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Date	May 2018	Marks available	1	Reference code	18M.2.SL.TZ1.T_2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Write down	Question number	T_2	Adapted from	N/A

Question

On one day 180 flights arrived at a particular airport. The distance travelled and the arrival status for each incoming flight was recorded. The flight was then classified as on time, slightly delayed, or heavily delayed.

The results are shown in the following table.

A χ² test is carried out at the 10 % significance level to determine whether the arrival status of incoming flights is independent of the distance travelled.

The critical value for this test is 7.779.

A flight is chosen at random from the 180 recorded flights.

State the alternative hypothesis.

[1]

Calculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.

[2]

Write down the number of degrees of freedom.

[1]

Write down the χ² statistic.

[2]

d.i.

Write down the associated p-value.

[1]

d.ii.

State, with a reason, whether you would reject the null hypothesis.

[2]

Write down the probability that this flight arrived on time.

[2]

Given that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.

[2]

Two flights are chosen at random from those which were slightly delayed.

Find the probability that each of these flights travelled at least 5000 km.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

The arrival status is dependent on the distance travelled by the incoming flight (A1)

Note: Accept “associated” or “not independent”.

[1 mark]

$\frac{{60 \times 45}}{{180}}$ OR $\frac{{60}}{{180}} \times \frac{{45}}{{180}} \times 180$ (M1)

Note: Award (M1) for correct substitution into expected value formula.

= 15 (A1) (G2)

[2 marks]

4 (A1)

Note: Award (A0) if “2 + 2 = 4” is seen.

[1 mark]

9.55 (9.54671…) (G2)

Note: Award (G1) for an answer of 9.54.

[2 marks]

d.i.

0.0488 (0.0487961…) (G1)

[1 mark]

d.ii.

Reject the Null Hypothesis (A1)(ft)

Note: Follow through from their hypothesis in part (a).

9.55 (9.54671…) > 7.779 (R1)(ft)

0.0488 (0.0487961…) < 0.1 (R1)(ft)

Note: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ²_calc > χ²_crit, provided the calculated value is explicitly seen in part (d)(i).

[2 marks]

$\frac{{52}}{{180}}\,\,\left( {0.289,\,\,\frac{{13}}{{45}},\,\,28.9\,{\text{% }}} \right)$ (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{{35}}{{97}}\,\,\left( {0.361,\,\,36.1\,{\text{% }}} \right)$ (A1)(A1) (G2)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

$\frac{{14}}{{45}} \times \frac{{13}}{{44}}$ (A1)(M1)

Note: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.

$= \frac{{182}}{{1980}}\,\,\left( {0.0919,\,\,\frac{{91}}{{990}},\,0.091919 \ldots ,\,9.19\,{\text{% }}} \right)$ (A1) (G2)

[3 marks]

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.2—Histograms, CF graphs, box plots

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Topic 4—Statistics and probability

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Markscheme

Examiners report

Syllabus sections

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