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Date	May Example question	Marks available	1	Reference code	EXM.1.AHL.TZ0.21
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Interpret	Question number	21	Adapted from	N/A

Question

Saloni wants to find a model for the temperature of a bottle of water after she removes it from the fridge. She uses a temperature probe to record the temperature of the water, every 5 minutes.

After graphing the data, Saloni believes a suitable model will be

$T = 28 - a{b^t}$ , where $a{\text{,}}\,\,b \in {\mathbb{R}^ + }$ .

Explain why $28 - T$ can be modeled by an exponential function.

[1]

Find the equation of the least squares exponential regression curve for $28 - T$ .

[3]

Write down the coefficient of determination, ${R^2}$ .

[1]

c.i.

Interpret what the value of ${R^2}$ tells you about the model.

[1]

c.ii.

Hence predict the temperature of the water after 3 minutes.

[2]

Markscheme

Rearranging the model gives $28 - T = a{b^t}$ A1

So $28 - T$ can be modeled by an exponential function. AG

[2 marks]

(A1)

$28 - T = 22.7{\left( {0.925} \right)^t}$ M1A1

[3 marks]

${R^2} = 0.974$ A1

[1 mark]

c.i.

Since the value of ${R^2}$ is close to +1, the model is a good fit for the data. R1

[1 mark]

c.ii.

$T = 28 - \left( {22.69 \ldots } \right){\left( {0.9250 \ldots } \right)^3} = 10.0$ minutes M1A1

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Syllabus sections

Topic 4—Statistics and probability » AHL 4.13—Non-linear regression

Show 56 related questions

22M.3.AHL.TZ1.1b.iii:
Write down the value of $R^{2}$ for this model.
EXM.1.AHL.TZ0.21a:
Explain why $28 - T$ can be modeled by an exponential function.
EXM.3.AHL.TZ0.9e:
Give two reasons why the prediction in part (b)(ii) might be lower than 14.
EXM.3.AHL.TZ0.9f.ii:
$c$ .
EXM.3.AHL.TZ0.9b.i:
the number of new people infected on day 6.
EXM.3.AHL.TZ0.7a.ii:
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.
EXM.1.AHL.TZ0.21b:
Find the equation of the least squares exponential regression curve for $28 - T$ .
EXM.2.AHL.TZ0.12c.ii:
Find the value of this area.
EXM.3.AHL.TZ0.9a:
Use an exponential regression to find the value of $a$ and of $b$ , correct to 4 decimal places.
EXM.3.AHL.TZ0.7b.ii:
Write down the coefficient of determination.
22M.3.AHL.TZ1.1b.iv:
Hence comment on the suitability of the model from (b)(ii) in comparison with the linear model found in part (a).
21M.2.AHL.TZ1.4b.ii:
By considering the gradient of this curve when $x = 4$ , explain why it may not be a good model.
21M.3.AHL.TZ2.1e.iv:
Find the coefficient of determination for each of the two models she considers.
EXM.2.AHL.TZ0.12b.ii:
Write down the coefficient of determination.
EXM.3.AHL.TZ0.7c.ii:
Find the value of this area.
EXM.3.AHL.TZ0.7d.ii:
Hence explain how a straight line graph could be drawn using the coordinates in the table.
EXM.3.AHL.TZ0.9b.ii:
the day when the total number of people infected will be greater than 1000.
EXM.3.AHL.TZ0.9f.i:
$L$ .
EXM.3.AHL.TZ0.9f.iii:
$k$ .
EXM.3.AHL.TZ0.9c:
Use your answer to part (a) to show that the model predicts 16.7 people will be infected on the first day.
EXM.2.AHL.TZ0.12b.i:
Use all the coordinates in the table to find the equation of the least squares cubic regression curve.
EXM.1.AHL.TZ0.21d:
Hence predict the temperature of the water after 3 minutes.
SPM.1.AHL.TZ0.12:
Product research leads a company to believe that the revenue ( $R$ ) made by selling its goods at a price ( $p$ ) can be modelled by the equation.

$R\left( p \right) = cp{{\text{e}}^{dp}}$ , $c$ , $d \in \mathbb{R}$

There are two competing models, A and B with different values for the parameters $c$ and $d$ .

Model A has $c$ = 3, $d$ = −0.5 and model B has $c$ = 2.5, $d$ = −0.6.

The company experiments by selling the goods at three different prices in three similar areas and the results are shown in the following table.

The company will choose the model with the smallest value for the sum of square residuals.

Determine which model the company chose.
EXM.2.AHL.TZ0.12a:
Use the trapezoidal rule to find an estimate for the area.
EXM.3.AHL.TZ0.9h:
Use the logistic model to find the day when the rate of increase of people infected is greatest.
EXM.3.AHL.TZ0.7a.i:
Use the trapezoidal rule to find an estimate for the area.
EXM.3.AHL.TZ0.7b.i:
Use all the coordinates in the table to find the equation of the least squares cubic regression curve.
EXM.3.AHL.TZ0.9d.ii:
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.
EXM.3.AHL.TZ0.9d.i:
Explain why the number of degrees of freedom is 2.
EXM.3.AHL.TZ0.9g:
Hence predict the total number of people infected by this disease after several months.
EXM.2.AHL.TZ0.12c.i:
Write down an expression for the area enclosed by the cubic regression curve, the $x$ -axis, the $y$ -axis and the line $x = 10$ .
EXM.3.AHL.TZ0.7d.i:
Show that ${\text{ln}}\,y = qx + {\text{ln}}\,p$ .
EXN.2.AHL.TZ0.4e:
Find the sum of the square residuals for Jorge’s model using the values $t = 1, 2, 3, 4$ .
EXM.3.AHL.TZ0.7d.iii:
By finding the equation of a suitable regression line, show that $p = 1.83$ and $q = 0.986$ .
EXM.3.AHL.TZ0.7d.iv:
Hence find the area enclosed by the exponential function, the $x$ -axis, the $y$ -axis and the line $x = 4.4$ .
EXM.1.AHL.TZ0.21c.i:
Write down the coefficient of determination, ${R^2}$ .
21M.2.AHL.TZ1.4b.i:
Find the equation of the least squares regression quadratic curve for these four points.
21M.2.AHL.TZ1.4c:
Find the equation of the new model.
EXM.3.AHL.TZ0.7c.i:
Write down an expression for the area enclosed by the cubic function, the $x$ -axis, the $y$ -axis and the line $x = 4.4$ .
21M.3.AHL.TZ2.1e.v:
Hence compare the two models.
21M.3.AHL.TZ2.1e.iii:
Find the value of $c$ , of $d$ and of $e$ .
21M.3.AHL.TZ2.1e.vi:
Juliet decides to use the coefficient of determination to choose between these two models.

Comment on the validity of her decision.
21N.1.AHL.TZ0.12a:
Use the data in the second table to find the value of $m$ and the value of $b$ for the regression line, $\ln x = m (\ln d) + b$ .
21N.1.AHL.TZ0.12b:
Assuming that the model found in part (a) remains valid, estimate the percentage of trees in stock when $d = 25$ .
21N.3.AHL.TZ0.2a.i:
Find the equation of the regression line of $h$ on $t$ .
21N.3.AHL.TZ0.2a.ii:
Interpret the meaning of parameter $a$ in the context of the model.
21N.3.AHL.TZ0.2a.iii:
Suggest why Eva’s use of the linear regression equation in this way could be unreliable.
21N.3.AHL.TZ0.2b.i:
Find the equation of the least squares quadratic regression curve.
21N.3.AHL.TZ0.2b.ii:
Find the value of $k$ .
21N.3.AHL.TZ0.2b.iii:
Hence, write down a suitable domain for Eva’s function $h (t) = p t^{2} + q t + r$ .
21N.3.AHL.TZ0.2c:
Show that $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ .
21N.3.AHL.TZ0.2d:
By solving the differential equation $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ , show that the general solution is given by $h = 17 640 {(c - R^{2} t)}^{2}$ , where $c \in ℝ$ .
21N.3.AHL.TZ0.2e:
Use the general solution from part (d) and the initial condition $h (0) = 3.2$ to predict the value of $T$ .
21N.3.AHL.TZ0.2f:
Find this new height.
21N.3.AHL.TZ0.2g.i:
Show that $\frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 \sqrt{H}$ , where $0 \leq t \leq T$ .
21N.3.AHL.TZ0.2g.ii:
Use Euler’s method with a step length of $0.5$ minutes to estimate the maximum value of $H$ .

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