EXM.3.AHL.TZ0.9

pestleMathematicsAIHLPaper 3EXM· sl-2-5-modelling-functionssource ↗

In this question you will explore possible models for the spread of an infectious disease

An infectious disease has begun spreading in a country. The National Disease Control Centre (NDCC) has compiled the following data after receiving alerts from hospitals.

A graph of $n$ against $d$ is shown below.

The NDCC want to find a model to predict the total number of people infected, so they can plan for medicine and hospital facilities. After looking at the data, they think an exponential function in the form $n = a{b^d}$ could be used as a model.

Use your answer to part (a) to predict

The NDCC want to verify the accuracy of these predictions. They decide to perform a ${\chi ^2}$ goodness of fit test.

The predictions given by the model for the first five days are shown in the table.

In fact, the first day when the total number of people infected is greater than 1000 is day 14, when a total of 1015 people are infected.

Based on this new data, the NDCC decide to try a logistic model in the form $n = \frac{L}{{1 + c{e^{ - kd}}}}$ .

Use the data from days 1–5, together with day 14, to find the value of

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

[3]

the number of new people infected on day 6.

[3]

b.i.

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

[2]

b.ii.

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

[1]

Explain why the number of degrees of freedom is 2.

[2]

d.i.

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.

[5]

d.ii.

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

[2]

$L$ .

[2]

f.i.

$c$ .

[1]

f.ii.

$k$ .

[1]

f.iii.

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

[2]

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

[3]

Markscheme / solution

$a = 9.7782{\text{,}}\,\,b = 1.7125$ M1A1A1

[3 marks]

$n\left( 6 \right) = 247$ A1

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

[3 marks]

b.i.

use of graph or table M1

day 9 A1

[2 marks]

b.ii.

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]

d.i.

${H_0}\,{\text{:}}$ 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]

d.ii.

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]

f.i.

108 A1

[1 mark]

f.ii.

0.560 A1

[1 mark]

f.iii.

As $d \to \infty$ 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]

Examiners’ report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

f.i.

[N/A]

f.ii.

[N/A]

f.iii.

[N/A]