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IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Session: November 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

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Questions

Questions 1–2 · 55 marks

Question 1

29 marks

Predicting football results: a binomial fit and a Markov model

Statistics and probability

(a)

2 marks

Mean of the 60 blocks.

\overset{x}{ˉ} = \frac{0 ( 0 ) + 1 ( 11 ) + 2 ( 21 ) + 3 ( 21 ) + 4 ( 7 )}{60} = \frac{144}{60} = 2.4

(b)

1 mark

The mean of a $B (4, p)$ is $4 p$ .

4 p = 2.4 \Rightarrow p = 0.6

(c)(i)

1 mark

No wins in four matches.

P (X = 0) = 0. 4^{4} = 0.0256

(c)(ii)

2 marks

Multiply by the number of blocks.

60 \times 0.0256 = 1.536

(d)(i)

1 mark

The null hypothesis says the model fits the data.

H_{0} : the number of wins in a block follows B (4, p)

(d)(ii)

1 mark

After combining rows there are 4 cells, and one parameter ( $p$ ) was estimated.

ν = 4 - 1 - 1 = 2

(d)(iii)

2 marks

Run a goodness of fit test with the observed $11, 21, 21, 7$ and expected $10.8, 20.7, 20.7, 7.8$ .

χ^{2} = 0.0945, p = 0.954

(d)(iv)

2 marks

Compare $p$ with 0.05.
Do not reject $H_{0}$ : the data are consistent with a binomial model.

0.954 > 0.05

(e)

2 marks

At least one win is the complement of no wins.

1 - 0. 4^{4} = 0.9744

(f)(i)

2 marks

After a win there were $85 + 60 = 145$ results, and 85 of them were wins.

\frac{85}{145} = \frac{17}{29}

(f)(ii)

2 marks

After a win: 85 wins, 60 not. After a non-win: 62 wins, 32 not ( $62 + 32 = 94$ ). Each column adds to 1.

T = (\frac{17}{29} \frac{12}{29} \frac{31}{47} \frac{16}{47})

(g)(i)

3 marks

The characteristic polynomial is $λ^{2} - (trace) λ + det T$ .

trace = \frac{17}{29} + \frac{16}{47} = \frac{1263}{1363}, det T = \frac{272 - 372}{1363} = - \frac{100}{1363} λ^{2} - \frac{1263}{1363} λ - \frac{100}{1363} = 0 \Rightarrow 1363 λ^{2} - 1263 λ - 100 = 0

(g)(ii)

1 mark

Solve the quadratic.

λ = 1, λ = - \frac{100}{1363}

(g)(iii)

3 marks

For $λ = 1$ : $- \frac{12}{29} x + \frac{31}{47} y = 0$ , so $y = \frac{564}{899} x$ . For $λ = - \frac{100}{1363}$ : $\frac{899}{1363} x + \frac{31}{47} y = 0$ , so $y = - x$ .

v_{1} = (899564), v_{2} = (1 - 1)

(h)

4 marks

Far into the future the state is the eigenvector for 1, scaled to total 1.
To win three in a row, win the first, then win twice more with probability $\frac{17}{29}$ each time.

P (A) = \frac{899}{899 + 564} = \frac{899}{1463} \approx 0.6145 \frac{899}{1463} (\frac{17}{29})^{2} \approx 0.211

Question 2

26 marks

A drug given repeatedly: exponential decay and a geometric sum

Calculus

(a)

3 marks

Separate the variables.
At $t = 0$ , $C = d$ .

\frac{d C}{d t} = - k C \Rightarrow \int \frac{d C}{C} = - k \int d t \Rightarrow ln C = - k t + c C = d e^{- k t}

(b)

2 marks

Set $C = 0.05 d$ with $k = 0.2$ .

e^{- 0.2 t} = 0.05 \Rightarrow t = \frac{ln 20}{0.2} = 15.0 hours

(c)

4 marks

After the first dose: $d$ . Just before the second: $d e^{- 0.2 T}$ , and after it $d + d e^{- 0.2 T}$ . Just before the third: $(d + d e^{- 0.2 T}) e^{- 0.2 T}$ .

d + (d + d e^{- 0.2 T}) e^{- 0.2 T} = d (1 + e^{- 0.2 T} + e^{- 0.4 T})

(d)

2 marks

The sum $1 + e^{- 0.2 T} + e^{- 0.4 T} + \dots + e^{- 0.2 (n - 1) T}$ is geometric with ratio $e^{- 0.2 T}$ and $n$ terms.

d \cdot \frac{1 - e ^{- 0.2 n T}}{1 - e ^{- 0.2 T}}

(e)(i)

2 marks

As $n \to \infty$ , $e^{- 0.2 n T} \to 0$ .

H_{\infty} = \frac{d}{1 - e ^{- 0.2 T}}

(e)(ii)

3 marks

The lowest concentration is just before the next dose, after a decay of $T$ hours from the highest.

L_{\infty} = H_{\infty} e^{- 0.2 T} = \frac{d e ^{- 0.2 T}}{1 - e ^{- 0.2 T}}

(f)(i)

2 marks

Subtract.

H_{\infty} - L_{\infty} = \frac{d ( 1 - e ^{- 0.2 T} )}{1 - e ^{- 0.2 T}} = d

(f)(ii)

3 marks

Divide $H_{\infty}$ by $L_{\infty}$ .

\frac{H _{\infty}}{L _{\infty}} = e^{0.2 T} \Rightarrow 0.2 T = ln \frac{H _{\infty}}{L _{\infty}} \Rightarrow T = 5 ln \frac{H _{\infty}}{L _{\infty}}

(g)(i)

1 mark

We need $L_{\infty} \geq 0.06$ and $H_{\infty} \leq 0.28$ , and $H_{\infty} - L_{\infty} = d$ , so $d \leq 0.22$ . A suitable choice is

d = 0.2

(g)(ii)

1 mark

Choose $L_{\infty} = 0.06$ and $H_{\infty} = 0.26$ , which add up correctly ( $0.26 - 0.06 = 0.2$ ).

T = 5 ln \frac{0.26}{0.06} \approx 7.33 hours

(h)

2 marks

After the first dose $C = 0.2 e^{- 0.2 t}$ . It stays at or above 0.06 until
As a proportion of the 7.33 hours between doses:

0.2 e^{- 0.2 t} = 0.06 \Rightarrow t = 5 ln \frac{0.2}{0.06} = 6.02 hours \frac{6.02}{7.33} \approx 0.821

(i)

1 mark

The label might say "every 8 hours" or "three times a day". A simple, regular time is easy to remember and fit into a person's day, whereas 7.33 hours is awkward.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Statistics and probability carries the most marks on this paper.

Statistics and probability29 marks · 53%
Q1
Calculus26 marks · 47%
Q2

Mentor insights

Traps to watch for

01
Degrees of freedom when p is estimated · Q1
There are 4 cells and the parameter $p$ was estimated from the data, so $ν = 4 - 1 - 1 = 2$ .
02
Columns of the transition matrix · Q1
Each column holds the probabilities of moving out of one state, so each column adds to 1.
03
Long-run probability · Q1
The long-run state is the eigenvector for eigenvalue 1 scaled to add to 1.
04
Instantaneous top-ups · Q2
Each dose adds $d$ at once. Between doses the concentration decays by the factor $e^{- 0.2 T}$ .
05
Limits of a geometric series · Q2
As $n \to \infty$ the term $e^{- 0.2 n T} \to 0$ , which gives $H_{\infty}$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Session: November 2022
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 Global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

29 marks

Predicting football results: a binomial fit and a Markov model

Statistics and probability

(a)

2 marks

Mean of the 60 blocks.

\overset{x}{ˉ} = \frac{0 ( 0 ) + 1 ( 11 ) + 2 ( 21 ) + 3 ( 21 ) + 4 ( 7 )}{60} = \frac{144}{60} = 2.4

(b)

1 mark

The mean of a $B (4, p)$ is $4 p$ .

4 p = 2.4 \Rightarrow p = 0.6

(c)(i)

1 mark

No wins in four matches.

P (X = 0) = 0. 4^{4} = 0.0256

(c)(ii)

2 marks

Multiply by the number of blocks.

60 \times 0.0256 = 1.536

(d)(i)

1 mark

The null hypothesis says the model fits the data.

H_{0} : the number of wins in a block follows B (4, p)

(d)(ii)

1 mark

After combining rows there are 4 cells, and one parameter ( $p$ ) was estimated.

ν = 4 - 1 - 1 = 2

(d)(iii)

2 marks

Run a goodness of fit test with the observed $11, 21, 21, 7$ and expected $10.8, 20.7, 20.7, 7.8$ .

χ^{2} = 0.0945, p = 0.954

(d)(iv)

2 marks

Compare $p$ with 0.05.
Do not reject $H_{0}$ : the data are consistent with a binomial model.

0.954 > 0.05

(e)

2 marks

At least one win is the complement of no wins.

1 - 0. 4^{4} = 0.9744

(f)(i)

2 marks

After a win there were $85 + 60 = 145$ results, and 85 of them were wins.

\frac{85}{145} = \frac{17}{29}

(f)(ii)

2 marks

After a win: 85 wins, 60 not. After a non-win: 62 wins, 32 not ( $62 + 32 = 94$ ). Each column adds to 1.

T = (\frac{17}{29} \frac{12}{29} \frac{31}{47} \frac{16}{47})

(g)(i)

3 marks

The characteristic polynomial is $λ^{2} - (trace) λ + det T$ .

trace = \frac{17}{29} + \frac{16}{47} = \frac{1263}{1363}, det T = \frac{272 - 372}{1363} = - \frac{100}{1363} λ^{2} - \frac{1263}{1363} λ - \frac{100}{1363} = 0 \Rightarrow 1363 λ^{2} - 1263 λ - 100 = 0

(g)(ii)

1 mark

Solve the quadratic.

λ = 1, λ = - \frac{100}{1363}

(g)(iii)

3 marks

For $λ = 1$ : $- \frac{12}{29} x + \frac{31}{47} y = 0$ , so $y = \frac{564}{899} x$ . For $λ = - \frac{100}{1363}$ : $\frac{899}{1363} x + \frac{31}{47} y = 0$ , so $y = - x$ .

v_{1} = (899564), v_{2} = (1 - 1)

(h)

4 marks

Far into the future the state is the eigenvector for 1, scaled to total 1.
To win three in a row, win the first, then win twice more with probability $\frac{17}{29}$ each time.

P (A) = \frac{899}{899 + 564} = \frac{899}{1463} \approx 0.6145 \frac{899}{1463} (\frac{17}{29})^{2} \approx 0.211

Question 2

26 marks

A drug given repeatedly: exponential decay and a geometric sum

Calculus

(a)

3 marks

Separate the variables.
At $t = 0$ , $C = d$ .

\frac{d C}{d t} = - k C \Rightarrow \int \frac{d C}{C} = - k \int d t \Rightarrow ln C = - k t + c C = d e^{- k t}

(b)

2 marks

Set $C = 0.05 d$ with $k = 0.2$ .

e^{- 0.2 t} = 0.05 \Rightarrow t = \frac{ln 20}{0.2} = 15.0 hours

(c)

4 marks

After the first dose: $d$ . Just before the second: $d e^{- 0.2 T}$ , and after it $d + d e^{- 0.2 T}$ . Just before the third: $(d + d e^{- 0.2 T}) e^{- 0.2 T}$ .

d + (d + d e^{- 0.2 T}) e^{- 0.2 T} = d (1 + e^{- 0.2 T} + e^{- 0.4 T})

(d)

2 marks

The sum $1 + e^{- 0.2 T} + e^{- 0.4 T} + \dots + e^{- 0.2 (n - 1) T}$ is geometric with ratio $e^{- 0.2 T}$ and $n$ terms.

d \cdot \frac{1 - e ^{- 0.2 n T}}{1 - e ^{- 0.2 T}}

(e)(i)

2 marks

As $n \to \infty$ , $e^{- 0.2 n T} \to 0$ .

H_{\infty} = \frac{d}{1 - e ^{- 0.2 T}}

(e)(ii)

3 marks

The lowest concentration is just before the next dose, after a decay of $T$ hours from the highest.

L_{\infty} = H_{\infty} e^{- 0.2 T} = \frac{d e ^{- 0.2 T}}{1 - e ^{- 0.2 T}}

(f)(i)

2 marks

Subtract.

H_{\infty} - L_{\infty} = \frac{d ( 1 - e ^{- 0.2 T} )}{1 - e ^{- 0.2 T}} = d

(f)(ii)

3 marks

Divide $H_{\infty}$ by $L_{\infty}$ .

\frac{H _{\infty}}{L _{\infty}} = e^{0.2 T} \Rightarrow 0.2 T = ln \frac{H _{\infty}}{L _{\infty}} \Rightarrow T = 5 ln \frac{H _{\infty}}{L _{\infty}}

(g)(i)

1 mark

We need $L_{\infty} \geq 0.06$ and $H_{\infty} \leq 0.28$ , and $H_{\infty} - L_{\infty} = d$ , so $d \leq 0.22$ . A suitable choice is

d = 0.2

(g)(ii)

1 mark

Choose $L_{\infty} = 0.06$ and $H_{\infty} = 0.26$ , which add up correctly ( $0.26 - 0.06 = 0.2$ ).

T = 5 ln \frac{0.26}{0.06} \approx 7.33 hours

(h)

2 marks

After the first dose $C = 0.2 e^{- 0.2 t}$ . It stays at or above 0.06 until
As a proportion of the 7.33 hours between doses:

0.2 e^{- 0.2 t} = 0.06 \Rightarrow t = 5 ln \frac{0.2}{0.06} = 6.02 hours \frac{6.02}{7.33} \approx 0.821

(i)

1 mark

The label might say "every 8 hours" or "three times a day". A simple, regular time is easy to remember and fit into a person's day, whereas 7.33 hours is awkward.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Statistics and probability carries the most marks on this paper.

Statistics and probability29 marks · 53%
Q1
Calculus26 marks · 47%
Q2

Mentor insights

Traps to watch for

01
Degrees of freedom when p is estimated · Q1
There are 4 cells and the parameter $p$ was estimated from the data, so $ν = 4 - 1 - 1 = 2$ .
02
Columns of the transition matrix · Q1
Each column holds the probabilities of moving out of one state, so each column adds to 1.
03
Long-run probability · Q1
The long-run state is the eigenvector for eigenvalue 1 scaled to add to 1.
04
Instantaneous top-ups · Q2
Each dose adds $d$ at once. Between doses the concentration decays by the factor $e^{- 0.2 T}$ .
05
Limits of a geometric series · Q2
As $n \to \infty$ the term $e^{- 0.2 n T} \to 0$ , which gives $H_{\infty}$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers