Find the equation of the line passing through these two points.

Find the equation of the least squares regression quadratic curve for these four points.

By considering the gradient of this curve when $x=4$, explain why it may not be a good model.

Find the equation of the new model.

Write down an expression for her estimate of the volume as a sum of two integrals.

Find the value of Charlotte’s estimate.

$y=\frac{5}{8}x+\frac{7}{2} \left(y=0.625x+3.5\right)$                  A1A1

Note: Award A1 for $0.625x$, A1 for $3.5$.
Award a maximum of A0A1 if not part of an equation.

[2 marks]

$y=-0.975x^2+9.56x-16.7$                  (M1)A1

$\left(y=-0.974630x^2+9.55919x-16.6569\dots \right)$

[2 marks]

gradient of curve is positive at $x=4$                 R1

Note: Accept a sensible rationale that refers to the gradient.

[1 mark]

METHOD 1

let $y=a{x}^2+bx+c$

differentiating or using $x=\frac{-b}{2a}$                       (M1)

$8a+b=0$

substituting in the coordinates
$7.5^{2}a+7.5b+c=0$                       (A1)
$4^{2}a+4b+c=6$                       (A1)

solve to get
$y=-\frac{24}{49}{x}^2+\frac{192}{49}x-\frac{90}{49}$  OR  $y=-0.490x^2+3.92x-1.84$                       A1

Note: Use of quadratic regression with points using the symmetry of the graph is a valid method.

METHOD 2

$y=a{\left(x-4\right)}^2+6$                       (M1)

$0=a{\left(7.5-4\right)}^2+6$                       (M1)

$a=-\frac{24}{49}$                       (A1)

$y=-\frac{24}{49}{\left(x-4\right)}^2+6$  OR  $y=-0.490{\left(x-4\right)}^2+6$                       A1

[4 marks]

$\mathrm{\pi }{\int }_0^4{\left(\frac{5}{8}x+3.5\right)}^{2}dx+\mathrm{\pi }{\int }_4^{7.5}{\left(-\frac{24}{49}{\left(x-4\right)}^2+6\right)}^{2}dx$                       (M1)(M1) (M1)A1

Note: Award (M1)(M1)(M1)A0 if $\mathrm{\pi }$ is omitted but response is otherwise correct. Award (M1) for an integral that indicates volume, (M1) for their part (a) within their volume integral, (M1) for their part (b)(i) within their volume integral, A1 for their correct two integrals with all correct limits.

[4 marks]

$501 {\text{cm}}^3 \left(501.189\dots \right)$                      A1

[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 $5.3 \text{kg}$.

Find the probability that exactly one of them weighs over $4.62 \text{kg}$.

Let $N$ be the number of cats weighing over $4.62 \text{kg}$.

Find the variance of $N$.

A cat is selected at random from all $160$ cats.

Find the probability that the cat was female, given that its weight was over $4.7 \text{kg}$.

                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]

$Y~\text{N}\left(4.5, 0.{45}^2\right)$,

$\left(\text{P}\left(Y>4.62\right)=\right) 0.394862\dots$                (A1)

use of binomial seen or implied                (M1)

using $\text{B}\left(10, 0.394862\dots \right)$                (M1)

$0.0430 \left(0.0429664\dots \right)$                A1

[4 marks]

$np\left(1-p\right)=2.39 \left(2.38946\dots \right)$                A1

[1 mark]

$\text{P}\left(F\cap \left(W>4.7\right)\right)=0.5\times 0.3284 \left(=0.1642\right)$                    (A1)

attempt use of tree diagram OR use of $\text{P}\left(F \overline{)W>4.7}\right)=\frac{\text{P}\left(F\cap \left(W>4.7\right)\right)}{\text{P}\left(W>4.7\right)}$                    (M1)

$\frac{0.5\times 0.3284}{0.5\times 0.9974+0.5\times 0.3284}$                    (A1)

$=0.248 \left(0.247669\dots \right)$                A1

[4 marks]

Write down a transition matrix T representing the movements between the two companies in a particular year.

Find the eigenvalues and corresponding eigenvectors of T.

Hence write down matrices P and D such that T = PDP⁻¹.

Find an expression for the number of customers company X has after $n$ years, where $n\in \mathbb{N}$.

Hence write down the number of customers that company X can expect to have in the long term.

      M1A1

[2 marks]

      M1

$\lambda =1$ and 0.7      A1

eigenvectors $\left(\begin{array}{c}1\\ 2\end{array}\right)$ and $\left(\begin{array}{c}1\\ -1\end{array}\right)$     (M1)A1

Note: Accept any scalar multiple of the eigenvectors.

[4 marks]

EITHER

P =   D =        A1A1

OR

P =   D =       A1A1

[2 marks]

P⁻¹ =        A1

       M1A1

attempt to multiply matrices         M1

so in company A, after $n$ years, $400\left(2+{0.7}^n\right)$         A1

[5 marks]

400 × 2 = 800        A1

[1 mark]

Write down a transition matrix $\mathit{T}$ indicating the annual population movement between clinics.

Find a prediction for the ratio of the number of patients Doctor Black will have, compared to Doctor Green, after two years.

Find a matrix $\mathit{P}$, with integer elements, such that $\mathit{T}=\mathit{P}\mathit{D}\mathbf{ }{\mathit{P}}^{-1}$, where $\mathit{D}$ is a diagonal matrix.

Hence, show that the long-term transition matrix ${\mathit{T}}^{\infty }$ is given by ${\mathit{T}}^{\infty }=\left(\begin{array}{cc}\frac{10}{17}& \frac{10}{17}\\ \frac{7}{17}& \frac{7}{17}\end{array}\right)$.

Hence, or otherwise, determine the expected ratio of the number of patients Doctor Black would have compared to Doctor Green in the long term.

$\mathit{T}=\left(\begin{array}{cc}0.965& 0.05\\ 0.035& 0.95\end{array}\right)$        M1A1

Note: Award M1A1 for $\mathit{T}=\left(\begin{array}{cc}0.95& 0.035\\ 0.05& 0.965\end{array}\right)$.
Award the A1 for a transposed $\mathit{T}$ if used correctly in part (b) i.e. preceded by $1\times 2$ matrix $\left(2100 3500\right)$ rather than followed by a $2\times 1$ matrix.

[2 marks]

${\left(\begin{array}{cc}0.965& 0.05\\ 0.035& 0.95\end{array}\right)}^2\left(\begin{array}{c}2100\\ 3500\end{array}\right)$        (M1)

$=\left(\begin{array}{c}2294\\ 3306\end{array}\right)$

so ratio is $2294:3306 \left(=1147:1653, 0.693889\dots \right)$        A1

[2 marks]

to solve $\mathit{A}x=\lambda x :$

$\left|\begin{array}{cc}0.965-\lambda & 0.05\\ 0.035& 0.95-\lambda \end{array}\right|=0$        (M1)

$\left(0.965-\lambda \right)\left(0.95-\lambda \right)-0.05\times 0.035=0$

$\lambda =0.915$  OR  $\lambda =1$        (A1)

attempt to find eigenvectors for at least one eigenvalue        (M1)

when $\lambda =0.915, x=\left(\begin{array}{c}1\\ -1\end{array}\right)$ (or any real multiple)        (A1)

when $\lambda =1, x=\left(\begin{array}{c}10\\ 7\end{array}\right)$ (or any real multiple)        (A1)

therefore $\mathit{P}=\left(\begin{array}{cc}1& 10\\ -1& 7\end{array}\right)$ (accept integer valued multiples of their eigenvectors and columns in either order)        A1

[6 marks]

${\mathit{P}}^{-1}={\left(\begin{array}{cc}1& 10\\ -1& 7\end{array}\right)}^{-1}=\frac{1}{17}\left(\begin{array}{cc}7& -10\\ 1& 1\end{array}\right)$            (A1)

Note: This mark is independent, and may be seen anywhere in part (d).

$\mathit{D}=\left(\begin{array}{cc}0.915& 0\\ 0& 1\end{array}\right)$            (A1)

${\mathit{T}}^n=\mathit{P}{\mathit{D}}^n\mathbf{ }{\mathit{P}}^{-1}=\left(\begin{array}{cc}1& 10\\ -1& 7\end{array}\right)\left(\begin{array}{cc}0.{915}^n& 0\\ 0& 1^n\end{array}\right)\frac{1}{17}\left(\begin{array}{cc}7& -10\\ 1& 1\end{array}\right)$            (M1)A1

Note: Award (M1)A0 for finding $\mathbf{ }{\mathit{P}}^{-1}{\mathit{D}}^n\mathit{P}$ correctly.

as $n\to \infty , {\mathit{D}}^n=\left(\begin{array}{cc}0.{915}^n& 0\\ 0& 1^n\end{array}\right)\to \left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$                  R1

so ${\mathit{T}}^n\to \frac{1}{17}\left(\begin{array}{cc}1& 10\\ -1& 7\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}7& -10\\ 1& 1\end{array}\right)$                 A1

$=\left(\begin{array}{cc}\frac{10}{17}& \frac{10}{17}\\ \frac{7}{17}& \frac{7}{17}\end{array}\right)$                   AG

Note: The AG line must be seen for the final A1 to be awarded.

[6 marks]

METHOD ONE

$\left(\begin{array}{cc}\frac{10}{17}& \frac{10}{17}\\ \frac{7}{17}& \frac{7}{17}\end{array}\right)\left(\begin{array}{c}2100\\ 3500\end{array}\right)=\left(\begin{array}{c}3294\\ 2306\end{array}\right)$            (M1)

so ratio is  $3294:2306 \left(1647:1153, 1.42844\dots , 0.700060\dots \right)$                A1

METHOD TWO

long term ratio is the eigenvector associated with the largest eigenvalue            (M1)

$10:7$                A1

[2 marks]

Find the probability that a melon selected at random will have a mass greater than $3.0 \text{kg}$.

both have a mass greater than $3.0 \text{kg}$.

have a total mass greater than $6.0 \text{kg}$.

Write down the null and alternative hypotheses for the test.

Find the critical region for this test.

Find the mean and standard deviation of the mass of the melons for this year.

Find the probability of a Type II error in the owner’s test.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

Let $X$ represent the mass of a melon

$\text{P}\left(X>3.0\right)=0.212 \left(0.2118\dots \right)$       (M1)A1

[2 marks]

$0.2118\dots \times 0.2118\dots$       (M1)

$=0.0449 \left(0.04488\dots \right)$         A1

[2 marks]

Let $T$ represent the total mass

$\text{E}\left(T\right)=5.2$         A1

$\text{Var}\left(T\right)=0.5^2+0.5^2=0.5$       (M1)A1

$T~\text{N}\left(5.2, 0.5\right)$

$\text{P}\left(T>6.0\right)=0.129 \left(0.1289\dots \right)$         A1

[4 marks]

Let $\mu$ be the mean mass of the melons produced by the farm.

${\text{H}}_0: \mu =2.6 \text{kg}$, ${\text{H}}_1: \mu >2.6 \text{kg}$ only         A1

Note: Accept ${\text{H}}_0:$ The mean mass of melons produced by the farm is equal to $2.6 \text{kg}$
${\text{H}}_1:$The mean mass of melons produced by the farm is greater than $2.6 \text{kg}$

Note: Award A0 if $2.6 \text{kg}$ does not appear in the hypothesis.

[1 mark]

Under ${\text{H}}_0$  $\overline{X}~\text{N}\left(2.6, \frac{0.5^2}{16}\right)$         A1

$\text{P}\left(\overline{X}>a\right)=0.05$          (M1)

$a=2.81 \left(2.805606\dots \right)$          (A1)

Critical region is $\overline{X}>2.81$         A1

[4 marks]

Let $W$ represent the new mass of the melons

$\text{E}\left(W\right)=1.1\times 2.6=2.86$         A1

Standard deviation of $W=1.1\times 0.5$          (M1)

$=0.55$         A1

Note: award M1A0 for $\text{Var}\left(W\right)=1.1^2\times 0.5^2=0.3025$

[3 marks]

$\mathrm{P}(\mathrm{Type} \mathrm{II} \mathrm{error})=$

$\text{P}\left(\overline{X}<2.81 \overline{) \mu =2.86, \sigma =\frac{0.55}{4}}\right)$          (M1)

$=0.346 \left(0.346204\dots \right)$         A1

Note: Accept $0.358$ from use of the three‐figure answer to part (d)

[2 marks]

State suitable hypotheses for testing this belief.

Calculate the mean of these data and hence estimate the value of $p$.

Calculate an appropriate value of ${\chi }^2$ and state your conclusion, using a 1% significance level.

Find the significance level of this procedure.

Some time later, the actual value of $\mu$ is 503. Find the probability of a Type II error.

H₀ : The data are B (4, $p$); H₁ : The data are not B (4, $p$)       A1

[1 mark]

Mean $=\frac{1\times 32+\dots +4\times 14}{100}$       M1A1

= 1.8       A1

$4\hat{p}=1.8\Rightarrow \hat{p}=0.45$       M1A1

[5 marks]

The expected frequencies are

   A1A1A1A1A1

The last two classes must be combined because the expected frequency for $x$ = 4 is less than 5.       R1

${\chi }^2=\frac{{17}^2}{9.15}+\frac{{32}^2}{29.95}+\frac{{19}^2}{36.75}+\frac{{32}^2}{24.15}-100$       M2

= 18.0        A2

DF = 2        (A1)

Critical value = 9.21        A1         

We conclude, at the 1% significance level, that X does not fit a binomial model.           R1         

Special case: award the following marks to candidates who do not combine classes.

${\chi }^2=\frac{{17}^2}{9.15}+\frac{{32}^2}{29.95}+\frac{{19}^2}{36.75}+\frac{{18}^2}{20.05}+\frac{{14}^2}{4.1}-100$       M2

= 39.6        A0

DF = 3        (A1)

Critical value = 11.345        A1         

We conclude, at the 1% significance level, that X does not fit a binomial model.           R1

[13 marks]

Under H₀, the distribution of $\overline{y}$ is N (500, 6.4).      (A1)

Significance level = P $\overline{y}$ < 495 or > 505 | H₀       M2

                             = 2 × 0.02405      (A1)

                             = 0.0481       A1 N5

[5 marks]

The distribution of $\overline{y}$ is now N (503, 6.4).      (A1)

P(Type ΙΙ error) = P(495 < $\overline{y}$ < 505)      (M1)

                       = 0.785       A1 N3

[3 marks]

Show that $\frac{dk}{dT}$ is always positive.

Given that $\underset{T\to \infty }{\lim }k=A$ and $\underset{T\to 0}{\lim }k=0$, sketch the graph of $k$ against $T$.

(i)   the gradient of this line in terms of $c$;

(ii)  the $y$-intercept of this line in terms of $A$.

Find the equation of the regression line for $\ln k$ on $\frac{1}{T}$.

$c$.

It is not required to state units for this value.

$A$.

It is not required to state units for this value.

attempt to use chain rule, including the differentiation of $\frac{1}{T}$          (M1)

$\frac{dk}{dT}=A\times \frac{c}{T^2}\times {\text{e}}^{-\frac{c}{T}}$          A1

this is the product of positive quantities so must be positive          R1

Note: The R1 may be awarded for correct argument from their derivative. R1 is not possible if their derivative is not always positive.

[3 marks]

         A1A1A1

Note: Award A1 for an increasing graph, entirely in first quadrant, becoming concave down for larger values of $T$, A1 for tending towards the origin and A1 for asymptote labelled at $k=A$.

[3 marks]

taking $\ln$ of both sides   OR   substituting $y=\ln x$  and  $x=\frac{1}{T}$           (M1)

$\ln k=\ln A-\frac{c}{T}$  OR  $y=-cx+\ln A$           (A1)

(i)   so gradient is $-c$         A1

(ii)  $y$-intercept is $\ln A$         A1

Note: The implied (M1) and (A1) can only be awarded if both correct answers are seen. Award zero if only one value is correct and no working is seen.

[4 marks]

an attempt to convert data to $\frac{1}{T}$ and $\ln k$           (M1)

e.g. at least one correct row in the following table

line is $\ln k=-13400\times \frac{1}{T}+15.0 \left(=-13383.1\dots \times \frac{1}{T}+15.0107\dots \right)$         A1

[2 marks]

$c=13400 \left(13383.1\dots \right)$         A1

[1 mark]

attempt to rearrange or solve graphically $\ln A=15.0107\dots$          (M1)

$A=3 300 000 \left(3 304 258\dots \right)$         A1

 Note: Accept an $A$ value of $3269017$… from use of $3\mathrm{sf}$ value.

[2 marks]

This question caused significant difficulties for many candidates and many did not even attempt the question. Very few candidates were able to differentiate the expression in part (a) resulting in difficulties for part (b). Responses to parts (c) to (e) illustrated a lack of understanding of linearizing a set of data. Those candidates that were able to do part (d) frequently lost a mark as their answer was given in x and y.

Using this distribution model, find .

Using this distribution model, find the standard deviation of $X$.

an estimate for the mean number of emails received per working day.

an estimate for the standard deviation of the number of emails received per working day.

Give one piece of evidence that suggests Willow’s Poisson distribution model is not a good fit.

Suppose that the probability of Archie receiving more than 10 emails in total on any one day is 0.99. Find the value of λ.

Now suppose that Archie received exactly 20 emails in total in a consecutive two day period. Show that the probability that he received exactly 10 of them on the first day is independent of λ.

$=\text{P}\left(X⩽59\right)$     (M1)

= 0.102     A1

[2 marks]

standard deviation = $\sqrt{70}$ (= 8.37)      (M1)A1

[2 marks]

use of midpoints (accept consistent use of 45, 55 etc.)     (M1)

$\frac{44.5\times 2+54.5\times 15+64.5\times 40+74.5\times 53+94.5+104.5\times 3+114.5\times 6}{2+15+40+53+0+1+3+6}$      (M1)

$=\frac{8530}{120}\left(=71.1\right)$      A1

Note: If 45, 55, etc. are used consistently instead of midpoints (implied by the answer 71.58…) award M1M1A0.

[3 marks]

13.9     (M1)A1

[2 marks]

valid reason given to include the examples below       R1

variance is 192 which is not close to the mean (accept not equal to) standard deviation too high (using parts (a)(ii) and (b)(ii))

relative frequency of $X$ ≤ 59 is 0.142 which is too high (using part (a)(i))

Poisson would give a frequency of roughly 14 for 80 ≤ $X$ ≤ 89

Note: Reasons which do not use values found in previous parts must be backed up with numerical evidence.

[1 mark]

$\text{P}\left(Y>10\right)=0.99$

$1-\text{P}\left(Y⩽10\right)=0.99\Rightarrow \text{P}\left(Y⩽10\right)=0.01$      (M1)

attempt to solve a correct equation       (M1)

λ = 20.1       A1

[3 marks]

in 1 day, no of emails is X ~ Po(λ)

in 2 days, no of emails is Y ~ Po(2λ)       (A1)

P(10 on first day | 20 in 2 days)        (M1)

$\frac{\text{P}\left(X=10\right)\times \text{P}\left(X=10\right)}{\text{P}\left(Y=20\right)}$        (M1)

$=\frac{{\left(\frac{{\lambda }^{10}{e}^{-\lambda }}{10\text{!}}\right)}^2}{\frac{{\left(2\lambda \right)}^{20}{e}^{-2\lambda }}{20\text{!}}}$     A1

$=\frac{{\lambda }^{20}{e}^{-2\lambda }}{2^{20}{\lambda }^{20}{e}^{-2\lambda }}\times \frac{20\text{!}}{{\left(10\text{!}\right)}^2}$      A1

$=\frac{20\text{!}}{2^{20}{\left(10\text{!}\right)}^2}$

which is independent of λ       AG

[5 marks]

It is sunny today. Find the probability that it will be sunny in three days’ time.

Find the eigenvalues and eigenvectors of $\mathit{T}$.

Write down the matrix $\mathit{P}$.

Write down the matrix $\mathit{D}$.

Hence find the long-term percentage of sunny days in Vokram.

finding ${\mathit{T}}^3$  OR  use of tree diagram                     (M1)

${\mathit{T}}^3=\left(\begin{array}{cc}0.65& 0.525\\ 0.35& 0.475\end{array}\right)$

the probability of sunny in three days’ time is $0.65$                     A1

[2 marks]

attempt to find eigenvalues                    (M1)

Note: Any indication that $\text{det}\left(\mathit{T}-\lambda \mathit{I}\right)=0$ has been used is sufficient for the (M1).

$\left|\begin{array}{cc}0.8-\lambda & 0.3\\ 0.2& 0.7-\lambda \end{array}\right|=\left(0.8-\lambda \right)\left(0.7-\lambda \right)-0.06=0$

$\left({\lambda }^2-1.5\lambda +0.5=0\right)$

$\lambda =1, \lambda =0.5$                     A1

attempt to find either eigenvector                   (M1)

$0.8x+0.3y=x\Rightarrow -0.2x+0.3y=0$ so an eigenvector is $\left(\begin{array}{c}3\\ 2\end{array}\right)$                     A1

$0.8x+0.3y=0.5x\Rightarrow 0.3x+0.3y=0$ so an eigenvector is $\left(\begin{array}{c}1\\ -1\end{array}\right)$                     A1

Note: Accept multiples of the stated eigenvectors.

[5 marks]

$\mathit{P}=\left(\begin{array}{cc}3& 1\\ 2& -1\end{array}\right)$  OR  $\mathit{P}=\left(\begin{array}{cc}1& 3\\ -1& 2\end{array}\right)$                    A1

Note: Examiners should be aware that different, correct, matrices $\mathit{P}$ may be seen.

[1 mark]

$\mathit{D}=\left(\begin{array}{cc}1& 0\\ 0& 0.5\end{array}\right)$  OR  $\mathit{D}=\left(\begin{array}{cc}0.5& 0\\ 0& 1\end{array}\right)$                    A1

Note: $\mathit{P}$ and $\mathit{D}$ must be consistent with each other.

[1 mark]

$0.5^n\to 0$                             (M1)

${\mathit{D}}^n=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$  OR  ${\mathit{D}}^n=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$                   (A1)

Note: Award A1 only if their ${\mathit{D}}^n$ corresponds to their $\mathit{P}$

$\mathit{P}{\mathit{D}}^n{\mathit{P}}^{-1}=\left(\begin{array}{cc}0.6& 0.6\\ 0.4& 0.4\end{array}\right)$                   (M1)

$60\%$                      A1

[4 marks]

Show that $a=32$ and $b=\frac{1}{12}$.

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

Find $\text{Var}(X)$.

Find the median of $X$.

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

Find $\text{P}(Y⩾3)$.

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

$\underset{0}{\overset{4}{\int }}\left(\frac{x^2}{a}+b\right)\text{d}x=1\Rightarrow {\left[\frac{x^3}{3a}+bx\right]}_0^4=1\Rightarrow \frac{64}{3a}+4b=1$    M1A1

$\underset{2}{\overset{4}{\int }}\left(\frac{x^2}{a}+b\right)\text{d}x=0.75\Rightarrow \frac{56}{3a}+2b=0.75$    M1A1

Note:    $\underset{0}{\overset{2}{\int }}\left(\frac{x^2}{a}+b\right)dx=0.25\Rightarrow \frac{8}{3a}+2b=0.25$ could be seen/used in place of either of the above equations.

evidence of an attempt to solve simultaneously (or check given a,b values are consistent)     M1

$a=32,b=\frac{1}{12}$     AG

[5 marks]

$\text{E}\left(X\right)=\underset{0}{\overset{4}{\int }}x\left(\frac{x^2}{32}+\frac{1}{12}\right)\text{d}x$    (M1)

$\text{E}(X)=\frac{8}{3}(=2.67)$     A1

[2 marks]

$\text{E}\left(X^2\right)=\underset{0}{\overset{4}{\int }}{x}^2\left(\frac{x^2}{32}+\frac{1}{12}\right)\text{d}x$     (M1)

$\text{Var}(X)=\text{E}(X^2)-[\text{E}(X){]}^2=\frac{16}{15}(=1.07)$     A1

[2 marks]

$\underset{0}{\overset{m}{\int }}\left(\frac{x^2}{32}+\frac{1}{12}\right)\text{d}x=0.5$    (M1)

$\frac{m^3}{96}+\frac{m}{12}=0.5(\Rightarrow m^3+8m-48=0)$     (A1)

$m=2.91$     A1

[3 marks]

$Y\sim B(8,0.75)$     (M1)

$\text{E}(Y)=8\times 0.75=6$     A1

[2 marks]

$\text{P}(Y⩾3)=0.996$     A1

[1 mark]

a dart lands less than $13 \text{cm}$ from $\text{O}$.

a dart lands more than $15 \text{cm}$ from $\text{O}$.

Find the probability that Arianne throws two consecutive darts that land more than $15 \text{cm}$ from $\text{O}$.

Find the probability that Arianne does not score a point on a turn of three darts.

Find Arianne’s expected score in the competition.

Find the probability that Arianne scores at least $5$ points in the competition.

Find the probability that Arianne scores at least $5$ points and less than $8$ points.

Given that Arianne scores at least $5$ points, find the probability that Arianne scores less than $8$ points.

Let $X$ be the random variable “distance from $\text{O}$”.

$X~\text{N}\left(10, 3^2\right)$

$\text{P}\left(X<13\right)=0.841 \left(0.841344\dots \right)$            (M1)(A1)

[2 marks]

$\left(\text{P}\left(X>15\right)=\right) 0.0478 \left(0.0477903\right)$            A1

[1 mark]

$\text{P}\left(X>15\right)\times \text{P}\left(X>15\right)$            (M1)

$=0.00228 \left(0.00228391\dots \right)$            A1

[2 marks]

$1-{\left(0.8143\right)}^3$            (M1)

$0.460 \left(0.460050\dots \right)$            A1

[2 marks]

let $Y$ be the random variable “number of points scored”

evidence of use of binomial distribution           (M1)

$Y~\text{B}\left(10, 0.539949\dots \right)$           (A1)

$\left(\text{E}\left(Y\right)=\right) 10\times 0.539949\dots$           (M1)

           $=5.40$            A1

[4 marks]

$\left(\text{P}\left(Y\ge 5\right)=\right) 0.717 \left(0.716650\dots \right)$            A1

[1 mark]

$\text{P}\left(5\le Y<8\right)$           (M1)

$=0.628 \left(0.627788\dots \right)$            A1

Note: Award M1 for a correct probability statement or indication of correct lower and upper bounds, $5$ and $7$.

[2 marks]

$\frac{\text{P}\left(5\le Y<8\right)}{\text{P}\left(Y\ge 5\right)} \left(=\frac{0.627788\dots }{0.716650\dots }\right)$           (M1)

$=0.876 \left(0.876003\dots \right)$            A1

[2 marks]

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Most candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.

Show that $\text{P}(X=3)=0.001$ and $\text{P}(X=4)=0.0027$.

Find the values of the constants $a$ and $b$.

Deduce that $\frac{\text{P}(X=n)}{\text{P}(X=n-1)}=\frac{0.9(n-1)}{n-3}$ for $n>3$.

(i)     Hence show that $X$ has two modes $m_1$ and $m_2$.

(ii)     State the values of $m_1$ and $m_2$.

Determine the minimum value of $x$ such that the probability Kati receives at least one free gift is greater than 0.5.

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

$\text{P}(X=3)=(0.1{)}^3$    A1

$=0.001$    AG

$\text{P}(X=4)=\text{P}(VV\overline{V}V)+\text{P}(V\overline{V}VV)+\text{P}(\overline{V}VVV)$    (M1)

$=3\times (0.1{)}^3\times 0.9$ (or equivalent)     A1

$=0.0027$    AG

[3 marks]

METHOD 1

attempting to form equations in $a$ and $b$     M1

$\frac{9+3a+b}{2000}=\frac{1}{1000}(3a+b=-7)$    A1

$\frac{16+4a+b}{2000}\times \frac{9}{10}=\frac{27}{10000}(4a+b=-10)$    A1

attempting to solve simultaneously     (M1)

$a=-3,b=2$    A1

METHOD 2

$\text{P}(X=n)=\left(\begin{array}{c}n-1\\ 2\end{array}\right)\times {0.1}^3\times {0.9}^{n-3}$    M1

$=\frac{(n-1)(n-2)}{2000}\times {0.9}^{n-3}$    (M1)A1

$=\frac{n^2-3n+2}{2000}\times {0.9}^{n-3}$    A1

$a=-3,b=2$    A1

Note: Condone the absence of ${0.9}^{n-3}$ in the determination of the values of $a$ and $b$.

[5 marks]

METHOD 1

EITHER

$\text{P}(X=n)=\frac{n^2-3n+2}{2000}\times {0.9}^{n-3}$    (M1)

OR

$\text{P}(X=n)=\left(\begin{array}{c}n-1\\ 2\end{array}\right)\times {0.1}^3\times {0.9}^{n-3}$    (M1)

THEN

$=\frac{(n-1)(n-2)}{2000}\times {0.9}^{n-3}$    A1

$\text{P}(X=n-1)=\frac{(n-2)(n-3)}{2000}\times {0.9}^{n-4}$    A1

$\frac{\text{P}(X=n)}{\text{P}(X=n-1)}=\frac{(n-1)(n-2)}{(n-2)(n-3)}\times 0.9$    A1

$=\frac{0.9(n-1)}{n-3}$    AG

METHOD 2

$\frac{\text{P}(X=n)}{\text{P}(X=n-1)}=\frac{\frac{n^2-3n+2}{2000}\times {0.9}^{n-3}}{\frac{{(n-1)}^2-3(n-1)+2}{2000}\times {0.9}^{n-4}}$    (M1)

$=\frac{0.9(n^2-3n+2)}{(n^2-5n+6)}$    A1A1

Note: Award A1 for a correct numerator and A1 for a correct denominator.

$=\frac{0.9(n-1)(n-2)}{(n-2)(n-3)}$    A1

$=\frac{0.9(n-1)}{n-3}$    AG

[4 marks]

(i)     attempting to solve $\frac{0.9(n-1)}{n-3}=1$ for $n$     M1

$n=21$    A1

   R1

   R1

$X$ has two modes     AG

Note: Award R1R1 for a clearly labelled graphical representation of the two inequalities (using $\frac{\text{P}(X=n)}{\text{P}(X=n-1)}$).

(ii)     the modes are 20 and 21     A1

[5 marks]

METHOD 1

$Y\sim \text{B}(x,0.1)$    (A1)

attempting to solve $\text{P}(Y⩾3)>0.5$ (or equivalent eg $1-\text{P}(Y⩽2)>0.5$) for $x$     (M1)

Note: Award (M1) for attempting to solve an equality (obtaining $x=26.4$).

$x=27$    A1

METHOD 2

$\sum _{n=0}^x\text{P}(X=n)>0.5$    (A1)

attempting to solve for $x$     (M1)

$x=27$    A1

[3 marks]

Find the probability that a randomly selected packet has a weight less than $100 \text{g}$.

The probability that a randomly selected packet has a weight greater than $w$ grams is $0.444$. Find the value of $w$.

A packet is randomly selected. Given that the packet has a weight greater than $105 \text{g}$, find the probability that it has a weight greater than $110 \text{g}$.

From a random sample of $500$ packets, determine the number of packets that would be expected to have a weight lying within $1.5$ standard deviations of the mean.

Packets are delivered to supermarkets in batches of $80$. Determine the probability that at least $20$ packets from a randomly selected batch have a weight less than $95 \text{g}$.

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

${\rm X}~\text{N}\left(102, 8^2\right)$

$\text{P}\left({\rm X}<100\right)=0.401$        (M1)A1

[2 marks]

$\text{P}\left({\rm X}>w\right)=0.444$       (M1)

$\Rightarrow w=103 \left(\text{g}\right)$        A1

[2 marks]

$\text{P}\left({\rm X}>100 \overline{) {\rm X}>105}\right)=\frac{\text{P}\left({\rm X}>100\cap {\rm X}>105\right)}{\text{P}\left({\rm X}>105\right)}$       (M1)

$=\frac{\text{P}\left({\rm X}>100\right)}{\text{P}\left({\rm X}>105\right)}$        (A1)

$=\frac{0.15865\dots }{0.35383\dots }$

$=0.448$       A1

[3 marks]

EITHER

$\text{P}\left(90<{\rm X}<114\right)=0.866\dots$        (A1)

OR

$\text{P}\left(-1.5<Z<1.5\right)=0.866\dots$        (A1)

THEN

$0.866\dots \times 500$       (M1)

$=433$         A1

[3 marks]

$p=\text{P}\left({\rm X}<95\right)=0.19078\dots$         (A1)

recognising  $Y~\text{B}\left(80, p\right)$        (M1)

now using  $Y~\text{B}\left(80, 0.19078\dots \right)$        (M1)

$\text{P}\left(Y\ge 20\right)=0.116$          A1

[4 marks]

Using Hank’s model, find the probability that Bill visits the garden on exactly four occasions during one particular month.

on exactly $12$ occasions.

during the first and third month only.

Find the probability that over a $12$-month period, there will be exactly $3$ months when Bill does not visit the garden.

After the first year, a number of baby magpies start to visit Hank’s garden. It may be assumed that each of these baby magpies visits the garden randomly and independently, and that the number of times each baby magpie visits the garden per month is modelled by a Poisson distribution with mean $2.1$.

Determine the least number of magpies required, including Bill, in order that the probability of Hank’s garden having at least $30$ magpie visits per month is greater than $0.2$.

$X_1~\text{Po}\left(3.1\right)$

$\text{P}\left(X_1=4\right)=0.173 \left(0.173349\dots \right)$                A1

[1 mark]

$X_2~\text{Po}\left(3\times 3.1\right)=\text{Po}\left(9.3\right)$               (M1)

$\text{P}\left(X_2=12\right)=0.0799 \left(0.0798950\dots \right)$                A1

[2 marks]

${\left(\text{P}\left(X_1>0\right)\right)}^2\times \text{P}\left(X_1=0\right)$               (M1)

$0.{95495}^2\times 0.04505$               (A1)

$=0.0411 \left(0.0410817\dots \right)$                A1

[3 marks]

$\text{P}\left(X_1=0\right)=0.04505$               (A1)

$X_1~\text{B}\left(12, 0.04505\right)$               (M1)(A1)

Note: Award M1 for recognizing binomial probability, and A1 for correct parameters.

$=0.0133 \left(0.013283\dots \right)$                A1

[4 marks]

METHOD ONE

               (M1)(A1)(A1)

Note: Award M1 for evidence of a cumulative Poisson with $\lambda =3.1+2.1n$, A1 for $0.136705$ and A1 for $0.253384$.

so require $12$ magpies (including Bill)               A1

METHOD TWO

evidence of a cumulative Poisson with $\lambda =3.1+2.1n$               (M1)

sketch of curve and $y=0.2$               (A1)

(intersect at) $10.5810\dots$               (A1)

rounding up gives $n=11$

so require $12$ magpies (including Bill)               A1

[4 marks]

It is given that one in five cups of coffee contain more than 120 mg of caffeine.
It is also known that three in five cups contain more than 110 mg of caffeine.

Assume that the caffeine content of coffee is modelled by a normal distribution.
Find the mean and standard deviation of the caffeine content of coffee.

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

let $X$ be the random variable “amount of caffeine content in coffee”

$\text{P}(X>120)=0.2,\text{ P}(X>110)=0.6$     (M1)

Note:     Award M1 for at least one correct probability statement.

$\frac{120-\mu }{\sigma }=0.84162\dots ,\frac{110-\mu }{\sigma }=-0.253347\dots$     (M1)(A1)(A1)

Note:     Award M1 for attempt to find at least one appropriate $z$-value.

$120-\mu =0.84162\sigma ,110-\mu =-0.253347\sigma$

attempt to solve simultaneous equations     (M1)

$\mu =112,\sigma =9.13$     A1

[6 marks]

Draw a tree diagram to represent this information for the first three days of July.

Find the probability that the 3rd July is calm.

Find the probability that the 1st July was calm given that the 3rd July is windy.

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

     M1A2

Note:     Award M1 for 3 stage tree-diagram, A2 for 0.8, 0.9, 0.3 probabilities correctly placed.

[3 marks]

$0.2\times 0.7\times 0.3+0.2\times 0.3\times 0.9+0.8\times 0.1\times 0.3+0.8\times 0.9\times 0.9=0.768$     (M1)A1

[2 marks]

$\text{P}(\text{1st July is calm | 3rd July is windy)}=\frac{\text{P}(\text{1st July is calm and 3rd July is windy)}}{\text{P}(\text{3rd July is windy)}}$     (M1)

$=\frac{0.8\times 0.1\times 0.7+0.8\times 0.9\times 0.1}{1-0.768}$

OR$$$\frac{0.8\times 0.1\times 0.7+0.8\times 0.9\times 0.1}{0.2\times 0.7\times 0.7+0.2\times 0.3\times 0.1+0.8\times 0.1\times 0.7+0.8\times 0.9\times 0.1}$