Find the equation of the regression line of $Q\left(t\right)$ on $t$.

Write down the value of $r$, Pearson’s product-moment correlation coefficient.

Explain why it would not be appropriate to conduct a hypothesis test on the value of $r$ found in (a)(ii).

Find the general solution of the differential equation $Q\text{'}\left(t\right)=\beta NQ\left(t\right)$.

Using the data in the table write down the equation for an appropriate non-linear regression model.

Write down the value of $R^2$ for this model.

Hence comment on the suitability of the model from (b)(ii) in comparison with the linear model found in part (a).

By considering large values of $t$ write down one criticism of the model found in (b)(ii).

Use your answer from part (b)(ii) to estimate the time taken for the number of infected computers to double.

Find in which city, X or Y, the computer virus is spreading more easily. Justify your answer using your results from part (b).

Determine the value of $a$ and of $b$. Give your answers correct to one decimal place.

Use linear regression to estimate the value of $k$ and of $L$.

The solution to the differential equation is given by

$Q\left(t\right)=\frac{L}{1+C{\text{e}}^{-kt}}$

where $C$ is a constant.

Using your answer to part (f)(i), estimate the percentage of computers in city X that are expected to have been infected by the virus over a long period of time.

$Q\left(t\right)=3090t-54000 \left(3094.27\dots t-54042.3\dots \right)$         A1A1

Note: Award at most A1A0 if answer is not an equation. Award A1A0 for an answer including either $x$ or $y$.

[2 marks]

$0.755 \left(0.754741\dots \right)$         A1

[1 mark]

$t$ is not a random variable OR it is not a (bivariate) normal distribution

OR data is not a sample from a population

OR data appears nonlinear

OR $r$ only measures linear correlation         R1

Note: Do not accept “$r$ is not large enough”.

[1 mark]

attempt to separate variables            (M1)

$\int \frac{1}{Q}dQ=\int \beta N dt$

$\ln \left|Q\right|=\beta Nt+c$           A1A1A1 

Note: Award A1 for LHS, A1 for $\beta Nt$, and A1 for $+c$.

Award full marks for $Q={\text{e}}^{\beta Nt+c}$  OR  $Q=A{\text{e}}^{\beta Nt}$.

Award M1A1A1A0 for $Q={\text{e}}^{\beta Nt}$

[4 marks]

attempt at exponential regression           (M1)

$Q=1.15{\text{e}}^{0.292t} \left(Q=1.14864\dots {\text{e}}^{0.292055\dots t}\right)$           A1

OR

attempt at exponential regression           (M1)

$Q=1.15\times 1.{34}^t \left(1.14864\dots \times 1.33917{\dots }^t\right)$           A1

Note: Condone answers involving $y$ or $x$. Condone absence of “$Q=$” Award M1A0 for an incorrect answer in correct format.

[2 marks]

$0.999 \left(0.999431\dots \right)$          A1

[1 mark]

comparing something to do with $R^2$ and something to do with $r$        M1

Note:   Examples of where the M1 should be awarded:

$R^2>r$
$R>r$
$0.999>0.755$
$0.999>0.{755}^2 \left(=0.563\right)$
The “correlation coefficient” in the exponential model is larger.
Model B has a larger $R^2$

Examples of where the M1 should not be awarded:

The exponential model shows better correlation (since not clear how it is being measured)
Model 2 has a better fit
Model 2 is more correlated

an unambiguous comparison between $R^2$ and $r^2$ or $R$ and $r$ leading to the conclusion that the model in part (b) is more suitable / better          A1

Note: Condone candidates claiming that $R$ is the “correlation coefficient” for the non-linear model.

[2 marks]

it suggests that there will be more infected computers than the entire population       R1

Note: Accept any response that recognizes unlimited growth. 

[1 mark]

$1.15{\text{e}}^{0.292t}=2.3$  OR  $1.15\times 1.{34}^t=2.3$  OR  $t=\frac{\ln 2}{0.292}$  OR using the model to find two specific times with values of $Q\left(t\right)$ which double          M1

$t=2.37$  (days)          A1

Note: Do not FT from a model which is not exponential. Award M0A0 for an answer of $2.13$ which comes from using $(10, 20)$ from the data or any other answer which finds a doubling time from figures given in the table.

[2 marks]

an attempt to calculate $\beta$ for city X          (M1)

$\beta =\frac{0.292055\dots }{2.6\times {10}^6}$  OR  $\beta =\frac{\ln 1.33917\dots }{2.6\times {10}^6}$

$=1.12328\dots \times {10}^{-7}$          A1

this is larger than $9.64\times {10}^{-8}$ so the virus spreads more easily in city X         R1

Note: It is possible to award M1A0R1.
Condone “so the virus spreads faster in city X” for the final R1.

[3 marks]

$a=38.3, b=3086.1$          A1A1

Note: Award A1A0 if values are correct but not to $1$ dp.

[2 marks]

$\frac{Q\text{'}}{Q}=0.42228-2.5561\times {10}^{-6}Q$          (A1)(A1)

Note: Award A1 for each coefficient seen – not necessarily in the equation. Do not penalize seeing in the context of $y$ and $x$.

identifying that the constant is $k$ OR that the gradient is $-\frac{k}{L}$          (M1)

therefore $k=0.422 \left(0.422228\dots \right)$          A1

$\frac{k}{L}=2.5561\times {10}^{-6}$

$L=165000 \left(165205\right)$          A1

Note: Accept a value of $L$ of $164843$ from use of $3$ sf value of $k$, or any other value from plausible pre-rounding.
Allow follow-through within the question part, from the equation of their line to the final two A1 marks.

[5 marks]

recognizing that their $L$ is the eventual number of infected        (M1)

$\frac{165205\dots }{2600000}=6.35\% \left(6.35403\dots \%\right)$          A1

Note: Accept any final answer consistent with their answer to part (f)(i) unless their $L$ is less than $120146$ in which case award at most M1A0.

[2 marks]

A significant minority were unable to attempt 1(a) which suggests poor preparation for the use of the GDC in this statistics-heavy course. Large numbers of candidates appeared to use $y, x, Q$ and $t$ interchangeably. Accurate use of notation is an important skill which needs to be developed.

1(a)(iii) was a question at the heart of the Applications and interpretations course. In modern statistics many of the calculations are done by a computer so the skill of the modern statistician lies in knowing which tests are appropriate and how to interpret the results. Very few candidates seemed familiar with the assumptions required for the use of the standard test on the correlation coefficient. Indeed, many candidates answered this by claiming that the value was either too large or too small to do a hypothesis test, indicating a major misunderstanding of the purpose of hypothesis tests.

1(b)(i) was done very poorly. It seems that perhaps adding parameters to the equation confused many candidates – if the equation had been $Q\text{'}\left(t\right)=5Q\left(t\right)$ many more would have successfully attempted this. However, the presence of parameters is a fundamental part of mathematical modelling so candidates should practise working with expressions involving them.

1(b)(ii) and (iii) were done relatively well, with many candidates using the data to recognize an exponential model was a good idea. Part (iv) was often communicated poorly. Many candidates might have done the right thing in their heads but just writing that the correlation was better did not show which figures were being compared. Many candidates who did write down the numbers made it clear that they were comparing an $R^2$ value with an $r$ value.

1(c) was not meant to be such a hard question. There is a standard formula for half-life which candidates were expected to adapt. However, large numbers of candidates conflated the data and the model, finding the time for one of the data points (which did not lie on the model curve) to double. Candidates also thought that the value of t found was equivalent to the doubling time, often giving answers of around 40 days which should have been obviously wrong.

1(d) was quite tough. Several candidates realized that $\beta$ was the required quantity to be compared but very few could calculate $\beta$ for city X using the given information.

1(e) was meant to be relatively straightforward but many candidates were unable to interpret the notation given to do the quite straightforward calculation.

1(f) was meant to be a more unusual problem-solving question getting candidates to think about ways of linearizing a non-linear problem. This proved too much for nearly all candidates.

Use an exponential regression to find the value of $a$ and of $b$, correct to 4 decimal places.

the number of new people infected on day 6.

the day when the total number of people infected will be greater than 1000.

Use your answer to part (a) to show that the model predicts 16.7 people will be infected on the first day.

Explain why the number of degrees of freedom is 2.

Perform a ${\chi }^2$ goodness of fit test at the 5% significance level. You should clearly state your hypotheses, the p-value, and your conclusion.

Give two reasons why the prediction in part (b)(ii) might be lower than 14.

$L$.

$c$.

$k$.

Hence predict the total number of people infected by this disease after several months.

Use the logistic model to find the day when the rate of increase of people infected is greatest.

$a=9.7782\text{,}b=1.7125$     M1A1A1

[3 marks]

$n\left(6\right)=247$       A1

number of new people infected = 247 – 140 = 107     M1A1

[3 marks]

use of graph or table      M1

day 9    A1

[2 marks]

9.7782(1.7125)¹      M1

= 16.7 people    AG

[1 mark]

2 parameters ($a$ and $b$) were estimated from the data.     R1

$\upsilon =5-1-2$     M1

= 2    AG

[2 marks]

$H_0\text{:}$ data is modeled by $n\left(d\right)=9.7782{\left(1.7125\right)}^d$ and $H_1\text{:}$ data is not modeled by $n\left(d\right)=9.7782{\left(1.7125\right)}^d$     A1

p-value = 0.893    A2

Since 0.893 > 0.05     R1

Insufficient evidence to reject $H_0$. So data is modeled by $n\left(d\right)=9.7782{\left(1.7125\right)}^d$    A1

[5 marks]

vaccine or medicine might slow down rate of infection     R1

People become more aware of disease and take precautions to avoid infection     R1

Accept other valid reasons

[2 marks]

1060      M1A1

[2 marks]

108      A1

[1 mark]

0.560     A1

[1 mark]

As $d\to \mathrm{\infty }$       M1

$n\to 1060$      A1

[2 marks]

sketch of $\frac{\text{d}n}{\text{d}d}$ or solve $\frac{{\text{d}}^{2}n}{\text{d}{d}^2}=0$        M1

$d=8.36$      A1

Day 8      A1

[3 marks]

Use this model to predict the number of fish in the lake when $t=8$.

Assuming the proportion of marked fish in the second sample is equal to the proportion of marked fish in the lake, show that the estate manager will estimate there are now $2000$ fish in the lake.

Write down the value of $n$ and the value of $p$.

State an assumption that is being made for $X$ to be considered as following a binomial distribution.

Show that an estimate for $\text{Var}\left(X\right)$ is $38.25$.

Hence show that the variance of the proportion of marked fish in the sample, $\text{Var}\left(\frac{X}{300}\right)$, is $0.000425$.

Taking the value for the variance given in (d) (ii) as a good approximation for the true variance, find the upper and lower bounds for the proportion of marked fish in the lake.

Hence find upper and lower bounds for the number of fish in the lake when $t=8$.

Given this result, comment on the validity of the linear model used in part (a).

Assuming a carrying capacity of $5000$ use the given values of $N\left(0\right)$ and $N\left(1\right)$ to calculate the parameters $C$ and $k$.

Use these parameters to calculate the value of $N\left(8\right)$ predicted by this model.

Comment on the likelihood of the fish population reaching $5000$.

* 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.

$N\left(8\right)=1000+200\times 8$        M1

$=2600$        A1

[2 marks]

$\frac{45}{300}=\frac{300}{N}$        M1A1

$N=2000$        AG

[2 marks]

$n=300, p=\frac{300}{2000}=0.15$     A1A1

[2 marks]

Any valid reason for example:          R1

Marked fish are randomly distributed, so $p$ constant.

Each fish caught is independent of previous fish caught

[1 mark]

$\text{Var}\left(X\right)=np\left(1-p\right)$          M1

$=300\times \frac{300}{2000}\times \frac{1700}{2000}$          A1

$=38.25$          AG

[2 marks]

$\text{Var}\left(\frac{X}{300}\right)\mathrm{=}\frac{\text{Var}\left(X\right)}{{300}^2}$          M1A1

$=0.000425$             AG

[2 marks]

$0.15\pm 1.5\sqrt{0.000425}$          (M1)

$0.181$ and $0.119$             A1

[2 marks]

$\frac{300}{N}=0.181\dots , \frac{300}{N}=0.119\dots$            M1

Lower bound $1658$ upper bound $2519$             A1

[2 marks]

Linear model prediction falls outside this range so unlikely to be a good model            R1A1

[2 marks]

$1000=\frac{5000}{1+C}$            M1

$C=4$            A1

$1200=\frac{5000}{1+4e^{-k}}$            M1

$e^{-k}=\frac{3800}{4\times 1200}$            (M1)

$k=-\ln \left(0.7916\dots \right)=0.2336\dots$         A1

[5 marks]

$N\left(8\right)=\frac{5000}{1+4e^{-0.2336\times 8}}=3090$         M1A1

Note: Accept any answer that rounds to $3000$.

[2 marks]

This is much higher than the calculated upper bound for $N\left(8\right)$ so the rate of growth of the fish is unlikely to be sufficient to reach a carrying capacity of $5000$.        M1R1

[2 marks]

Find $\left(f\circ g\right)\left(\left(x\text{,}y\right)\right)$.

Find $\left(g\circ f\right)\left(\left(x\text{,}y\right)\right)$.

State with a reason whether or not $f$ and $g$ commute.

Find the inverse of $f$.

$\left(f\circ g\right)\left(\left(x\text{,}y\right)\right)=f\left(g\left(\left(x\text{,}y\right)\right)\right)$  ($=f\left(\left(xy,x+y\right)\right)$)       (M1)

$=\left(xy+x+y,xy-x-y\right)$       A1A1

[3 marks]

$\left(g\circ f\right)\left(\left(x\text{,}y\right)\right)=g\left(f\left(\left(x\text{,}y\right)\right)\right)$

$=g\left(\left(x+y,x-y\right)\right)$

$=\left(\left(x+y\right)\left(x-y\right),x+y+x-y\right)$

$=\left(x^2-y^2\text{,}2x\right)$       A1A1

[2 marks]

no because $f\circ g\ne g\circ f$        R1

Note: Accept counter example.

[1 mark]

$f\left(\left(x\text{,}y\right)\right)=\left(a\text{,}b\right)\Rightarrow \left(x+y,x-y\right)=\left(a\text{,}b\right)$       (M1)

$\{\begin{array}{c}x=\frac{a+b}{2}\\ y=\frac{a-b}{2}\end{array}$       (M1)

$f^{-1}\left(\left(x\text{,}y\right)\right)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$        A1

[3 marks]

Show that $b\approx 0.00939$.

Find the angle through which Mars rotates on its axis each hour.

Show that the maximum value of $\omega =1.98$, correct to three significant figures.

Find the minimum value of $\omega$.

the maximum value of $R\left(t\right)$.

the minimum value of $R\left(t\right)$.

Hence show that $a=1.6$, correct to two significant figures.

Find the value of $c$.

Find the value of $d$.

Write down $z_1$ and $z_2$ in exponential form, with a constant modulus.

Hence or otherwise find an equation for $L$ in the form $L\left(t\right)=p \sin (qt+r)+d$, where $p, q, r, d\in \mathrm{ℝ}$.

Find, in hours, the shortest time from sunrise to sunset at point $\text{A}$ that is predicted by this model.

recognition that period $=669$                  (M1)

$b=\frac{2\mathrm{\pi }}{669}$  OR  $b=0.00939190\dots$                A1

Note: Award A1 for a correct expression leading to the given value or for a correct value of $b$ to 4 sf or greater accuracy.

$b\approx 0.00939$             AG

[2 marks]

length of day$=24\frac{2}{3}$ hours                (A1)

Note: Award A1 for $\frac{2}{3}, 0.666\dots , 0.\overline{6}$ or $0.667$.

$\frac{2\mathrm{\pi }}{24{\displaystyle \frac{2}{3}}}$               (M1)

Note: Accept $\left(\frac{360}{24{\displaystyle \frac{2}{3}}}\right)$.

$=0.255$ radians $\left(0.254723\dots , \frac{3\mathrm{\pi }}{37}, 14.5945\dots °\right)$               A1

[3 marks]

substitution of either value of $\delta$ into equation              (M1)

correct use of arccos to find a value for $\omega$              (M1)

Note: Both (M1) lines may be seen in either part (c)(i) or part (c)(ii).

$\cos \omega =0.839 \tan \left(-0.440\right)$               A1

$\omega =1.97684\dots$

$\approx 1.98$               AG

Note: For substitution of $1.98$ award M0A0.

[3 marks]

$\delta =0.440$

$\omega =1.16 (1.16474\dots )$           A1

[1 mark]

$R_{\text{max}}=\frac{1.97684\dots }{0.25472\dots }$           (M1)

$=7.76 \text{hours} (7.76075\dots )$           A1

Note: Accept $7.70$ from use of $1.98$.

[2 marks]

$R_{\text{min}}=\frac{1.16474\dots }{0.25472\dots }$

$=4.57 \text{hours} (4.57258\dots )$           A1

Note: Accept $4.55$ and $4.56$ from use of rounded values.

[1 mark]

$a=\frac{7.76075\dots -4.57258\dots }{2}$           M1

$\approx 1.59408\dots$           A1

Note: Award M1 for substituting their values into a correct expression. Award A1 for a correct value of $a$ from their expression which has at least 3 significant figures and rounds correctly to $1.6$.

$\approx 1.6$ (correct to $2$ sf)          AG

[2 marks]

EITHER

$c=\frac{7.76075\dots +4.57258\dots }{2} \left(=\frac{12.333\dots }{2}\right)$           (M1)

OR

$c=4.57258\dots +1.59408\dots$  or  $c=7.76075\dots -1.59408\dots$

THEN

$=6.17 \left(6.16666\dots \right)$           A1

Note: Accept $6.16$ from use of rounded values. Follow through on their answers to part (d) and $1.6$.


[2 marks]

$d=18.65-6.16666\dots$           (M1)

$=12.5 \left(12.4833\dots \right)$           A1

Note: Follow through for $18.65$ minus their answer to part (f).


[2 marks]

at least one expression in the form $r{\text{e}}^{g\left(t\right)\text{i}}$           (M1)

$z_1=1.5{\text{e}}^{\left(0.00939t+2.83\right)\text{i}}\mathrm{,} {\mathrm{z}}_2=1.6{\text{e}}^{\left(0.00939t\right)\text{i}}$           A1A1

[3 marks]

EITHER

$z_1-z_2=1.5{\text{e}}^{\left(0.00939t+2.83\right)\text{i}}-1.6{\text{e}}^{\left(0.00939t\right)\text{i}}$

$={\text{e}}^{0.00939t\text{i}}\left(1.5{\text{e}}^{2.83\text{i}}-1.6\right)$               (M1)

$={\text{e}}^{0.00939t\text{i}}\left(3.06249\dots {\text{e}}^{2.99086\dots \text{i}}\right)$            (A1)(A1)

OR

graph of $L$ or $f$

$p=3.06249...$            (A1)

$r=-0.150729...$  OR  $r=2.99086...$            (M1)(A1)

Note: The $p$ and $r$ variables (or equivalent) must be seen.


THEN

$L\left(t\right)=3.06 \sin (0.00939t+2.99)+12.5$                 A1

$\left(L\left(t\right)=3.06248\dots \sin (0.00939t+2.99086\dots )+12.4833\dots \right)$

Note: Accept equivalent forms, e.g. $L\left(t\right)=3.06 \sin (0.00939t-0.151)+12.5$.
Follow through on their answer to part (g) replacing $12.5$.

 
[4 marks]

shortest time between sunrise and sunset

$12.4833\dots -3.06249\dots$               (M1)

$=9.42$ hours  $\left(9.420843\dots \right)$                 A1

Note: Accept $9.44$ from use of $3$ sf values.

[2 marks]

Use the trapezoidal rule to find an estimate for the area.

With reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.

Use all the coordinates in the table to find the equation of the least squares cubic regression curve.

Write down the coefficient of determination.

Write down an expression for the area enclosed by the cubic function, the $x$-axis, the $y$-axis and the line $x=4.4$.

Find the value of this area.

Show that $\text{ln}y=qx+\text{ln}p$.

Hence explain how a straight line graph could be drawn using the coordinates in the table.

By finding the equation of a suitable regression line, show that $p=1.83$ and $q=0.986$.

Hence find the area enclosed by the exponential function, the $x$-axis, the $y$-axis and the line $x=4.4$.

Area $=\frac{1.1}{2}\left(2+2\left(5+15+47\right)+148\right)$         M1A1

Area = 156 units²          A1

[3 marks]

The graph is concave up,         R1

so the trapezoidal rule will give an overestimate.         A1

[2 marks]

$f\left(x\right)=3.88x^3-12.8x^2+14.1x+1.54$         M1A2

[3 marks]

$R^2=0.999$        A1

[1 mark]

Area $=\underset{0}{\overset{4.4}{\int }}\left(3.88x^3-12.8x^2+14.1x+1.54\right)dx$        A1A1

[2 marks]

Area = 145 units²    (Condone 143–145 units², using rounded values.)      A2

[2 marks]

$\text{ln}y=\text{ln}\left(p{\text{e}}^{qx}\right)$      M1

$\text{ln}y=\text{ln}p+\text{ln}\left({\text{e}}^{qx}\right)$      A1

$\text{ln}y=qx+\text{ln}p$      AG

[2 marks]

Plot $\text{ln}y$ against $p$.      R1

[1 mark]

Regression line is $\text{ln}y=0.986x+0.602$       M1A1

So $q=$ gradient = 0.986    R1

$p=e^{0.602}=1.83$       M1A1

[5 marks]

Area $=\underset{0}{\overset{4.4}{\int }}1.83e^{0.986x}dx=140$ units²     M1A1

[2 marks]