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Date May Example question Marks available 2 Reference code EXM.1.AHL.TZ0.58
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number 58 Adapted from N/A

Question

The number of telephone calls received by a helpline over 80 one-minute periods are summarized in the table below.

Find the exact value of the mean of this distribution.

[2]
a.

Test, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution.

[12]
b.

Markscheme

Mean  λ = ( 9 × 0 + 12 × 1 + 22 × 2 + 10 × 3 + 11 × 4 + 8 × 5 + 8 × 6 ) 80         (M1)

              = 2.725 = ( 109 40 )         A1

Note: Do not accept 2.73.

[2 marks]

a.

H0: the data can be modelled by a Poisson distribution            A1

H1: the data cannot be modelled by a Poisson distribution            A1

        A3

Note:  Award A2 for one error, A1 for two errors, A0 for three or more errors.

Combining last two columns                (M1)

Note:  Allow FT from not combining the last two columns and / or getting 2.98 for the last expected frequency.

EITHER

χ 2 = 9 2 5.244 + 12 2 14.289 + 22 2 19.469 + 10 2 17.684 + 11 2 12.047 + 16 2 11.267 80         (M1)(A1)

               = 8.804  (accept 8.8)            A1

v = 6 2 = 4 ,   χ 5 2 = 9.488             A1A1

Hence 8.804 is not significant since 8.804 < 9.488 and we accept H0            R1

OR

p-value = 0.0662    (accept 0.066) which is not significant since              A5

0.0662 > 0.05 and we accept H0          R1  N0

[12 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 4—Statistics and probability » SL 4.11—Expected, observed, hypotheses, chi squared, gof, t-test
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Topic 4—Statistics and probability » AHL 4.12—Data collection, reliability and validity tests
Topic 4—Statistics and probability » AHL 4.17—Poisson distribution
Topic 4—Statistics and probability

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