IB DP Maths Topic 5.4 Mathematical and contextual interpretation SL Paper 2

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IB DP Maths SL Paper 2 - All Topics

 

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Question

The following table shows the amount of fuel (\(y\) litres) used by a car to travel certain distances (\(x\) km).

 

Distance (x km)4075120150195
Amount of fuel (y litres)3.66.59.913.116.2

This data can be modelled by the regression line with equation \(y = ax + b\).

Write down the value of \(a\) and of \(b\).

[2]
a(i).

Explain what the gradient \(a\) represents.

[1]
a(ii).

Use the model to estimate the amount of fuel the car would use if it is driven \(110\) km.

[2]
b.
Answer/Explanation

Markscheme

\(a = 0.0823604{\text{, }}b = 0.306186\)

\(a = 0.0824{\text{, }}b = 0.306\)     A1A1     N2

[2 marks]

a(i).

correct explanation with reference to number of litres

required for \(1\) km     A1     N1

eg     \(a\) represents the (average) amount of fuel (litres) required to drive \(1\) km, (average) litres per kilometre, (average) rate of change in fuel used for each km travelled

[1 marks]

a(ii).

valid approach     (M1)

eg     \(y = 0.0824(110) + 0.306\), sketch

\(9.36583\)

\(9.37\) (litres)     A1 N2

[2 marks]

b.

Question

The following table shows the Diploma score \(x\) and university entrance mark \(y\) for seven IB Diploma students.

Find the correlation coefficient.

[2]
a.

The relationship can be modelled by the regression line with equation \(y = ax + b\).

Write down the value of \(a\) and of \(b\).

[2]
b.

Rita scored a total of \(26\) in her IB Diploma.

Use your regression line to estimate Rita’s university entrance mark.

[2]
c.
Answer/Explanation

Markscheme

evidence of set up     (M1)

eg\(\;\;\;\)correct value for \(r\) (or for \(a\) or \(r\), seen in (b))

\(0.996010\)

\(r = 0.996\;\;\;[0.996,{\text{ }}0.997]\)     A1     N2

[2 marks]

a.

\(a = 3.15037,{\text{ }}b =  – 15.4393\)

\(a = 3.15{\text{ }}[3.15,{\text{ }}3.16],{\text{ }}b =  – 15.4{\text{ }}[ – 15.5,{\text{ }} – 15.4]\)     A1A1     N2

[2 marks]

b.

substituting \(26\) into their equation     (M1)

eg\(\;\;\;\)\(y = 3.15(26) – 15.4\)

\(66.4704\)

\(66.5{\text{ }}[66.4,{\text{ }}66.5]\)     A1     N2

[2 marks]

Total [6 marks]

c.

Question

An environmental group records the numbers of coyotes and foxes in a wildlife reserve after \(t\) years, starting on 1 January 1995.

Let \(c\) be the number of coyotes in the reserve after \(t\) years. The following table shows the number of coyotes after \(t\) years.

The relationship between the variables can be modelled by the regression equation \(c = at + b\).

Find the value of \(a\) and of \(b\).

[3]
a.

Use the regression equation to estimate the number of coyotes in the reserve when \(t = 7\).

[3]
b.

Let \(f\) be the number of foxes in the reserve after \(t\) years. The number of foxes can be modelled by the equation \(f = \frac{{2000}}{{1 + 99{{\text{e}}^{ – kt}}}}\), where \(k\) is a constant.

Find the number of foxes in the reserve on 1 January 1995.

[3]
c.

After five years, there were 64 foxes in the reserve. Find \(k\).

[3]
d.

During which year were the number of coyotes the same as the number of foxes?

[4]
e.
Answer/Explanation

Markscheme

evidence of setup     (M1)

eg\(\;\;\;\)correct value for \(a\) or \(b\)

\(13.3823\), \(137.482\)

\(a{\rm{ }} = {\rm{ }}13.4\), \(b{\rm{ }} = {\rm{ }}137\)     A1A1     N3

[3 marks]

a.

correct substitution into their regression equation

eg\(\;\;\;13.3823 \times 7 + 137.482\)     (A1)

correct calculation

\(231.158\)     (A1)

\(231\) (coyotes) (must be an integer)     A1     N2

[3 marks]

b.

recognizing \(t = 0\)     (M1)

eg\(\;\;\;f(0)\)

correct substitution into the model

eg\(\;\;\;\frac{{2000}}{{1 + 99{{\text{e}}^{ – k(0)}}}},{\text{ }}\frac{{2000}}{{100}}\)     (A1)

\(20\) (foxes)     A1     N2

[3 marks]

c.

recognizing \((5,{\text{ }}64)\) satisfies the equation     (M1)

eg\(\;\;\;f(5) = 64\)

correct substitution into the model

eg\(\;\;\;64 = \frac{{2000}}{{1 + 99{{\text{e}}^{ – k(5)}}}},{\text{ }}64(1 + 99\(e\)^{ – 5k}}) = 2000\)     (A1)

\(0.237124\)

\(k =  – \frac{1}{5}\ln \left( {\frac{{11}}{{36}}} \right){\text{ (exact), }}0.237{\text{ }}[0.237,{\text{ }}0.238]\)     A1     N2

[3 marks]

d.

valid approach     (M1)

eg\(\;\;\;c = f\), sketch of graphs

correct working     (A1)

eg\(\;\;\;\frac{{2000}}{{1 + 99{{\text{e}}^{ – 0.237124t}}}} = 13.382t + 137.482\), sketch of graphs, table of values

\(t = 12.0403\)     (A1)

\(2007\)     A1     N2

Note:     Exception to the FT rule. Award A1FT on their value of \(t\).

[4 marks]

Total [16 marks]

e.

Question

The following table shows values of ln x and ln y.

The relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.

Find the value of a and of b.

[3]
a.

Use the regression equation to estimate the value of y when x = 3.57.

[3]
b.

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.

[7]
c.
Answer/Explanation

Markscheme

valid approach      (M1)

eg  one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14      A1A1 N3

[3 marks]

a.

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]

b.

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]

c.
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