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Date May 2021 Marks available 4 Reference code 21M.2.AHL.TZ2.2
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number 2 Adapted from N/A

Question

It is known that the weights of male Persian cats are normally distributed with mean 6.1kg and variance 0.52kg2.

A group of 80 male Persian cats are drawn from this population.

The male cats are now joined by 80 female Persian cats. The female cats are drawn from a population whose weights are normally distributed with mean 4.5kg and standard deviation 0.45kg.

Ten female cats are chosen at random.

Sketch a diagram showing the above information.

[2]
a.

Find the proportion of male Persian cats weighing between 5.5kg and 6.5kg.

[2]
b.

Determine the expected number of cats in this group that have a weight of less than 5.3kg.

[3]
c.

Find the probability that exactly one of them weighs over 4.62kg.

[4]
d.i.

Let N be the number of cats weighing over 4.62kg.

Find the variance of N.

[1]
d.ii.

A cat is selected at random from all 160 cats.

Find the probability that the cat was female, given that its weight was over 4.7kg.

[4]
e.

Markscheme

                A1A1


Note: Award A1 for a normal curve with mean labelled 6.1 or μ, A1 for indication of SD (0.5): marks on horizontal axis at 5.6 and/or 6.6 OR μ-0.5 and/or μ+0.5 on the correct side and approximately correct position.

[2 marks]

a.

X~N6.1, 0.52

P5.5<X<6.5  OR  labelled sketch of region                (M1)

=0.673  0.673074                A1


[2 marks]

b.

PX<5.3= 0.0547992                (A1)

0.0547992×80                (M1)

=4.38   4.38393                A1


[3 marks]

c.

Y~N4.5, 0.452,

PY>4.62= 0.394862                (A1)

use of binomial seen or implied                (M1)

using B10, 0.394862                (M1)

0.0430  0.0429664                A1


[4 marks]

d.i.

np1-p=2.39  2.38946                A1


[1 mark]

d.ii.

PFW>4.7=0.5×0.3284  =0.1642                    (A1)

attempt use of tree diagram OR use of PFW>4.7=PFW>4.7PW>4.7                    (M1)

0.5×0.32840.5×0.9974+0.5×0.3284                    (A1)

=0.248  0.247669                A1


[4 marks]

e.

Examiners report

[N/A]
a.
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b.
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c.
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d.i.
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d.ii.
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e.

Syllabus sections

Topic 4—Statistics and probability » SL 4.8—Binomial distribution
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Topic 4—Statistics and probability

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