrease in temperature for 2 minutes eg $\frac{105-98.4}{2}=3.3$.

[2 marks]

Copy and complete the following tree diagram.

Find the probability that Pablo leaves home before 07:00 and is late for work.

Find the probability that Pablo is late for work.

Given that Pablo is late for work, find the probability that he left home before 07:00.

Two days next week Pablo will drive to work. Find the probability that he will be late at least once.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

A1A1A1 N3

Note: Award A1 for each bold fraction.

[3 marks]

multiplying along correct branches      (A1)
eg  $\frac{3}{4}\times \frac{1}{8}$

P(leaves before 07:00 ∩ late) = $\frac{3}{32}$    A1 N2

[2 marks]

multiplying along other “late” branch      (M1)
eg  $\frac{1}{4}\times \frac{5}{8}$

adding probabilities of two mutually exclusive late paths      (A1)
eg  $\left(\frac{3}{4}\times \frac{1}{8}\right)+\left(\frac{1}{4}\times \frac{5}{8}\right),\frac{3}{32}+\frac{5}{32}$

$\text{P}\left(L\right)=\frac{8}{32}\left(=\frac{1}{4}\right)$    A1 N2

[3 marks]

recognizing conditional probability (seen anywhere)      (M1)
eg  $\text{P}\left(A|B\right),\text{P}\left(\text{before 7}|\text{late}\right)$

correct substitution of their values into formula      (A1)
eg $\frac{\frac{3}{32}}{\frac{1}{4}}$

$\text{P}\left(\text{left before 07:00}|\text{late}\right)=\frac{3}{8}$    A1 N2

[3 marks]

valid approach      (M1)
eg  1 − P(not late twice), P(late once) + P(late twice)

correct working      (A1)
eg  $1-\left(\frac{3}{4}\times \frac{3}{4}\right),2\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}$

$\frac{7}{16}$    A1 N2

[3 marks]

Write down the modal length of the rods.

Find the median length of the rods.

Calculate the lower quartile.

Calculate the interquartile range.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3     (A1) (C1)
 

[1 mark]

median is 13th position      (M1)

CF: 2, 6, 14, 20, 23, 25       (M1)

median = 3      (A1) (C3)

[3 marks]

2.5       (A1) (C1)

Note: Award (A1)(ft) if the sum of their parts (c)(i) and (c)(ii) is 4.

[1 mark]

1.5       (A1)(ft)  (C1)

Note: Award (A1)(ft) if the sum of their parts (c)(i) and (c)(ii) is 4.

[1 mark]

$p$.

$q$.

On the same grid, complete the cumulative frequency curve for these data.

Use the cumulative frequency curve to find an estimate for the number of students who worked at most 35 hours per month.

$p=10$     (A1)   (C1)  

Note: Award (A1) for each correct value.

[1 mark]

$q=56$     (A1)   (C1)  

Note: Award (A1) for each correct value.

[1 mark]

   (A1)(A1)   (C2)  

Note: Award (A1)(ft) for their 3 correctly plotted points; award (A1)(ft) for completing diagram with a smooth curve through their points. The second (A1)(ft) can follow through from incorrect points, provided the gradient of the curve is never negative. Award (C2) for a completely correct smooth curve that goes through the correct points.

[2 marks]

a straight vertical line drawn at 35 (accept 35 ± 1)    (M1)

26 (students)      (A1)   (C2)  

Note: Accept values between 25 and 27 inclusive.

[2 marks]

State the number of boys who answered questions in Portuguese.

Find the probability that the boy answered questions in Hindi.

Two girls are selected at random.

Calculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

20     (A1) (C1)

[1 mark]

$\frac{5}{43}\left(0.11627\dots ,11.6279\dots \text{\% }\right)$     (A1)(A1) (C2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{7}{37}\times \frac{12}{36}+\frac{12}{37}\times \frac{7}{36}$     (A1)(M1)

Note: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.

$=\frac{14}{111}\left(0.12612\dots ,12.6126\text{\% }\right)$     (A1) (C3)

[3 marks]

Write down the modal group for these data.

Use your graphic display calculator to find an estimate of the standard deviation of the weights of mangoes from this harvest.

On the grid below, draw a histogram for the data in the table.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.

$400\le w<500$        (A1)   (C1)

Note: Accept alternative notation $[400, 500)$ or $[400, 500[.$
Do not accept "$400\text{-}500$".

[1 mark]

$115 \left(115.265\dots \left(g\right)\right)$        (A2)   (C2)

Note: Award (A1)(A0) for an answer of $116 \left(116.459\dots \right)$.

[2 marks]

        (A2)(A1)   (C3)

Note: Award (A2) for all correct heights of bars or (A1) for three or four correct heights of bars.
Award (A1) for rectangular bars all with correct left and right end points ($100, 200, 300, 400, 500$ and $600$) and for no gaps; the bars do not have to be shaded.
Award at most (A2)(A0) if a ruler is not used for all lines.

[3 marks]

Complete the tree diagram below.

Find the value of $p$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

     (A1)(A1)     (C2)

Note:     Award (A1) for each correct pair of probabilities.

[2 marks]

$\frac{4}{5}p+\frac{1}{5}\times \frac{1}{4}=\frac{3}{5}$     (A1)(ft)(M1)(M1)

Note:     Award (A1)(ft) for two correct products from part (a), (M1) for adding their products, (M1) for equating the sum of any two probabilities to $\frac{3}{5}$.

$(p=)\frac{11}{16}(0.688,0.6875)$    (A1)(ft)     (C4)

Note:     Award the final (A1)(ft) only if $0⩽p⩽1$. Follow through from part (a).

[4 marks]

Find the largest value of $c$ that would not be considered an outlier.

One of the adults surveyed is $42$ years old. Estimate the age of their eldest child.

Find the mean age of all the adults surveyed.

$\text{IQR}=10-6\left(=4\right)$             (A1)

attempt to find $Q_3+1.5\times \text{IQR}$             (M1)

$10+6$

$16$             A1

[3 marks]

choosing $c=\frac{1}{2}a-9$             (M1)

$\frac{1}{2}\times 42-9$

$=12$ (years old)            A1

[2 marks]

attempt to solve system by substitution or elimination            (M1)

$34$ (years old)            A1

[2 marks]

Many candidates correctly found the value of 16. Some then incorrectly went on to state that 15 was therefore the minimum value that was not an outlier. For part (b) students needed to choose the appropriate rule to use to estimate the child's age. It was clear that many did not know there was a choice to be made and used both equations. As the mean point (𝑐̅,𝑎̅ ) lies on both regression lines, in part (c) candidates needed to solve the system of equations to find the mean adult age, 𝑎̅. Few candidates seemed to be aware of this.

Write down the probability that the mass of one of these corncobs is greater than 400 grams.

Find the value of $a$.

Estimate the interquartile range of the distribution.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$0.5\left(50\mathrm{\%},\frac{1}{2}\right)$     (A1)     (C1)

[1 mark]

$\text{P}(X>a)=0.25$$$OR$$     (M1)

Note:     Award (M1) for a sketch of approximate normal curve with a vertical line drawn to the right of the mean with the area to the right of this line shaded.

$a=434\text{ (g) }\left(433.724\dots \text{ (g)}\right)$     (A1)     (C2)

[2 marks]

$33.7244\dots \times 2$     (A1)(ft)(M1)

Note:     Award (A1)(ft) for $33.7244\dots (\text{or }433.7244\dots -400)$ seen, award (M1) for multiplying their 33.7244… by 2. Follow through from their answer to part (b).

OR

$434-366.275\dots$     (A1)(ft)(M1)

Note:     Award (A1)(ft) for their $366.275\dots (366)$ seen, (M1) for difference between their answer to (b) and their 366.

OR

     (A1)(ft)(M1)

Note:     Award (A1)(ft) for their $366.275\dots (366)$ seen. Award (M1) for correct symmetrical region indicated on labelled normal curve.

67.4 (g)     (A1)(ft)     (C3)

Note:     Accept an answer of 68 from use of rounded values. Follow through from part (b).

[3 marks]

Write down the probability that a light bulb produced by Home Shine is not defective.

Find the probability that both light bulbs are not defective.

Find the probability that at least one of Francesco’s light bulbs is defective.

Write down an expression, in terms of $a$, for the probability that at least one of Deborah’s three light bulbs is defective.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

0.93 (93%)     (A1)     (C1)

[1 mark]

$0.93\times 0.93$     (M1)

Note:     Award (M1) for squaring their answer to part (a).

0.865 (0.8649; 86.5%)     (A1)(ft)     (C2)

Notes:     Follow through from part (a).

Accept $0.86\left(\text{unless it follows }\frac{93}{100}\times \frac{92}{99}\right)$.

[2 marks]

$1-0.8649$     (M1)

Note:     Follow through from their answer to part (b)(i).

OR

$0.07\times 0.07+2\times (0.07\times 0.93)$     (M1)

Note:     Follow through from part (a).

0.135 (0.1351; 13.5%)     (A1)(ft)     (C2)

[2 marks]

$1-a^3$     (A1)     (C1)

Note:     Accept $3a^2(1-a)+3a(1-a{)}^2+(1-a{)}^3$ or equivalent.

[1 mark]

Find the median number of hours per week these Year 12 students spend doing homework.

Given that $10\%$ of these Year 12 students spend more than $k$ hours per week doing homework, find the value of $k$.

Find the value of $p$ and the value of $q$.

Estimate the number of Year 12 students that spend more than $15$ hours each week doing homework.

Explain why this sampling method might not provide an accurate representation of the amount of time all of the students in the school spend doing homework.

Suggest a more appropriate sampling method.

evidence of median position             (M1)

$40$ students

median $=14$ (hours)           A1

[2 marks]

recognizing there are $8$ students in the top $10\%$             (M1)

$72$ students spent less than $k$ hours             (A1)

$k=18$ (hours)           A1

[3 marks]

$15$ hours is $60$ students  OR  $p=60-4$             (M1)

$p=56$           A1

$21$ hours is $76$ students  OR  $q=80-76$  OR  $q=80-4-56-16$             (A1)

$q=4$           A1

[4 marks]

$20$ of the $80$ students OR $\frac{1}{4}$ spend more than $15$ hours doing homework             (A1)

$\frac{20}{80}=\frac{x}{320}$  OR  $\frac{1}{4}\times 320$  OR  $4\times 20$             (A1)

$80$ (students)           A1

[3 marks]

only year 12 students surveyed OR amount of homework might be different for different year levels        R1

[1 mark]

stratified sampling OR survey students in all years     R1

[1 mark]

Find the probability, in terms of $n$, that the game will end on her first draw.

Find the probability, in terms of $n$, that the game will end on her second draw.

third draw.

fourth draw.

Hayley plays the game when $n$ = 5. She pays $20 to play and can earn money back depending on the number of draws it takes to obtain a blue marble. She earns no money back if she obtains a blue marble on her first draw. Let M be the amount of money that she earns back playing the game. This information is shown in the following table.

Find the value of $k$ so that this is a fair game.

$\frac{2}{n}$     A1 N1

[1 mark]

correct probability for one of the draws      A1

eg   P(not blue first) = $\frac{n-2}{n}$,   blue second = $\frac{2}{n-1}$

valid approach      (M1)

eg   recognizing loss on first in order to win on second, P(B' then B),  P(B') × P(B | B'),  tree diagram

correct expression in terms of $n$       A1 N3

eg   $\frac{n-2}{n}\times \frac{2}{n-1}$,  $\frac{2n-4}{n^2-n}$,  $\frac{2\left(n-2\right)}{n\left(n-1\right)}$

[3 marks]

correct working      (A1)

eg   $\frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}$

$\frac{12}{60}\left(=\frac{1}{5}\right)$     A1  N2

[2 marks]

correct working      (A1)

eg  $\frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times \frac{2}{2}$

$\frac{6}{60}\left(=\frac{1}{10}\right)$    A1  N2

[2 marks]

correct probabilities (seen anywhere)      (A1)(A1)

eg   $\text{P}\left(\text{1}\right)=\frac{2}{5}$,  $\text{P}\left(\text{2}\right)=\frac{6}{20}$  (may be seen on tree diagram)

valid approach to find E (M) or expected winnings using their probabilities      (M1)

eg   $\text{P}\left(\text{1}\right)\times \left(0\right)+\text{P}\left(\text{2}\right)\times \left(20\right)+\text{P}\left(\text{3}\right)\times \left(8k\right)+\text{P}\left(\text{4}\right)\times \left(12k\right)$,

$\text{P}\left(\text{1}\right)\times \left(-20\right)+\text{P}\left(\text{2}\right)\times \left(0\right)+\text{P}\left(\text{3}\right)\times \left(8k-20\right)+\text{P}\left(\text{4}\right)\times \left(12k-20\right)$

correct working to find E (M) or expected winnings      (A1)

eg   $\frac{2}{5}\left(0\right)+\frac{3}{10}\left(20\right)+\frac{1}{5}\left(8k\right)+\frac{1}{10}\left(12k\right)$,

$\frac{2}{5}\left(-20\right)+\frac{3}{10}\left(0\right)+\frac{1}{5}\left(8k-20\right)+\frac{1}{10}\left(12k-20\right)$

correct equation for fair game      A1

eg   $\frac{3}{10}\left(20\right)+\frac{1}{5}\left(8k\right)+\frac{1}{10}\left(12k\right)=20$,  $\frac{2}{5}\left(-20\right)+\frac{1}{5}\left(8k-20\right)+\frac{1}{10}\left(12k-20\right)=0$

correct working to combine terms in $k$       (A1)

eg   $-8+\frac{14}{5}k-4-2=0$,  $6+\frac{14}{5}k=20$,  $\frac{14}{5}k=14$

$k$ = 5    A1 N0

Note: Do not award the final A1 if the candidate’s FT probabilities do not sum to 1.

[7 marks]

Find the probability that a student, chosen at random arrives at least 60 minutes after the school opens.

Find the probability that a student, chosen at random arrives between 45 minutes and 55 minutes after the school opens.

A second school, Mulberry Park, also opens at 08:00 every morning. The arrival times of the students at this school follows exactly the same distribution as Malthouse school.

Given that, on one morning, 15 students arrive at least 60 minutes after the school opens, estimate the number of students at Mulberry Park school.

0.0548  (0.054799…, 5.48%)     (A2) (C2)

[2 marks]

0.645  (0.6449900…, 64.5%)     (A2) (C2)

[2 marks]