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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 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 May 2022 TZ2 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

27 marks

Rice sales: Poisson fit, chi-squared test and whether advertising paid off

Statistics and probability

(a)(i)

2 marks

Use the statistics mode on the frequency table. There are $n = 90$ days.

\overset{x}{ˉ} = 4.23, s_{n - 1}^{2} = 4.32

(a)(ii)

1 mark

For a Poisson distribution the mean equals the variance. These two are close, which is why Aimmika believes the data are Poisson.

(b)

1 mark

Sales must occur independently of each other and at a constant average rate each day. Customers do not influence each other, and the average daily rate does not change over the period.

(c)

5 marks

Use the Poisson distribution with mean 4.2 and multiply by 90.

a = 90 P (X \leq 1) = 7.018 b = 90 P (X = 4) = 17.498 c = 90 P (X \geq 8) = 5.755

(d)(i)

1 mark

There are 8 cells and no parameters are estimated from the data.

ν = 8 - 1 = 7

(d)(ii)

7 marks

The observed frequencies for the cells $\leq 1, 2, 3, 4, 5, 6, 7, \geq 8$ are $9, 12, 11, 19, 14, 13, 8, 4$ .
Use a goodness of fit test with $ν = 7$ .
Since $0.728 > 0.05$ , there is insufficient evidence to reject $H_{0}$ . The data are consistent with a Poisson distribution with mean 4.2.

H_{0} : the data follow Po (4.2), H_{1} : the data do not follow Po (4.2) χ^{2} = \sum \frac{( O - E ) ^{2}}{E} \approx 4.44 p \approx 0.728

(e)(i)

6 marks

Over 60 days the mean under $H_{0}$ is $4.2 \times 60 = 252$ , so $X \sim Po (252)$ .
Find the smallest $c$ with $P (X \geq c) \leq 0.05$ .
The critical value is 279, so the critical region is $X \geq 279$ .
The observed 282 is in the critical region, so reject $H_{0}$ : there is evidence that the advertising increased the number of bags sold.

H_{0} : μ = 252, H_{1} : μ > 252 P (X \geq 278) = 0.0558, P (X \geq 279) = 0.0493

(e)(ii)

1 mark

The probability of a Type I error is the probability of landing in the critical region when $H_{0}$ is true.

P (X \geq 279) = 0.0493

(f)

3 marks

Extra bags sold: $282 - 252 = 30$ . The extra profit is
The advertising cost for 60 days is
Since $14850 < 18000$ , the advertising did not increase profit overall (an average of 247.50 THB extra per day against 300 THB per day). So Nichakarn is right on money, but Aimmika is right that sales rose. There may be other benefits, such as new customers who buy other things or return later.

30 \times 495 = 14850 300 \times 60 = 18000

Question 2

28 marks

School fibre-optic network: spanning trees, bounds and reliability

Geometry and trigonometry
Number and algebra
Statistics and probability

(a)

3 marks

Use the cosine rule in triangle ADF.

A F^{2} = 89. 2^{2} + 104. 9^{2} - 2 (89.2) (104.9) cos 8 3^{\circ} A F \approx 129 m

(b)

2 marks

Divide the cost by the length.

\frac{21310}{129.15} \approx $165 per metre

(c)

1 mark

The route from A to F crosses the lake, which makes cabling more difficult and more expensive, so the cost per metre is higher than the $110 per metre elsewhere.

(d)(i)

3 marks

Sort the edges and add the cheapest edge that does not form a cycle.
The edges GH, EG, AD and BE would each make a cycle, so they are skipped. Seven edges join the eight buildings.

C E (8239), D G (8668), B D (8778), A B (8811), D E (8833), E H (9251), D F (11539)

(d)(ii)

2 marks

Add the weights of the seven edges.

8239 + 8668 + 8778 + 8811 + 8833 + 9251 + 11539 = $64119

(e)

1 mark

A Hamiltonian cycle visits every vertex once but uses only some of the edges. An Eulerian circuit has to use every edge of the graph exactly once, so a Hamiltonian cycle is not an Eulerian circuit when the graph has other edges.

(f)

5 marks

Start at D and always take the cheapest edge to an unvisited building.
Return from F to D along the only edge back.
Add them all to get the upper bound.

D \to G (8668) \to H (9603) \to E (9251) \to C (8239) \to B (13156) \to A (8811) \to F (21310) F \to D (11539) 8668 + 9603 + 9251 + 8239 + 13156 + 8811 + 21310 + 11539 = $90577

(g)

6 marks

Delete D and find the minimum spanning tree of the other seven buildings with Kruskal's algorithm.
Now put D back using its two cheapest edges, DG (8668) and BD (8778).
So the cycle costs at least $76109.

C E (8239), A B (8811), E H (9251), G H (9603), B E (10153), F G (12606) total 58663 58663 + 8668 + 8778 = $76109

(h)

5 marks

A connection fails with probability 0.014. A cable has two ends, so it works only if both ends hold.
The cables form a single cycle through the 8 buildings. Such a network stays connected if at most one cable fails, and fails when two or more cables fail. Let $Y \sim B (8, 0.027804)$ .
Since $1.94% < 2%$ , the design meets the requirement.

P (a cable fails) = 1 - 0.98 6^{2} = 0.027804 P (network fails) = P (Y \geq 2) = 1 - P (Y \leq 1) = 0.0194

Mentor tip

Check with a different route: 16 connections give $P (\geq 2 fail) = 0.0206$ , but that is an overestimate because two failures at both ends of the same cable do not break the cycle.

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 probability32 marks · 58%
Q1, Q2
Number and algebra20 marks · 36%
Q2
Geometry and trigonometry3 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Group small expected counts · Q1
The last cell (8 or more) is built from the tail, and the first cell (0 or 1) from two values, so every expected frequency is at least 5.
02
Degrees of freedom · Q1
There are 8 cells and the mean 4.2 was given in advance, not estimated from the data, so $ν = 8 - 1 = 7$ .
03
Kruskal adds the cheapest edge that does not make a cycle · Q2
Sort the edges by weight and skip any edge that joins two vertices already connected.
04
Lower bound by deleting a vertex · Q2
Find the minimum spanning tree of the graph without D, then add the two cheapest edges at D.
05
Binomial for connections, then for cables · Q2
A cable fails if either of its two ends fails. First find that probability, then count failed cables.

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

Timezone

TZ1TZ2

Session: May 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 May 2022 TZ2 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

27 marks

Rice sales: Poisson fit, chi-squared test and whether advertising paid off

Statistics and probability

(a)(i)

2 marks

Use the statistics mode on the frequency table. There are $n = 90$ days.

\overset{x}{ˉ} = 4.23, s_{n - 1}^{2} = 4.32

(a)(ii)

1 mark

For a Poisson distribution the mean equals the variance. These two are close, which is why Aimmika believes the data are Poisson.

(b)

1 mark

Sales must occur independently of each other and at a constant average rate each day. Customers do not influence each other, and the average daily rate does not change over the period.

(c)

5 marks

Use the Poisson distribution with mean 4.2 and multiply by 90.

a = 90 P (X \leq 1) = 7.018 b = 90 P (X = 4) = 17.498 c = 90 P (X \geq 8) = 5.755

(d)(i)

1 mark

There are 8 cells and no parameters are estimated from the data.

ν = 8 - 1 = 7

(d)(ii)

7 marks

The observed frequencies for the cells $\leq 1, 2, 3, 4, 5, 6, 7, \geq 8$ are $9, 12, 11, 19, 14, 13, 8, 4$ .
Use a goodness of fit test with $ν = 7$ .
Since $0.728 > 0.05$ , there is insufficient evidence to reject $H_{0}$ . The data are consistent with a Poisson distribution with mean 4.2.

H_{0} : the data follow Po (4.2), H_{1} : the data do not follow Po (4.2) χ^{2} = \sum \frac{( O - E ) ^{2}}{E} \approx 4.44 p \approx 0.728

(e)(i)

6 marks

Over 60 days the mean under $H_{0}$ is $4.2 \times 60 = 252$ , so $X \sim Po (252)$ .
Find the smallest $c$ with $P (X \geq c) \leq 0.05$ .
The critical value is 279, so the critical region is $X \geq 279$ .
The observed 282 is in the critical region, so reject $H_{0}$ : there is evidence that the advertising increased the number of bags sold.

H_{0} : μ = 252, H_{1} : μ > 252 P (X \geq 278) = 0.0558, P (X \geq 279) = 0.0493

(e)(ii)

1 mark

The probability of a Type I error is the probability of landing in the critical region when $H_{0}$ is true.

P (X \geq 279) = 0.0493

(f)

3 marks

Extra bags sold: $282 - 252 = 30$ . The extra profit is
The advertising cost for 60 days is
Since $14850 < 18000$ , the advertising did not increase profit overall (an average of 247.50 THB extra per day against 300 THB per day). So Nichakarn is right on money, but Aimmika is right that sales rose. There may be other benefits, such as new customers who buy other things or return later.

30 \times 495 = 14850 300 \times 60 = 18000

Question 2

28 marks

School fibre-optic network: spanning trees, bounds and reliability

Geometry and trigonometry
Number and algebra
Statistics and probability

(a)

3 marks

Use the cosine rule in triangle ADF.

A F^{2} = 89. 2^{2} + 104. 9^{2} - 2 (89.2) (104.9) cos 8 3^{\circ} A F \approx 129 m

(b)

2 marks

Divide the cost by the length.

\frac{21310}{129.15} \approx $165 per metre

(c)

1 mark

The route from A to F crosses the lake, which makes cabling more difficult and more expensive, so the cost per metre is higher than the $110 per metre elsewhere.

(d)(i)

3 marks

Sort the edges and add the cheapest edge that does not form a cycle.
The edges GH, EG, AD and BE would each make a cycle, so they are skipped. Seven edges join the eight buildings.

C E (8239), D G (8668), B D (8778), A B (8811), D E (8833), E H (9251), D F (11539)

(d)(ii)

2 marks

Add the weights of the seven edges.

8239 + 8668 + 8778 + 8811 + 8833 + 9251 + 11539 = $64119

(e)

1 mark

(f)

5 marks

Start at D and always take the cheapest edge to an unvisited building.
Return from F to D along the only edge back.
Add them all to get the upper bound.

D \to G (8668) \to H (9603) \to E (9251) \to C (8239) \to B (13156) \to A (8811) \to F (21310) F \to D (11539) 8668 + 9603 + 9251 + 8239 + 13156 + 8811 + 21310 + 11539 = $90577

(g)

6 marks

Delete D and find the minimum spanning tree of the other seven buildings with Kruskal's algorithm.
Now put D back using its two cheapest edges, DG (8668) and BD (8778).
So the cycle costs at least $76109.

C E (8239), A B (8811), E H (9251), G H (9603), B E (10153), F G (12606) total 58663 58663 + 8668 + 8778 = $76109

(h)

5 marks

A connection fails with probability 0.014. A cable has two ends, so it works only if both ends hold.
The cables form a single cycle through the 8 buildings. Such a network stays connected if at most one cable fails, and fails when two or more cables fail. Let $Y \sim B (8, 0.027804)$ .
Since $1.94% < 2%$ , the design meets the requirement.

P (a cable fails) = 1 - 0.98 6^{2} = 0.027804 P (network fails) = P (Y \geq 2) = 1 - P (Y \leq 1) = 0.0194

Mentor tip

Check with a different route: 16 connections give $P (\geq 2 fail) = 0.0206$ , but that is an overestimate because two failures at both ends of the same cable do not break the cycle.

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 probability32 marks · 58%
Q1, Q2
Number and algebra20 marks · 36%
Q2
Geometry and trigonometry3 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Group small expected counts · Q1
The last cell (8 or more) is built from the tail, and the first cell (0 or 1) from two values, so every expected frequency is at least 5.
02
Degrees of freedom · Q1
There are 8 cells and the mean 4.2 was given in advance, not estimated from the data, so $ν = 8 - 1 = 7$ .
03
Kruskal adds the cheapest edge that does not make a cycle · Q2
Sort the edges by weight and skip any edge that joins two vertices already connected.
04
Lower bound by deleting a vertex · Q2
Find the minimum spanning tree of the graph without D, then add the two cheapest edges at D.
05
Binomial for connections, then for cables · Q2
A cable fails if either of its two ends fails. First find that probability, then count failed cables.

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