Write down a formula for $A$, the surface area to be coated.

Express this volume in $\text{c}{\text{m}}^3$.

Write down, in terms of $r$ and $h$, an equation for the volume of this water container.

Show that $A=\pi r^2+\frac{1000000}{r}$.

Find $\frac{\text{d}A}{\text{d}r}$.

Using your answer to part (e), find the value of $r$ which minimizes $A$.

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

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

$(A=)\pi r^2+2\pi rh$    (A1)(A1)

Note:     Award (A1) for either $\pi r^2$ OR $2\pi rh$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500000$    (A1)

Notes:     Units not required.

[1 mark]

$500000=\pi r^{2}h$    (A1)(ft)

Notes:     Award (A1)(ft) for $\pi r^{2}h$ equating to their part (b).

Do not accept unless $V=\pi r^{2}h$ is explicitly defined as their part (b).

[1 mark]

$A=\pi r^2+2\pi r\left(\frac{500000}{\pi r^2}\right)$    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their $\frac{500000}{\pi r^2}$ seen.

Award (M1) for correctly substituting only $\frac{500000}{\pi r^2}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $\pi rh=\frac{500000}{r}$ and substituting for $\pi rh$ in expression for $A$.

$A=\pi r^2+\frac{1000000}{r}$    (AG)

Notes:     The conclusion, $A=\pi r^2+\frac{1000000}{r}$, must be consistent with their working seen for the (A1) to be awarded.

Accept ${10}^6$ as equivalent to $1000000$.

[2 marks]

$2\pi r-\frac{\text{1}\text{000}\text{000}}{r^2}$    (A1)(A1)(A1)

Note:     Award (A1) for $2\pi r$, (A1) for $\frac{1}{r^2}$ or $r^{-2}$, (A1) for $-\text{1}\text{000}\text{000}$.

[3 marks]

$2\pi r-\frac{1000000}{r^2}=0$    (M1)

Note:     Award (M1) for equating their part (e) to zero.

$r^3=\frac{1000000}{2\pi }$ OR $r=\sqrt[3]{\frac{1000000}{2\pi }}$     (M1)

Note:     Award (M1) for isolating $r$.

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

$(r=)54.2(\text{cm})(54.1926\dots )$    (A1)(ft)(G2)

[3 marks]

$\pi (54.1926\dots {)}^2+\frac{1000000}{(54.1926\dots )}$    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

$=27700(\text{c}{\text{m}}^2)(27679.0\dots )$    (A1)(ft)(G2)

[2 marks]

$\frac{27679.0\dots }{2000}$    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

$=13.8395\dots$    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their $13.8395\dots$ to the next integer.

[3 marks]

Find $q$.

Find $p$.

Write down the probability of drawing three blue marbles.

Explain why the probability of drawing three white marbles is $\frac{1}{6}$.

The bag contains a total of ten marbles of which $w$ are white. Find $w$.

Jill plays the game nine times. Find the probability that she wins exactly two prizes.

Grant plays the game until he wins two prizes. Find the probability that he wins his second prize on his eighth attempt.

correct substitution into $\text{E}(X)$ formula     (A1)

eg$$$0(p)+1(0.5)+2(0.3)+3(q)=1.2$

$q=\frac{1}{30}$, 0.0333     A1     N2

[2 marks]

evidence of summing probabilities to 1     (M1)

eg$$$p+0.5+0.3+q=1$

$p=\frac{1}{6},0.167$     A1     N2

[2 marks]

$\text{P (3 blue)}=\frac{1}{30},0.0333$     A1     N1

[1 mark]

valid reasoning     R1

eg$$$\text{P (3 white)}=\text{P(0 blue)}$

$\text{P(3 white)}=\frac{1}{6}$     AG     N0

[1 mark]

valid method     (M1)

eg$$$\text{P(3 white)}=\frac{w}{10}\times \frac{w-1}{9}\times \frac{w-2}{8},\frac{{}_w{C}_3}{{}_{10}{C}_3}$

correct equation     A1

eg$$$\frac{w}{10}\times \frac{w-1}{9}\times \frac{w-2}{8}=\frac{1}{6},\frac{{}_w{C}_3}{{}_{10}{C}_3}=0.167$

$w=6$     A1     N2

[3 marks]

valid approach     (M1)

eg$$$\text{B}(n,p),\left(\begin{array}{c}n\\ r\end{array}\right)p^r{q}^{n-r},(0.167{)}^2(0.833{)}^7,\left(\begin{array}{c}9\\ 2\end{array}\right)$

0.279081

0.279     A1     N2

[2 marks]

recognizing one prize in first seven attempts     (M1)

eg$$$\left(\begin{array}{c}7\\ 1\end{array}\right),{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^6$

correct working     (A1)

eg$$$\left(\begin{array}{c}7\\ 1\end{array}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^6,0.390714$

correct approach     (A1)

eg$$$\left(\begin{array}{c}7\\ 1\end{array}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^6\times \frac{1}{6}$

0.065119

0.0651     A1     N2

[4 marks]

Show that $b=0.3-a$.

Find the difference between the greatest possible expected value and the least possible expected value.

correct approach  A1

eg       $0.2+0.5+b+a=1$,  $0.7+a+b=1$

$b=0.3-a$       AG  N0

[1 mark]

correct substitution into $\text{E}\left(X\right)$        (A1)

eg       $0.2+4\times 0.5+a\times b+\left(a+b-0.5\right)\times a$,  $0.2+2+a\times b-0.2a$

valid attempt to express $\text{E}\left(X\right)$ in one variable       M1

eg       $0.2+4\times 0.5+a\times \left(0.3-a\right)+\left(-0.2\right)\times a$,  $2.2+0.1a-a^2$,

           $0.2+4\times 0.5+\left(0.3-b\right)\times b+\left(-0.2\right)\times \left(0.3-b\right)$,   $2.14+0.5b-b^2$

correct value of greatest $\text{E}\left(X\right)$        (A1)

$2.2025$  (exact)

valid attempt to find least value        (M1)

eg       graph with minimum indicated, $\text{E}\left(0\right)$  and  $\text{E}\left(0.3\right)$

          $\left(0\text{, }2.2\right)$  and  $\left(0.3\text{, }2.14\right)$ if $\text{E}\left(X\right)$ in terms of $a$

          $\left(0\text{, }2.14\right)$  and  $\left(0.3\text{, }2.2\right)$ if $\text{E}\left(X\right)$ in terms of $b$

correct value of least $\text{E}\left(X\right)$        (A1)

eg       $2.14$  (exact)

difference $=0.0625$ (exact)       A1  N2

[6 marks]

Find the value of $\sigma$.

On a randomly selected day, find the probability that Suzi’s drive to work will take longer than $45$ minutes.

Find the probability that she will be late to work at least one day next week.

Given that Suzi will be late to work at least one day next week, find the probability that she will be late less than three times.

Find the probability that Suzi will receive a bonus.

METHOD 1

$T~\text{N}\left(35,{\sigma }^2\right)$

$\text{P}\left(T>40\right)=0.25$  or  $\text{P}\left(T<40\right)=0.75$          (M1)

attempt to solve for $\sigma$ graphically or numerically using the GDC          (M1)

graph of normal curve $T~\text{N}\left(35,{\sigma }^2\right)$ for $\text{P}\left(T>40\right)$ and $y=0.25$  OR  $\text{P}\left(T<40\right)$ and $y=0.75$
OR table of values for $\text{P}\left(T<40\right)$ or $\text{P}\left(T>40\right)$

$\sigma =7.413011\dots$

$\sigma =7.41$ (min)          A2

METHOD 2

$T~\text{N}\left(35,{\sigma }^2\right)$

$\text{P}\left(T>40\right)=0.25$  or  $\text{P}\left(T<40\right)=0.75$          (M1)

$z=0.674489\dots$          (A1)

valid equation using their $z$-score (clearly identified as $z$-score and not a probability)          (M1)

$\frac{40-35}{\sigma }=0.674489\dots$  OR  $5=0.674489\dots \sigma$

$7.413011\dots$

$\sigma =7.41$ (min)          A1

[4 marks]

$\text{P}\left(T>45\right)$          (M1)

$=0.0886718\dots$

$=0.0887$          A1

[2 marks]

recognizing binomial probability            (M1)

$L~\text{B}\left(5, 0.0886718\dots \right)$

$\text{P}\left(L\ge 1\right)=1-\text{P}\left(L=0\right)$  OR

$\text{P}\left(L\ge 1\right)=\text{P}\left(L=1\right)+\text{P}\left(L=2\right)+\text{P}\left(L=3\right)+\text{P}\left(L=4\right)+\text{P}\left(L=5\right)$            (M1)

$0.371400\dots$

$\text{P}\left(L\ge 1\right)=0.371$            A1

[3 marks]

recognizing conditional probability in context            (M1)

finding $\left\{L<3\right\}\cap \left\{L\ge 1\right\}=\left\{L=1, L=2\right\}$   (may be seen in conditional probability)            (A1)

$\text{P}\left(L=1\right)+\text{P}\left(L=2\right)=0.36532\dots$  (may be seen in conditional probability)            (A1)

$P\left(L<3 \overline{) L\ge 1}\right)=\frac{0.36532\dots }{0.37140\dots }$            (A1)

$0.983636\dots$

$0.984$            A1

[5 marks]

METHOD 1

recognizing that Suzi can be late no more than once (in the remaining six days)            (M1)

$\text{X}~\text{B}\left(6, 0.0886718\dots \right)$, where $X$ is the number of days late            (A1)

$\text{P}\left(X\le 1\right)=\text{P}\left(X=0\right)+\text{P}\left(X=1\right)$            (M1)

$=0.907294\dots$

$\text{P}\left(\text{Suzi gets a bonus}\right)=0.907$            A1

Note: The first two marks may be awarded independently.

METHOD 2

recognizing that Suzi must be on time at least five times (of the remaining six days)            (M1)

$\text{X}~\text{B}\left(6, 0.911328\dots \right)$, where $X$ is the number of days on time           (A1)

$\text{P}\left(X\ge 5\right)=1-\text{P}\left(X\le 4\right)$  OR  $1-0.0927052\dots$  OR $\text{P}\left(X=5\right)\text{+}\left(X=6\right)$ OR $0.334434\dots +0.572860\dots$            (M1)

$=0.907294\dots$

$\text{P}\left(\text{Suzi gets a bonus}\right)=0.907$            A1

Note: The first two marks may be awarded independently.

[4 marks]

In part (a) many candidates did not know to use inverse normal to find a $z$ value. Some did find $z$, then rounded it to 3 sf and got an incorrect value for sigma.

Part (b) was mostly well done.

In (c) most recognised the binomial and handled 'at least one' correctly.

In (d) many recognised conditional probability, but most candidates were not able to find the intersection of the events as P(1) + P(2).

In part (e), those candidates who did understand what to do often misunderstood that they needed to look at 1 or no more lates and just considered one more late. Something similar happened to those who approached the question by considering the times Suzi was on time. 

This question was only correctly answered by a few, and students tended to perform either very well or very poorly.

Find the value of a and of b.

Use the regression equation to estimate the value of y when x = 3.57.

The relationship between x and y can be modelled using the formula y = kxⁿ, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.

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

valid approach      (M1)

eg  one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14      A1A1 N3

[3 marks]

correct substitution     (A1)

eg   −0.454 ln 3.57 + 6.14

correct working     (A1)

eg  ln y = 5.56484

261.083 (260.409 from 3 sf)

y = 261, (y = 260 from 3sf)       A1 N3

Note: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.

[3 marks]

METHOD 1

valid approach for expressing ln y in terms of ln x      (M1)

eg  $\text{ln}y=\text{ln}\left(k{x}^n\right),\text{ln}\left(k{x}^n\right)=a\text{ln}x+b$

correct application of addition rule for logs      (A1)

eg  $\text{ln}k+\text{ln}\left(x^n\right)$

correct application of exponent rule for logs       A1

eg  $\text{ln}k+n\text{ln}x$

comparing one term with regression equation (check FT)      (M1)

eg  $n=a,b=\text{ln}k$

correct working for k      (A1)

eg  $\text{ln}k=6.14210,k=e^{6.14210}$

465.030

$n=-0.454,k=465$ (464 from 3sf)     A1A1 N2N2

METHOD 2

valid approach      (M1)

eg  $e^{\text{ln}y}=e^{a\text{ln}x+b}$

correct use of exponent laws for $e^{a\text{ln}x+b}$     (A1)

eg  $e^{a\text{ln}x}\times e^b$

correct application of exponent rule for $a\text{ln}x$     (A1)

eg  $\text{ln}{x}^a$

correct equation in y      A1

eg  $y=x^a\times e^b$

comparing one term with equation of model (check FT)      (M1)

eg  $k=e^b,n=a$

465.030

$n=-0.454,k=465$ (464 from 3sf)     A1A1 N2N2

METHOD 3

valid approach for expressing ln y in terms of ln x (seen anywhere)      (M1)

eg  $\text{ln}y=\text{ln}\left(k{x}^n\right),\text{ln}\left(k{x}^n\right)=a\text{ln}x+b$

correct application of exponent rule for logs (seen anywhere)      (A1)

eg  $\text{ln}\left(x^a\right)+b$

correct working for b (seen anywhere)      (A1)

eg  $b=\text{ln}\left(e^b\right)$

correct application of addition rule for logs      A1

eg  $\text{ln}\left(e^b{x}^a\right)$

comparing one term with equation of model (check FT)     (M1)

eg  $k=e^b,n=a$

465.030

$n=-0.454,k=465$ (464 from 3sf)     A1A1 N2N2

[7 marks]

Write down the number of students in the group who take art class.

the student does not take art class.

the student takes either art class or music class, but not both.

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

valid approach      (M1)

$\left(A\cap M^{\prime }\right)+\left(A\cap M\right)$,  $\frac{17}{35}$,   11 + 6

number of students taking art class = 17        A1 N2

[2 marks]

valid approach      (M1)

13 + 5,   35 − 17,   18,   1 − P(A)

0.514285

P(A') = $\frac{18}{35}$  (exact), 0.514         A1 N2

[2 marks]

valid approach      (M1)

11 + 13,   35 − 6 − 5,   24

0.685714

P(A or M but not both) = $\frac{24}{35}$  (exact), 0.686         A1 N2

[2 marks]

Use your graphic display calculator to write down $\overline{x}$, the mean project mark.

Use your graphic display calculator to write down $\overline{y}$, the mean examination score.

Use your graphic display calculator to write down r , Pearson’s product–moment correlation coefficient.

Find the exact value of m and of c for these data.

Show that the point M ($\overline{x}$, $\overline{y}$) lies on the regression line y on x.

Use the regression line y on x to estimate Jerome’s examination score.

Justify whether it is valid to use the regression line y on x to estimate Jerome’s examination score.

In his final IB examination Jerome scored 65.

Calculate the percentage error in Jerome’s estimated examination score.

14      (G1)

[1 mark]

54     (G1)

[1 mark]

0.5     (G2)

[2 marks]

m = 0.875, c = 41.75  $\left(m=\frac{7}{8}\text{,}c=\frac{167}{4}\right)$        (A1)(A1)

Note: Award (A1) for 0.875 seen. Award (A1) for 41.75 seen. If 41.75 is rounded to 41.8 do not award (A1).

[2 marks]

y = 0.875(14) + 41.75      (M1)

Note: Award (M1) for their correct substitution into their regression line. Follow through from parts (a)(i) and (b)(i).

= 54

and so the mean point lies on the regression line      (A1)

(accept 54 is $\overline{y}$, the mean value of the y data)

Note: Do not award (A1) unless the conclusion is explicitly stated and the 54 seen. The (A1) can be awarded only if their conclusion is consistent with their equation and it lies on the line.

The use of 41.8 as their c value precludes awarding (A1).

OR

54 = 0.875(14) + 41.75      (M1)

54 = 54

Note: Award (M1) for their correct substitution into their regression line. Follow through from parts (a)(i) and (b)(i).

and so the mean point lies on the regression line     (A1)

Note: Do not award (A1) unless the conclusion is explicitly stated. Follow through from part (a). 

The use of 41.8 as their c value precludes the awarding of (A1).

[2 marks]

y = 0.875(17) + 41.75      (M1)

Note: Award (M1) for correct substitution into their regression line.

= 56.6   (56.625)      (A1)(ft)(G2)

Note: Follow through from part (b)(i).

[2 marks]

the estimate is valid      (A1)

since this is interpolation and the correlation coefficient is large enough      (R1)

OR

the estimate is not valid      (A1)

since the correlation coefficient is not large enough      (R1)

Note: Do not award (A1)(R0). The (R1) may be awarded for reasoning based on strength of correlation, but do not accept “correlation coefficient is not strong enough” or “correlation is not large enough”.

Award (A0)(R0) for this method if no numerical answer to part (a)(iii) is seen.

[2 marks]

$\left|\frac{56.6-65}{65}\right|\times 100$      (M1)

Note: Award (M1) for correct substitution into percentage error formula. Follow through from part (c)(i).

= 12.9  (%)(12.9230…)      (A1)(ft)(G2)

Note: Follow through from part (c)(i). Condone use of percentage symbol.
Award (G0) for an answer of −12.9 with no working.

[2 marks]

Sketch a diagram to represent this information.

Show that $m=15.7$.

Find the probability that Emlyn plays between $13 \text{minutes}$ and $18 \text{minutes}$ in a game.

Find the probability that Emlyn plays more than $20 \text{minutes}$ in a game.

Find the value of $x$.

Find the probability he plays between $13 \text{minutes}$ and $18 \text{minutes}$ in one game and more than $20 \text{minutes}$ in the other game.

Find the expected number of successful shots Emlyn will make on Monday, based on the results from Saturday and Sunday.

Emlyn claims the results from Saturday and Sunday show that his expected number of successful shots will be more than Johan’s.

Determine if Emlyn’s claim is correct. Justify your reasoning.

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

         (A1)(A1)

Note: Award (A1) for bell shaped curve with mean $m$ or $13.6$ indicated. Award (A1) for approximately correct shaded region.

[2 marks]

$\text{P}\left(T>17.8\right)=0.3$         (M1)

OR

        (M1)

Note: Award (M1) for correct probability equation using $0.3$ OR correctly shaded diagram indicating $17.8$. Strict or weak inequalities are accepted in parts (b), (c) and (d).

$\frac{13.6+17.8}{2}$   $\left(17.8-\frac{17.8-13.6}{2}\right)$  OR  $\left(13.6+\frac{17.8-13.6}{2}\right)$         (M1)

Note: Award (M0)(M1) for unsupported $\frac{13.6+17.8}{2}$ OR $\left(17.8-\frac{17.8-13.6}{2}\right)$ OR $\left(13.6+\frac{17.8-13.6}{2}\right)$ OR the midpoint of $13.6$ and $17.8$ is $15.7$.
Award at most (M1)(M0) if the final answer is not seen. Award (M0)(M0) for using known values $m=15.7$ and $\sigma =4$ to validate $\text{P}\left(T<17.8\right)=0.7$ or $\text{P}\left(T<13.6\right)=0.3$.

$15.7$         (AG)

[2 marks]

$\text{P}\left(13\le T\le 18\right)$         (M1)

OR

        (M1)

Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating $13$ and $18$.

$0.468 \left(46.8\%, 0.467516\dots \right)$         (A1)(G2)

[2 marks]

$\text{P}\left(T\ge 20\right)$         (M1)

OR

        (M1)

Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating $20$.

$0.141 \left(14.1\%, 0.141187\dots \right)$         (A1)(G2)

[2 marks]

$\text{P}\left(T<t\right)=0.6$         (M1)

OR

        (M1)

Note: Award (M1) for correct probability equation OR for a correctly shaded region with $x$ indicated to the right-hand side of the mean.

$16.7 \left(16.7133\dots \right)$         (A1)(G2)

[2 marks]

$0.467516\dots \times 0.141187\dots \times 2$         (M1)(M1)

OR

$\left(0.467516\dots \times 0.141187\dots \right)+\left(0.141187\dots \times 0.467516\dots \right)$        (M1)(M1)

Note: Award (M1) for the multiplication of their parts (c)(i) and (c)(ii), (M1) for multiplying their product by $2$ or for adding their products twice. Follow through from part (c).

$0.132 \left(13.2\%, 0.132014\dots \right)$         (A1)(ft)(G2)

Note: Award (G0) for an unsupported final answer of $0.066007\dots$

[3 marks]

$\frac{69}{102}\times 200$         (M1)

Note: Award (M1) for correct probability multiplied by $200$.

$135 \left(135.294\dots \right)$         (A1)(G2)

[2 marks]

$\left(\frac{67}{98}\times 200=\right) 136.734\dots$         (A1)

Note: Award (M1) for $137$ or $136.734\dots$ seen.

Emlyn is incorrect, $135<137 \left(135.294\dots <136.734\dots \right)$         (R1)

Note: To award the final (R1), both the conclusion and the comparison must be seen. Award at most (A0)(R1)(ft) for consistent incorrect methods in parts (f) and (g).

OR

$\left(\frac{67}{98}=\right) 0.684 \left(0.683673\dots \right) \left(\frac{69}{102}=\right) 0.676 \left(0.676470\dots \right)$         (A1)

Note: Award (A1) for both correct probabilities seen.

Emlyn is incorrect, $0.676<0.684$         (R1)

Note: To award the final (R1), both the conclusion and the comparison must be seen. Award at most (A0)(R1)(ft) for consistent incorrect methods in parts (f) and (g).

[2 marks]

Find the value of $a$.

Find the value of $b$.

Find the number of students who visited at least two types of main attraction.

Write down the value of $n( R\cap W)$.

Find the probability that a randomly selected student visited the rollercoasters.

Find the probability that a randomly selected student visited the virtual reality rides.

Hence determine whether the events in parts (d)(i) and (d)(ii) are independent. Justify your reasoning. 

$74-\left(32+12+10+9+5\right)$  OR  $74-68$     (M1)

Note: Award (M1) for setting up a correct expression.

$\left(a=\right) 6$       (A1)(G2)

[2 marks]

$100-\left(74+18\right)$     (M1)

OR

$100-92$     (M1)

OR

$100-\left(32+9+5+12+10+18+6\right)$     (M1)

Note: Award (M1) for setting up a correct expression. Follow through from part (a)(i) but only for $a\ge 0$.

$\left(b=\right) 8$       (A1)(ft)(G2)

Note: Follow through from part(a)(i). The value of $b$ must be greater or equal to zero for the (A1)(ft) to be awarded.

[2 marks]

$9+5+12+10$     (M1)

Note: Award (M1) for adding $9, 5, 12$ and $10$.

$36$       (A1)(G2)

[2 marks]

$14$     (A1)

[1 mark]

$\frac{58}{100} \left(\frac{29}{50}, 0.58, 58\%\right)$     (A1)(A1)(G2)

Note: Award (A1) for correct numerator. Award(A1) for the correct denominator. Award (A0) for $58$ only.

[2 marks]

$\frac{45}{100} \left(\frac{9}{20}, 0.45, 45\%\right)$     (A1)(ft)

Note: Follow through from their denominator from part (d)(i).

[1 mark]

they are not independent     (A1)(ft)

$\frac{58}{100}\times \frac{45}{100}\ne \frac{17}{100}$  OR  $0.261\ne 0.17$     (R1)

Note: Comparison of numerical values must be seen for (R1) to be awarded.
Do not award (A1)(R0). Follow through from parts (d)(i) and (d)(ii).

[2 marks]

Find the probability that the student plays a sport and is involved in theatre.

Find the probability that the student is involved in theatre, but does not play a sport.

Find $\text{P}\left(G\cap T\right)$.

Determine if the events $G$ and $T$ are independent. Justify your answer.

EITHER

$\text{P}\left(S\right)+\text{P}\left(T\right)+\text{P}\left(S\text{'}\cap T\text{'}\right)-\text{P}\left(S\cap T\right)=1$  OR  $\text{P}\left(S\cup T\right)=\text{P}\left(\left(S\text{'}\cap T\text{'}\right)\mathrm{\text{'}}\right)$          (M1)

$0.7+0.2+0.18-\text{P}\left(S\cap T\right)=1$  OR  $\text{P}\left(S\cup T\right)=1-0.18$

OR

a clearly labelled Venn diagram        (M1)

THEN

$P\left(S\cap T\right)=0.08$  (accept $8\%$)              A1

Note: To obtain the M1 for the Venn diagram all labels must be correct and in the correct sections. For example, do not accept $0.7$ in the area corresponding to $S\cap T\text{'}$.

[2 marks]

EITHER

$\text{P}\left(T\cap S\text{'}\right)=\text{P}\left(T\right)-\text{P}\left(T\cap S\right)\left(=0.2-0.08\right)$  OR

$\text{P}\left(T\cap S\text{'}\right)=\text{P}\left(T\cup S\right)-\text{P}\left(S\right)\left(=0.82-0.7\right)$         (M1)

 
OR

a clearly labelled Venn diagram including $\text{P}\left(S\right)$, $\text{P}\left(T\right)$ and $\text{P}\left(S\cap T\right)$         (M1)

THEN

$=0.12$  (accept $12\%$)              A1

[2 marks]

$\text{P}\left(G\cap T\right)=\text{P}\left(T/G\right)\text{P}\left(G\right) \left(0.25\times 0.48\right)$         (M1)

$=0.12$             A1

[2 marks]

METHOD 1

$\text{P}\left(G\right)\times \text{P}\left(T\right)\left(=0.48\times 0.2\right)=0.096$             A1

$\text{P}\left(G\right)\times \text{P}\left(T\right)\mathrm{\ne }\text{P}\left(G\cap T\right)\mathrm{\Rightarrow }$  $G$ and $T$ are not independent             R1

METHOD 2

$\text{P}\left(T \overline{) G}\right)=0.25$             A1

$\text{P}\left(T \overline{) G}\right)\ne \text{P}\left(T\right)\mathrm{\Rightarrow }$  $G$ and $T$ are not independent             R1

Note: Do not award A0R1.

[2 marks]

The heights of adult males in a country are normally distributed with a mean of 180 cm and a standard deviation of $\sigma \text{ cm}$. 17% of these men are shorter than 168 cm. 80% of them have heights between $(192-h)\text{ cm}$ and 192 cm.

Find the value of $h$.

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

finding the $z$-value for 0.17     (A1)

eg$$$z=-0.95416$

setting up equation to find $\sigma$,     (M1)

eg$$$z=\frac{168-180}{\sigma },-0.954=\frac{-12}{\sigma }$

$\sigma =12.5765$     (A1)

EITHER (Properties of the Normal curve)

correct value (seen anywhere)     (A1)

eg$$

correct working     (A1)

eg$$

$\text{P}(X>192-h)=0.8+0.17$

correct equation in $h$

eg$$$\frac{(192-h)-180}{12.576}=-1.88079,192-h=156.346$     (A1)

35.6536

$h=35.7$     A1     N3

OR (Trial and error using different values of h)

two correct probabilities whose 2 sf will round up and down, respectively, to 0.8     A2

eg$$

$h=35.7$     A2

[7 marks]

Find the range of the average body weights for these seven species of mammal.

For the data from these seven species calculate $r$, the Pearson’s product–moment correlation coefficient;

For the data from these seven species describe the correlation between the average body weight and the average weight of the brain.

Write down the equation of the regression line $y$ on $x$, in the form $y=mx+c$.

Use your regression line to estimate the average weight of the brain of grey wolves.

Find the percentage error in your estimate in part (d).

State whether it is valid to use the regression line to estimate the average weight of the brain of mice. Give a reason for your answer.

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

$529-3$     (M1)

$=526\text{ (kg)}$     (A1)(G2)

[2 marks]

$0.922(0.921857\dots )$     (G2)

[2 marks]

(very) strong, positive     (A1)(ft)(A1)(ft)

Note:     Follow through from part (b)(i).

[2 marks]

$y=0.000986x+0.0923(y=0.000985837\dots x+0.0923391\dots )$     (A1)(A1)

Note:     Award (A1) for $0.000986x$, (A1) for 0.0923.

Award a maximum of (A1)(A0) if the answer is not an equation in the form $y=mx+c$.

[2 marks]

$0.000985837\dots (36)+0.0923391\dots$     (M1)

Note:     Award (M1) for substituting 36 into their equation.

$0.128\text{ (kg) }\left(0.127829\dots \text{ (kg)}\right)$     (A1)(ft)(G2)

Note:     Follow through from part (c). The final (A1) is awarded only if their answer is positive.

[2 marks]

$\left|\frac{0.127829\dots -0.120}{0.120}\right|\times 100$     (M1)

Note:     Award (M1) for their correct substitution into percentage error formula.

$6.52(\mathrm{\%})\left(\mathrm{6.52442...}(\mathrm{\%})\right)$     (A1)(ft)(G2)

Note: Follow through from part (d). Do not accept a negative answer.

[2 marks]

Not valid     (A1)

the mouse is smaller/lighter/weighs less than the cat (lightest mammal)     (R1)

OR

as it would mean the mouse’s brain is heavier than the whole mouse     (R1)

OR

0.023 kg is outside the given data range.     (R1)

OR

Extrapolation     (R1)

Note:     Do not award (A1)(R0). Do not accept percentage error as a reason for validity.

[2 marks]

Show that event A and event D are not independent.

Find $\text{P}\left(A\cap D^{\prime }\right)$.

 Given that all passengers for a flight arrive on time, find the probability that the flight does not depart on time.

Find the value of $\sigma$.

All flights have two pilots. Find the percentage of flights where both pilots flew more than 30 hours last week.

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

METHOD 1

multiplication of P(A) and P(D)     (A1)

eg   0.70 × 0.85,  0.595

correct reasoning for their probabilities       R1

eg   $0.595\ne 0.65$,   $0.70\times 0.85\ne \text{P}\left(A\cap D\right)$

A and D are not independent      AG N0

METHOD 2

calculation of $\text{P}\left(D|A\right)$       (A1)

eg   $\frac{13}{14}$,  0.928

correct reasoning for their probabilities       R1

eg   $0.928\ne 0.85$,   $\frac{0.65}{0.7}\text{P}\left(D\right)$

A and D are not independent      AG N0

[2 marks]

correct working       (A1)

eg   $\text{P}\left(A\right)-\text{P}\left(A\cap D\right)$ ,  0.7 − 0.65 , correct shading and/or value on Venn diagram

$\text{P}\left(A\cap D^{\prime }\right)=0.05$       A1  N2

[2 marks]

recognizing conditional probability (seen anywhere)       (M1)

eg   $\frac{\text{P}\left(D^{\prime }\cap A\right)}{\text{P}\left(A\right)}$,  $\text{P}\left(A|B\right)$

correct working       (A1)

eg    $\frac{0.05}{0.7}$

0.071428

$\text{P}\left(D^{\prime }|A\right)=\frac{1}{14}$ , 0.0714     A1  N2

[3 marks]

finding standardized value for 28 hours (seen anywhere)       (A1)

eg   $z=1.28155$

correct working to find $\sigma$       (A1)

eg    $1.28155=\frac{28-25}{\sigma }$ , $\frac{28-25}{1.28155}$

2.34091

$\sigma =2.34$     A1  N2

[3 marks]

$\text{P}\left(X>30\right)=0.0163429$       (A1)

valid approach (seen anywhere)        (M1)

eg   ${\left[\text{P}\left(X>30\right)\right]}^2$ ,  (0.01634)² ,  B(2, 0.0163429) , 2.67E-4 , 2.66E-4

0.0267090

0.0267 %    A2  N3

[4 marks]

Calculate the mean test grade of the students;

Calculate the standard deviation.

Find the median test grade of the students.

Find the interquartile range.

Find the probability that this student scored a grade 5 or higher.

Given that the first student chosen at random scored a grade 5 or higher, find the probability that both students scored a grade 6.

Calculate the probability that a student chosen at random spent at least 90 minutes preparing for the test.

Calculate the expected number of students that spent at least 90 minutes preparing for the test.

$\frac{1(1)+3(2)+7(3)+13(4)+11(5)+10(6)+5(7)}{50}=\frac{230}{50}$     (M1)

Note:     Award (M1) for correct substitution into mean formula.

$=4.6$     (A1)     (G2)

[2 marks]

$1.46(1.45602\dots )$     (G1)

[1 mark]

5     (A1)

[1 mark]

$6-4$     (M1)

Note:     Award (M1) for 6 and 4 seen.

$=2$     (A1)     (G2)

[2 marks]

$\frac{11+10+5}{50}$     (M1)

Note:     Award (M1) for $11+10+5$ seen.

$=\frac{26}{50}\left(\frac{13}{25},0.52,52\mathrm{\%}\right)$     (A1)     (G2)

[2 marks]

$\frac{10}{\text{their }26}\times \frac{9}{49}$     (M1)(M1)

Note:     Award (M1) for $\frac{10}{\text{their }26}$ seen, (M1) for multiplying their first probability by $\frac{9}{49}$.

OR

$\frac{\frac{10}{50}\times \frac{9}{49}}{\frac{26}{50}}$

Note:     Award (M1) for $\frac{10}{50}\times \frac{9}{49}$ seen, (M1) for dividing their first probability by $\frac{\text{their }26}{50}$.

$=\frac{45}{637}\text{ (}0.0706,0.0706436\dots ,7.06436\dots \mathrm{\%})$     (A1)(ft)     (G3)

Note:     Follow through from part (d).

[3 marks]

$\text{P}(X⩾90)$     (M1)

OR

     (M1)

Note:     Award (M1) for a diagram showing the correct shaded region $(>0.5)$.

$0.773(0.773372\dots )0.773(0.773372\dots ,77.3372\dots \mathrm{\%})$     (A1)     (G2)

[2 marks]

$0.773372\dots \times 50$     (M1)

$=38.7(38.6686\dots )$     (A1)(ft)     (G2)

Note:     Follow through from part (f)(i).

[2 marks]

On graph paper, draw a scatter diagram for these data. Use a scale of 2 cm to represent 5 hours on the $x$-axis and 2 cm to represent 10 points on the $y$-axis.

(i)     $\overline{x}$, the mean number of hours spent on social media;

(ii)     $\overline{y}$, the mean number of IB Diploma points.

Plot the point $(\overline{x},\overline{y})$ on your scatter diagram and label this point M.

Write down the value of $r$, the Pearson’s product–moment correlation coefficient, for these data.

Write down the equation of the regression line $y$ on $x$ for these eight male students.

Draw the regression line, from part (e), on your scatter diagram.

Use the given equation of the regression line to estimate the number of IB Diploma points that this girl obtained.

Write down a reason why this estimate is not reliable.

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

     (A4)

Notes:     Award (A1) for correct scale and labelled axes.

Award (A3) for 7 or 8 points correctly plotted,

(A2) for 5 or 6 points correctly plotted,

(A1) for 3 or 4 points correctly plotted.

Award at most (A0)(A3) if axes reversed.

Accept $x$ and $y$ sufficient for labelling.

If graph paper is not used, award (A0).

If an inconsistent scale is used, award (A0). Candidates’ points should be read from this scale where possible and awarded accordingly.

A scale which is too small to be meaningful (ie mm instead of cm) earns (A0) for plotted points.

[4 marks]

(i)     $\overline{x}=21$     (A1)

(ii)    $\overline{y}=31$     (A1)

[2 marks]

$(\overline{x},\overline{y})$ correctly plotted on graph     (A1)(ft)

this point labelled M     (A1)

Note:     Follow through from parts (b)(i) and (b)(ii).

Only accept M for labelling.

[2 marks]

$-0.973(-0.973388\dots )$    (G2)

Note:     Award (G1) for 0.973, without minus sign.

[2 marks]

$y=-0.761x+47.0(y=-0.760638\dots x+46.9734\dots )$    (A1)(A1)(G2)

Notes:     Award (A1) for $-0.761x$ and (A1) $+47.0$. Award a maximum of (A1)(A0) if answer is not an equation.

[2 marks]

line on graph     (A1)(ft)(A1)(ft)

Notes:     Award (A1)(ft) for straight line that passes through their M, (A1)(ft) for line (extrapolated if necessary) that passes through $(0,47.0)$.

If M is not plotted or labelled, follow through from part (e).

[2 marks]

$y=-\frac{2}{3}(34)+\frac{125}{3}$    (M1)

Note:     Award (M1) for correct substitution.

19 (points)     (A1)(G2)

[2 marks]

extrapolation     (R1)

OR

34 hours is outside the given range of data     (R1)

Note:     Do not accept ‘outlier’.

[1 mark]

Draw a diagram that shows this information.

(i)     Find the probability that a box of cereal, chosen at random, is sold.

(ii)     Calculate the manufacturer’s expected daily income from these sales.

Calculate the manufacturer’s expected daily recycling cost.

Calculate the value of $w$.

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

(A1)(A1)

Notes:     Award (A1) for bell shape with mean of 502.

Award (A1) for an indication of standard deviation eg 500 and 504.

[2 marks]

(i)     $0.921(0.920968\dots ,92.0968\dots \mathrm{\%})$     (G2)

Note:     Award (M1) for a diagram showing the correct shaded region.

(ii)     $1500\times 2\times 0.920968\dots$     (M1)

$=(\mathrm{\$})2760(2762.90\dots )$    (A1)(ft)(G2)

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

[4 marks]

$1500\times 0.16\times 0.079031\dots$    (M1)

Notes:     Award (A1) for $1500\times 0.16\times \text{ their }(1-0.920968\dots )$.

OR

$(1500-1381.45)\times 0.16$    (M1)

Notes:     Award (M1) for $(1500-\text{their }1381.45)\times 0.16$.

$=(\mathrm{\$})19.0\text{ (}18.9676\dots )$    (A1)(ft)(G2)

[2 marks]

$347(\text{grams})(346.614\dots )$    (G3)

Notes:     Award (G2) for an answer that rounds to 346.

Award (G1) for $353.385\dots$ seen without working (for finding the top 3%).

[3 marks]

Find the median of the examination results.

Find the interquartile range.

Find the final examination result required to obtain the highest possible grade.

Write down the modal class.

Write down the mid-interval value of the modal class.

Calculate an estimate of the mean examination result.

Calculate an estimate of the standard deviation, giving your answer correct to three decimal places.

The teacher sets a grade boundary that is one standard deviation below the mean.

Use the cumulative frequency graph to estimate the number of students whose final examination result was below this grade boundary.

60     (A2)

[2 marks]

68 − 48     (A1)(M1)

Note: Award (A1) for two correct quartiles seen, (M1) for finding the difference between their two quartiles.

= 20      (A1)(ft)(G3)

[3 marks]

3200 − 350 = 2850      (M1)

Note: Award (M1) for 2850 seen. Follow through from their 3200.

(Top grade boundary =) 76        (A1)(ft)(G2)

[2 marks]

60 < x ≤ 80     (A1)(A1)

Note: Award (A1) for 60, 80 seen, (A1) for correct strict and weak inequalities.

[2 marks]

70     (A1)(ft)

Note: Follow through from part (c)(i).

[1 mark]

57.2   (57.1875)     (A2)(ft)

Note: Follow through from part (c)(ii).

[2 marks]

18.496     (A1)

Note: Award (A0) for 18.499.

[1 mark]

57.2 − 18.5      (M1)

= 38.7  (38.6918…)       (A1)(ft)

Note: Award (M1) for subtracting their standard deviation from their mean. Follow through from part (d) even if no working is shown.

450 (students)       (A1)(ft)(G2)

Note: Accept any answer within the range of 450 to 475, inclusive. Follow through from part (d), adjusting the acceptable range as necessary.

[3 marks]

Find the probability that both people chosen are not allergic to nuts.

Copy and complete the tree diagram.

Find the probability that this adult is allergic to nuts and the liquid turns blue.

Find the probability that the liquid turns blue.

Find the probability that the tested adult is allergic to nuts given that the liquid turned blue.

Estimate the number of employees, from this 38, who are allergic to nuts.

$\frac{34}{60}\times \frac{33}{59}$     (M1)

Note:    Award (M1) for their correct product.

$=0.317\left(\frac{187}{590},0.316949\dots ,31.7\mathrm{\%}\right)$     (A1)(ft)(G2)

Note:    Follow through from part (a).

[2 marks]

     (A1)(A1)(A1)

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

[3 marks]

$0.006\times 0.98$     (M1)

Note:     Award (M1) for multiplying 0.006 by 0.98.

$=0.00588\left(\frac{147}{25000},0.588\mathrm{\%}\right)$     (A1)(G2)

[2 marks]

$0.006\times 0.98+0.994\times 0.05(0.00588+0.994\times 0.05)$     (A1)(ft)(M1)

Note:     Award (A1)(ft) for their two correct products, (M1) for adding two products.

$=0.0556\left(0.05558,5.56\mathrm{\%},\frac{2779}{50000}\right)$     (A1)(ft)(G3)

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

[3 marks]

$\frac{0.006\times 0.98}{0.05558}$     (M1)(M1)

Note:     Award (M1) for their correct numerator, (M1) for their correct denominator.

$=0.106\left(0.105793\dots ,10.6\mathrm{\%},\frac{42}{397}\right)$     (A1)(ft)(G3)

Note:     Follow through from parts (d) and (e).

[3 marks]

$0.105793\dots \times 38$     (M1)

Note:     Award (M1) for multiplying 38 by their answer to part (f).

$=4.02(4.02015\dots )$     (A1)(ft)(G2)

Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).

[2 marks]

Write down the value of a.

Write down the value of b.

Use the tree diagram to find the probability that an employee encountered traffic and was late for work.

Use the tree diagram to find the probability that an employee was late for work.

Use the tree diagram to find the probability that an employee encountered traffic given that they were late for work.

Find the value of x.

Find the value of y.

Find the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.

Find $n\left(\left(C\cup B\right)\cap P^{\prime }\right)$.

a = 0.2     (A1)

[1 mark]

b = 0.85     (A1)

[1 mark]

0.25 × 0.8     (M1)

Note: Award (M1) for a correct product.

$=0.2\left(\frac{1}{5},20\mathrm{\%}\right)$     (A1)(G2)

[2 marks]

0.25 × 0.8 + 0.75 × 0.15     (A1)(ft)(M1)

Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.

$=0.313\left(0.3125,\frac{5}{16},31.3\mathrm{\%}\right)$    (A1)(ft)(G3)

Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).

[3 marks]

$\frac{0.25\times 0.8}{0.25\times 0.8+0.75\times 0.15}$    (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).