maximum value of $h$.

minimum value of $h$.

Find the time, in seconds, it takes for the blade $\text{[BC]}$ to make one complete rotation under these conditions.

Calculate the angle, in degrees, that the blade $\text{[BC]}$ turns through in one second.

Write down the amplitude of the function.

Find the period of the function.

Sketch the function $h\left(t\right)$ for $0\le t\le 5$, clearly labelling the coordinates of the maximum and minimum points.

Find the height of $\text{C}$ above the ground when $t=2$.

Find the time, in seconds, that point $\text{C}$ is above a height of $100 \text{m}$, during each complete rotation.

At any given instant, find the probability that point $\text{C}$ is visible from Tim’s window.

The wind speed increases. The blades rotate at twice the speed, but still at a constant rate.

At any given instant, find the probability that Tim can see point $\text{C}$ from his window. Justify your answer.

maximum $h=130$ metres             A1

[1 mark]

minimum $h=50$ metres             A1

[1 mark]

$\left(60÷12=\right) 5 \text{seconds}$             A1

[1 mark]

$360÷5$            (M1)

Note: Award (M1) for $360$ divided by their time for one revolution.

$=72°$             A1

[2 marks]

(amplitude =)  $40$         A1

[1 mark]

(period $=\frac{360}{72}=$)  $5$         A1

[1 mark]

Maximum point labelled with correct coordinates.         A1

At least one minimum point labelled. Coordinates seen for any minimum points must be correct.         A1

Correct shape with an attempt at symmetry and “concave up" evident as it approaches the minimum points. Graph must be drawn in the given domain.         A1

[3 marks]

$h=90-40 \cos \left(144°\right)$           (M1)

$\left(h=\right) 122 \text{m} \left(122.3606\dots \right)$           A1

[2 marks]

evidence of $h=100$ on graph  OR  $100=90-40 \cos \left(72t\right)$           (M1)

$t$ coordinates $3.55 (3.54892...)$  OR  $1.45 (1.45107...)$ or equivalent           (A1)

Note: Award A1 for either $t$-coordinate seen.

$=2.10$ seconds  $\left(2.09784\dots \right)$           A1

[3 marks]

$5-2.09784\dots$           (M1)

$\frac{\left(2.902153\dots \right)}{5}$           (M1)

$0.580 \left(0.580430\dots \right)$           A1

[3 marks]

METHOD 1

changing the frequency/dilation of the graph will not change the proportion of time that point $\text{C}$ is visible.         A1

$0.580 (0.580430...)$           A1

METHOD 2

correct calculation of relevant found values

$\frac{\left(2.902153\dots \right)/2}{5/2}$           A1

$0.580 (0.580430...)$           A1

Note: Award A0A1 for an unsupported correct probability.

[2 marks]

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

Judging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of $h\left(t\right)$ reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the "trace" feature in their GDC. Most were able to find the height of point $\mathrm{C}$ when $t=2$ and make an attempt to find a time at which point $\mathrm{C}$ is at a height of $100 \mathrm{m}$. It was pleasing to see a number of candidates draw $h=100$ on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.

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]

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 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]

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

State the sampling method being used.

$a$.

$b$.

$c$.

$d$.

will not have the disease and will test positive.

will test negative.

has the disease given that they tested negative.

The medical centre finds the actual number of positive results in their sample is different than predicted by the tree diagram. Explain why this might be the case.

Draw a Venn diagram to illustrate this information, placing all relevant information on the diagram.

Find the total number of patients who visited the centre during this day.

convenience sampling                 (A1)

[1 mark]

$95\%$                A1

[1 mark]

$1\%$                A1

[1 mark]

$2\%$                A1

[1 mark]

$98\%$                A1

[1 mark]

$0.95\times 0.02$               (M1)

$0.019$                A1

[2 marks]

$0.05\times 0.01+0.95\times 0.98$               (M1)(M1)

Note: Award M1 for summing two products and M1 for correct products seen.

$0.932 (0.9315)$                A1

[3 marks]

recognition of conditional probability             (M1)

$\frac{0.05\times 0.01}{0.05\times 0.01+0.95\times 0.98}$                A1

$0.000537 (0.000536768\dots )$                A1

Note: Accept $0.000536$ if $0.932$ used.

[3 marks]

EITHER
sample may not be representative of population           A1

OR
sample is not randomly selected           A1

OR
unrealistic to think expected and observed values will be exactly equal            A1

[1 mark]

                A1A1A1   

Note: Award A1 for rectangle and $3$ labelled circles and $9$ in centre region; A1 for $2, 40, 24$; A1 for $18, 1,$ and $11$.

[3 marks]

$18+9+1+11+2+40+24$              (M1)   

$105$                        A1

Note: Follow through from the entries on their Venn diagram in part (e). Working required for FT.

[2 marks]

State whether $N$ is a discrete or a continuous variable.

Write down, for $N$, the modal class;

Write down, for $N$, the mid-interval value of the modal class.

Use your graphic display calculator to estimate the mean of $N$;

Use your graphic display calculator to estimate the standard deviation of $N$.

Find the expected frequency of students choosing the Science category and obtaining 31 to 40 correct answers.

Write down the null hypothesis for this test;

Write down the number of degrees of freedom.

Write down the $p$-value for the test;

Write down the ${\chi }^2$ statistic.

State the result of the test. 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.

discrete     (A1)

[1 mark]

$11⩽N⩽20$     (A1)

[1 mark]

15.5     (A1)(ft)

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

[1 mark]

$21.2(21.2125)$     (G2)

[2 marks]

$9.60(9.60428\dots )$     (G1)

[1 marks]

$\frac{260}{800}\times \frac{157}{800}\times 800$$$OR$$$\frac{260\times 157}{800}$     (M1)

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

$=51.0(51.025)$     (A1)(G2)

[2 marks]

choice of category and number of correct answers are independent     (A1)

Notes:     Accept “no association” between (choice of) category and number of correct answers. Do not accept “not related” or “not correlated” or “influenced”.

[1 mark]

6     (A1)

[1 mark]

$0.0644(0.0644123\dots )$     (G1)

[1 mark]

$11.9(11.8924\dots )$     (G2)

[2 marks]

the null hypothesis is not rejected (the null hypothesis is accepted)     (A1)(ft)

OR

(choice of) category and number of correct answers are independent     (A1)(ft)

as $$OR$$$0.0644>0.05$     (R1)

Notes:     Award (R1) for a correct comparison of either their ${\chi }^2$ statistic to the ${\chi }^2$ critical value or their $p$-value to the significance level. Award (A1)(ft) from that comparison.

Follow through from part (f). Do not award (A1)(ft)(R0).

[2 marks]

Find the probability that it will take Fiona between $15$ minutes and $30$ minutes to walk to the bus stop.

Find $\sigma$.

Find the probability that the bus journey takes less than $45$ minutes.

Find the probability that Fiona will arrive on time.

This year, Fiona will go to school on $183$ days.

Calculate the number of days Fiona is expected to arrive on time.

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

$0.158655$

$\text{P}\left(15<W<30\right)=0.159$    A2   N2

[2 marks]

finding standardized value for $60$       (A1)

eg       $z=1.56322$

correct substitution using their $z$-value       (A1)

eg       $\frac{60-50}{\sigma }=1.56322, \frac{60-50}{1.56322}=\sigma$

$6.39703$

$\sigma =6.40$    A1   N3

[3 marks]

$0.217221$

$\text{P}\left(B<45\right)=0.217$    A2   N2

[2 marks]

valid attempt to find one possible way of being on time (do not penalize incorrect use of strict inequality signs)       (M1)

eg       $W\le 15$ and $B<60$, $15<W\le 30$ and $B<45$

correct calculation for $\text{P}\left(W\le 15 \text{and} B<60\right)$ (seen anywhere)       (A1)

eg       $0.841\times 0.941, 0.7917$

correct calculation for $\text{P}\left(15<W\le 30 \text{and} B<45\right)$ (seen anywhere)       (A1)

eg       $0.159\times 0.217, 0.03446$

correct working       (A1)

eg       $0.841\times 0.941+0.159\times 0.217, 0.7917+0.03446$

$0.826168$

$\text{P\hspace{0.05em}}$(on time) $=0.826$    A1   N2

[5 marks]

recognizing binomial with $n=183, p=0.826168$       (M1)

eg       $X~\text{B}\left(183, 0.826\right)$

$151.188$   ($151.158$ from $3 \text{sf\hspace{0.05em}}$)

$151$    A1   N2

[2 marks]

Find the probability that this person is not allergic to nuts.

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.

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

$\frac{34}{60}\left(\frac{17}{30},0.567,0.566666\dots ,56.7\mathrm{\%}\right)$     (A1)(A1)

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

[2 marks]

$\frac{34}{60}\times \frac{33}{59}$     (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 total number of people, from this group, who are pet owners.

Write down the modal number of pets.

For these data, write down the median number of pets.

For these data, write down the lower quartile.

For these data, write down the upper quartile.

Write down the ratio of teenagers to non-teenagers in its simplest form.

State the null hypothesis.

State the alternative hypothesis.

Write down the number of degrees of freedom for this test.

Calculate the expected number of teenagers that prefer cats.

State the conclusion for this test. 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.

140       (A1)

[1 mark]

1       (A1)

[1 mark]

2       (A1)

[1 mark]

1       (A1)

[1 mark]

3       (A1)

[1 mark]

17:15  OR  $\frac{17}{15}$      (A1)

Note: Award (A0) for 85:75 or 1.13:1.

[1 mark]

preferred pet is independent of “whether or not the respondent was a teenager" or "age category”     (A1)

Note: Accept there is no association between pet and age. Do not accept “not related” or “not correlated” or “influenced”.

[1 mark]

preferred pet is not independent of age    (A1)(ft)

Note: Follow through from part (e)(i) i.e. award (A1)(ft) if their alternative hypothesis is the negation of their null hypothesis. Accept “associated” or “dependent”.

[1 mark]

3    (A1)

[1 mark]

$\frac{85\times 55}{160}$  OR  $\frac{85}{160}\times \frac{55}{160}\times 160$     (M1)

29.2 (29.2187…)      (A1)(G2)

[2 marks]

0.208 > 0.1      (R1)

accept null hypothesis  OR  fail to reject null hypothesis      (A1)(ft)

Note: Award (R1) for a correct comparison of their $p$-value to the significance level, award (A1)(ft) for the correct result from that comparison. Accept “$p$-value > 0.1” as part of the comparison but only if their $p$-value is explicitly seen in part (h). Follow through from their answer to part (h). Do not award (R0)(A1).

[2 marks]

Write down the null hypothesis, H₀ , for this test.

State the number of degrees of freedom.

the expected frequency of female students who chose to take the Chinese class.

State whether or not H₀ should be rejected. Justify your statement.

Find the probability that the student does not take the Spanish class.

Find the probability that neither of the two students take the Spanish class.

Find the probability that at least one of the two students is female.

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

(H₀:) (choice of) language is independent of gender       (A1)

Note: Accept “there is no association between language (choice) and gender”. Accept “language (choice) is not dependent on gender”. Do not accept “not related” or “not correlated” or “not influenced”.

[1 mark]

2       (AG)

[1 mark]

16.4  (16.4181…)      (G1)

[1 mark]

(we) reject the null hypothesis      (A1)(ft)

8.68507… > 5.99     (R1)(ft)

Note: Follow through from part (c)(ii). Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).

OR

(we) reject the null hypothesis       (A1)

0.0130034 < 0.05       (R1)

Note: Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).

[2 marks]

$\frac{88}{110}\left(\frac{4}{5}\text{,}0.8\text{,}80\text{\% }\right)$   (A1)(A1)(G2)

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

[2 marks]

$\frac{88}{110}\times \frac{87}{109}$    (M1)(M1)

Note: Award (M1) for multiplying two fractions. Award (M1) for multiplying their correct fractions.

OR

$\left(\frac{46}{110}\right)\left(\frac{45}{109}\right)+2\left(\frac{46}{110}\right)\left(\frac{42}{109}\right)+\left(\frac{42}{110}\right)\left(\frac{41}{109}\right)$    (M1)(M1)

Note: Award (M1) for correct products; (M1) for adding 4 products.

$0.639\left(0.638532\dots \text{,}\frac{348}{545}\text{,}63.9\text{\% }\right)$       (A1)(ft)(G2)

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

[3 marks]

$1-\frac{67}{110}\times \frac{66}{109}$   (M1)(M1)

Note: Award (M1) for multiplying two correct fractions. Award (M1) for subtracting their product of two fractions from 1.

OR

$\frac{43}{110}\times \frac{42}{109}+\frac{43}{110}\times \frac{67}{109}+\frac{67}{110}\times \frac{43}{109}$   (M1)(M1)

Note: Award (M1) for correct products; (M1) for adding three products.

$0.631\left(0.631192\dots \text{,}63.1\text{\% ,}\frac{344}{545}\right)$      (A1)(G2)

[3 marks]

Find the probability of rolling exactly one red face.

Find the probability of rolling two or more red faces.

Show that, after a turn, the probability that Ted adds exactly $10 to his winnings is $\frac{1}{3}$.

Write down the value of $x$.

Hence, find the value of $y$.

Ted will always have another turn if he expects an increase to his winnings.

Find the least value of $w$ for which Ted should end the game instead of having another turn.

valid approach to find P(one red)     (M1)

eg  ${}_n{C}_a\times p^a\times q^{n-a}$,  $\text{B}\left(n\text{, }p\right)$,  $3\left(\frac{1}{3}\right){\left(\frac{2}{3}\right)}^2$,  $\left(\begin{array}{c}3\\ 1\end{array}\right)$

listing all possible cases for exactly one red (may be indicated on tree diagram)

P(1 red) = 0.444 $\left(=\frac{4}{9}\right)$   [0.444, 0.445]           A1  N2

 [3 marks] [5 maximum for parts (a.i) and (a.ii)]

valid approach     (M1)

eg  P($X=2$) + P($X=3$), 1 − P($X$ ≤ 1),  binomcdf$\left(3\text{,}\frac{1}{3}\text{,}2\text{,}3\right)$

correct working       (A1)

eg   $\frac{2}{9}+\frac{1}{27}$,   0.222 + 0.037 ,  $1-{\left(\frac{2}{3}\right)}^3-\frac{4}{9}$

0.259259

P(at least two red) = 0.259 $\left(=\frac{7}{27}\right)$          A1  N3

[3 marks]  [5 maximum for parts (a.i) and (a.ii)]

recognition that winning $10 means rolling exactly one green        (M1)

recognition that winning $10 also means rolling at most 1 red        (M1)

eg “cannot have 2 or more reds”

correct approach        A1

eg  P(1G ∩ 0R) + P(1G ∩ 1R),  P(1G) − P(1G ∩ 2R),

      “one green and two yellows or one of each colour”

Note: Because this is a “show that” question, do not award this A1 for purely numerical expressions.

one correct probability for their approach        (A1)

eg   $3\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^2$,  $\frac{6}{27}$, $3\left(\frac{1}{3}\right){\left(\frac{2}{3}\right)}^2$,  $\frac{1}{9}$,  $\frac{2}{9}$

correct working leading to $\frac{1}{3}$      A1

eg   $\frac{3}{27}+\frac{6}{27}$,  $\frac{12}{27}-\frac{3}{27}$,  $\frac{1}{9}+\frac{2}{9}$

probability = $\frac{1}{3}$      AG N0

[5 marks]

$x=\frac{7}{27}$,  0.259 (check FT from (a)(ii))      A1 N1

[1 mark]

evidence of summing probabilities to 1       (M1)

eg   $\sum =1$,  $x+y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}=1$,  $1-\frac{7}{27}-\frac{9}{27}-\frac{6}{27}-\frac{1}{27}$

0.148147  (0.148407 if working with their $x$ value to 3 sf)

$y=\frac{4}{27}$  (exact), 0.148     A1 N2

[2 marks]

correct substitution into the formula for expected value      (A1)

eg   $-w\cdot \frac{7}{27}+10\cdot \frac{9}{27}+20\cdot \frac{6}{27}+30\cdot \frac{1}{27}$

correct critical value (accept inequality)       A1

eg   $w$ = 34.2857  $\left(=\frac{240}{7}\right)$,  $w$ > 34.2857

$40      A1 N2

[3 marks]

On graph paper, draw a scatter diagram to show the results of Don’s investigation. Use a scale of $1 \text{cm}$ to represent $2$ units on the $x$-axis, and $1 \text{cm}$ to represent $5$ units on the $y$-axis.

Calculate $\overline{x}$, the mean wind speed.

Calculate $\overline{y}$, the mean time to fully charge the robot.

Plot and label the point $\text{M}$ on your scatter diagram.

Calculate $r$, Pearson’s product–moment correlation coefficient.

Describe the correlation between the wind speed and the time to fully charge the robot.

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

Draw this regression line on your scatter diagram.

Hence or otherwise estimate the charging time when the wind speed is $27 {\text{km\hspace{0.05em}h}}^{-1}$.

Don concluded from his investigation: “There is no causation between wind speed and the time to fully charge the robot”.

In the context of the question, briefly explain the meaning of “no causation”.

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

       (A4)

Note: Award (A1) for correct scales and labels.
Award (A3) for all six points correctly plotted.
Award (A2) for four or five points correctly plotted.
Award (A1) for two or three points correctly plotted.
Award at most (A0)(A3) if axes reversed.
If graph paper is not used, award at most (A1)(A0)(A0)(A0).

[4 marks]

$19 \left({\text{km\hspace{0.05em}h}}^{-1}\right)$       (A1)

[1 mark]

$32$  (minutes)      (A1)

[1 mark]

point in correct position, labelled $\text{M}$     (A1)(ft)(A1)

Note: Award (A1)(ft) for point plotted in correct position, (A1) for point labelled $\text{M}$ Follow through from their part (b).

[2 marks]

$\left(r=\right) 0.944 \left(0.943733\dots \right)$     (G2)

Note: Award (G1) for $0.943$ (incorrect rounding).

[2 marks]

(very) strong positive correlation   (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for (very) strong. Award (A1)(ft) for positive. Follow though from their part (d)(i). If there is no answer to part (d)(i), award at most (A0)(A1) for a correct direction.

[2 marks]

$y=0.465x+23.2 \left(y=0.465020\dots x+23.1646\dots \right)$   (A1)(A1)(G2)

Note: Award (A1) for $0.465x$. Award (A1) for $23.2$. If the answer is not an equation, award at most (A1)(A0).

[2 marks]

regression line through their $\text{M}$        (A1)(ft)

regression line through their $(0, 23.2)$        (A1)(ft)

Note: Award a maximum of (A1)(A0) if the line is not straight/ruler not used. Award (A0)(A0) if the points are connected.
Follow through from their point $\text{M}$ in part (b) and their $y$-intercept in part (e)(i).
If $\text{M}$ is not plotted or labelled, then follow through from part (b).

[2 marks]

$\left(y=\right) 0.465020\dots \left(27\right)+23.1646\dots$        (M1)

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

$35.7$ (minutes) $\left(35.7201\dots \right)$        (A1)(ft)(G2)

Note: Follow through from their equation in part (e)(i).

OR

an attempt to use their regression line to find the $y$ value at $x=27$

Note: Award (M1) for an indication of using their regression line. This must be illustrated by vertical and horizontal lines or marks at the correct place(s) on their scatter diagram.

$35.7$ (minutes)        (A1)(ft)

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

[2 marks]

wind speed does not cause a change in the time to charge (the robot)      (A1)

Note: Award (A1) for a statement that communicates the meaning of a non-causal relationship between the two variables.

[1 mark]

Sketch a diagram showing the above information.

Find the proportion of male Persian cats weighing between $5.5 \text{kg}$ and $6.5 \text{kg}$.

Determine the expected number of cats in this group that have a weight of less than $5.3 \text{kg}$.

It is found that $12$ of the cats weigh more than $x \text{kg}$. Estimate the value of $x$.

Ten of the cats are chosen at random. Find the probability that exactly one of them weighs over $6.25 \text{kg}$.

                A1A1

Note: Award A1 for a normal curve with mean labelled $6.1$ or $\mu$, A1 for indication of SD $(0.5):$ marks on horizontal axis at $5.6$ and/or $6.6$ OR $\mu -0.5$ and/or $\mu +0.5$ on the correct side and approximately correct position.

[2 marks]

$X~\text{N}\left(6.1, 0.5^2\right)$

$\text{P}\left(5.5<X<6.5\right)$  OR  labelled sketch of region                (M1)

$=0.673 \left(0.673074\dots \right)$                A1

[2 marks]

$\left(\text{P}\left(X<5.3\right)=\right) 0.0547992\dots$                (A1)

$0.0547992\dots \times 80$                (M1)

$=4.38 \left(4.38393\dots \right)$                A1

[3 marks]

$0.15$  OR  $0.85$               (A1)

$\text{P}\left(X>x\right)=0.15$  OR  $\text{P}\left(X<x\right)=0.85$  OR  labelled sketch of region                (M1)

$6.62 \left(6.61821\dots \right)$                A1

[3 marks]

$\left(\text{P}\left(X>6.25\right)=\right) 0.382088\dots$               (A1)

recognition of binomial                (M1)

e.g. $\text{B}(10, 0.382088\dots )$

$0.0502 \left(0.0501768\dots \right)$                A2

[4 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).

$=0.64\left(\frac{16}{25},64\text{\% }\right)$     (A1)(ft)(G3)

Note: Award final (A1)(ft) only if answer does not exceed 1.

[3 marks]

(x =) 3     (A1)

[1 Mark]

(y =) 10     (A1)(ft)

Note: Following through from part (c)(i) but only if their x is less than or equal to 13.

[1 Mark]

54 − (10 + 3 + 4 + 2 + 6 + 8 + 13)     (M1)

Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).

= 8      (A1)(ft)(G2)

Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).

[2 marks]

6 + 8 + 13     (M1)

Note: Award (M1) for summing 6, 8 and 13.

27     (A1)(G2)

[2 marks]

If the teacher chooses a response at random, find the probability that it is a response to the Calculus question;

If the teacher chooses a response at random, find the probability that it is a satisfactory response to the Calculus question;

If the teacher chooses a response at random, find the probability that it is a satisfactory response, given that it is a response to the Calculus question.

The teacher groups the responses by topic, and chooses two responses to the Logic question. Find the probability that both are not satisfactory.

State the null hypothesis for this test.

Show that the expected frequency of satisfactory Calculus responses is 12.

Write down the number of degrees of freedom for this test.

Use your graphic display calculator to find the ${\chi }^2$ statistic for this data.

State the conclusion of this ${\chi }^2$ test. Give a reason for your answer.

$\frac{1}{5}\left(\frac{18}{90};0.2;20\mathrm{\%}\right)$     (A1)(A1)(G2)

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

[2 marks]

$\frac{1}{9}\left(\frac{10}{90};0\text{.}\overline{1};0.111111\dots ;11.1\mathrm{\%}\right)$     (A1)(A1)(G2)

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

[2 marks]

$\frac{5}{9}\left(\frac{10}{18};0.\overline{5};0.555556\dots ;55.6\mathrm{\%}\right)$     (A1)(A1)(G2)

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

[2 marks]

$\frac{6}{20}\times \frac{5}{19}$     (A1)(M1)

Note:     Award (A1) for two correct fractions seen, (M1) for multiplying their two fractions.

$\frac{3}{38}\left(\frac{30}{380};0.0789473\dots ;7.89\mathrm{\%}\right)$     (A1)(G2)

[3 marks]

${\text{H}}_0$: quality (of response) and topic (from the syllabus) are independent     (A1)

Note:     Accept there is no association between quality (of response) and topic (from the syllabus). Do not accept “not related” or “not correlated” or “influenced”.

[1 mark]

$\frac{18}{90}\times \frac{60}{90}\times 90$$$OR$$$\frac{18\times 60}{90}$     (M1)

Note:     Award (M1) for correct substitution in expected value formula.

$(=)12$     (AG)

Note:     The conclusion, $(=)12$, must be seen for the (A1) to be awarded.

[1 mark]

3     (A1)

[1 mark]

$({\chi }_{calc}^2=)1.46(1.46\overline{36};1.46363\dots )$     (G2)

[2 marks]

$$OR$$$0.690688\dots >0.05$     (R1)

the null hypothesis is not rejected     (A1)(ft)

OR

the quality of the response and the topic are independent     (A1)(ft)

Note:     Award (R1) for a correct comparison of either their ${\chi }^2$ statistic to the ${\chi }^2$ critical value or the correct $p$-value 0.690688… to the test level, award (A1)(ft) for the correct result from that comparison. Accept “” for the comparison, but only if their ${\chi }_{\text{calc}}^2$ value is explicitly seen in part (f). Follow through from their answers to part (f) and part (c). Do not award (R0)(A1).

[2 marks]

median reaction time.

interquartile range of the reaction times.

Find the estimated number of teenagers who have a reaction time greater than $0.4$ seconds.

Determine the $90\text{th}$ percentile of the reaction times from the cumulative frequency graph.

Write down the value of $a$.

Write down the value of $b$.

Write down the modal class from the table.

Use your graphic display calculator to find an estimate of the mean reaction time.

Suggest how, if at all, the estimated mean and estimated median reaction times will change if the errors are corrected. Justify your response.

$0.58 \left(\text{s}\right)$          A1

[1 mark]

$0.7-0.42$           (A1)(M1)

Note: Award A1 for correct quartiles seen, M1 for subtraction of their quartiles.

$0.28 \left(\text{s}\right)$          A1

[3 marks]

$9$ (people have reaction time $\le 0.4$)           (A1)

$31$ (people have reaction time $>0.4$)          A1

[2 marks]

$\left(90\%\times 40=\right) 36$   OR   $4$           (A1)

$0.8 s$          A1

[2 marks]

$\left(a=\right) 6$         A1

[1 mark]

$\left(b=\right) 4$         A1

[1 mark]

$0.6<t\le 0.8$         A1

[1 mark]

$0.55 \text{s}$         A2

[2 marks]

the mean will increase         A1

because the incorrect reaction times are moving from a lower interval to a higher interval which will increase the numerator of the mean calculation         R1

the median will stay the same         A1

because the median or middle of the data is greater than both intervals being changed         R1

Note: Do not award A1R0.

[4 marks]

Most candidates were able to determine the median and interquartile range from the given graph. Some lost marks due to use of one significant figure values or because of incorrectly reading the quartiles as 0.75 and 0.25. Candidates were also able to find the estimated number of teenagers with reaction time greater than 0.4s in part (b), but determining the 90th percentile in part (c) proved to be more challenging. Most made a good attempt at completing the frequency table in part (d), but some used cumulative values from the graph incorrectly. Candidates who lost marks in part (d), were able to get “follow through” marks in parts (e) and (f). In part (e), most candidates were able to determine the modal class correctly. Not all candidates used the correct formula to find an estimate for the mean. Candidates who used their calculators usually obtained the correct answer. In part (g), few candidates were able to produce correct statements related to the changes of the mean and the median, and even fewer were able to support these statements with well-articulated reasons.

State the name for this type of sampling technique.

Calculate the number of volunteers in the sample under the age of 30.

The new drug.

The current drug.

State an assumption that the company is making, in order to use a t-test.

State the hypotheses for this t-test.

Find the p-value for this t-test.

State the conclusion of this test, in context, giving a reason.

stratified sampling        A1

[1 mark]

$0.2\times 18=3.6$       M1A1

so 4 volunteers need to be chosen       A1

[3 marks]

34.8 mg/dL      A1

[1 mark]

24.7 mg/dL      A1

[1 mark]

EITHER

The decreases in cholesterol are distributed normally    A1

OR

The variance of the two groups of volunteers is equal.    A1

[1 mark]

$H_0\text{:}\overline{N}=\overline{C}$ and $H_1\text{:}\overline{N}>\overline{C}$         A1

where N and C represent the decreases of the new and current drug

[1 mark]

df = 16, t = 2.77        (M1)

p-value = 0.00683        A2

[3 marks]

Since 0.00683 < 0.01        R1

Reject H₀. There is evidence, at the 1% level, that the new drug is more effective.       A1

[2 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 $w$.

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]

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]

Find P(A ∩ B′ ).

Given that P((A ∪ B)′ ) = 0.19, find P(A | B′ ).

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

valid approach

eg  Venn diagram, P(A) − P (A ∩ B), 0.62 − 0.18      (M1)

P(A ∩ B' ) = 0.44      A1 N2

[2 marks]

valid approach to find either P(B′ ) or P(B)      (M1)

eg   (seen anywhere), 1 − P(A ∩ B′ ) − P((A ∪ B)′ )

correct calculation for P(B′ ) or P(B)      (A1)

eg  0.44 + 0.19, 0.81 − 0.62 + 0.18

correct substitution into $\frac{\text{P}\left(A\cap B^{\prime }\right)}{\text{P}\left(B^{\prime }\right)}$      (A1)

eg  $\frac{0.44}{0.19+0.44},\frac{0.44}{1-0.37}$

0.698412

P(A | B′ ) = $\frac{44}{63}$  (exact), 0.698     A1 N3

[4 marks]

Find the probability that a basketball player has a weight that is less than 61 kg.

In a training session there are 40 basketball players.

Find the expected number of players with a weight less than 61 kg in this training session.

Sketch a normal curve to represent this probability.

Find the value of q.

Given that P(W > k) = 0.225 , find the value of k.

For this test state the null hypothesis.

For this test find the p-value.

State a conclusion for this test. Justify your answer.

P(W < 61)    (M1)

Note: Award (M1) for correct probability statement.

OR

 (M1)

Note: Award (M1) for correct region labelled and shaded on diagram.

= 0.212 (0.21185…, 21.2%)     (A1)(G2)

[2 marks]

40 × 0.21185…     (M1)

Note: Award (M1) for product of 40 and their 0.212.

= 8.47 (8.47421...)     (A1)(ft)(G2)

Note: Follow through from their part (a)(i) provided their answer to part (a)(i) is less than 1.

[2 marks]

    (A1)(M1)

Note: Award (A1) for two correctly labelled vertical lines in approximately correct positions. The values 57.5 and 72.5, or μ − 1.5σ and μ + 1.5σ are acceptable labels. Award (M1) for correctly shaded region marked by their two vertical lines.

[2 marks]

0.866 (0.86638…, 86.6%)      (A1)(ft)

Note: Follow through from their part (b)(i) shaded region if their values are clear.

[1 mark]

P(W < k) = 0.775     (M1)

OR

  (M1)

Note: Award (A1) for correct region labelled and shaded on diagram.

(k =) 68.8  (68.7770…)     (A1)(G2)

[2 marks]

(H₀:) performance (of players) and (their) weight are independent.     (A1)

Note: Accept “there is no association between performance (of players) and (their) weight”. Do not accept "not related" or "not correlated" or "not influenced".

[1 mark]

0.287  (0.287436…)     (G2)

[2 marks]

accept/ do not reject null hypothesis/H₀     (A1)(ft)

OR

performance (of players) and (their) weight are independent. (A1)(ft)

0.287 > 0.05     (R1)(ft)

Note: Accept p-value>significance level provided their p-value is seen in b(ii). Accept 28.7% > 5%. Do not award (A1)(R0). Follow through from part (d).

[2 marks]

State the alternative hypothesis.

Calculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.

Write down the number of degrees of freedom.

Write down the χ² statistic.

Write down the associated p-value.

State, with a reason, whether you would reject the null hypothesis.

Write down the probability that this flight arrived on time.

Given that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.

Two flights are chosen at random from those which were slightly delayed.

Find the probability that each of these flights travelled at least 5000 km.

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

The arrival status is dependent on the distance travelled by the incoming flight     (A1)

Note: Accept “associated” or “not independent”.

[1 mark]

$\frac{60\times 45}{180}$  OR  $\frac{60}{180}\times \frac{45}{180}\times 180$     (M1)

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

= 15     (A1) (G2)

[2 marks]

4     (A1)

Note: Award (A0) if “2 + 2 = 4” is seen.

[1 mark]

9.55 (9.54671…)    (G2)

Note: Award (G1) for an answer of 9.54.

[2 marks]

0.0488 (0.0487961…)     (G1)

[1 mark]

Reject the Null Hypothesis     (A1)(ft)

Note: Follow through from their hypothesis in part (a).

9.55 (9.54671…) > 7.779     (R1)(ft)

OR

0.0488 (0.0487961…) < 0.1     (R1)(ft)

Note: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ²_calc > χ²_crit , provided the calculated value is explicitly seen in part (d)(i).

[2 marks]

$\frac{52}{180}\left(0.289,\frac{13}{45},28.9\text{\% }\right)$     (A1)(A1) (G2)

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

[2 marks]

$\frac{35}{97}\left(0.361,36.1\text{\% }\right)$     (A1)(A1) (G2)

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

[2 marks]

$\frac{14}{45}\times \frac{13}{44}$     (A1)(M1)

Note: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.

$=\frac{182}{1980}\left(0.0919,\frac{91}{990},0.091919\dots ,9.19\text{\% }\right)$     (A1) (G2)

[3 marks]