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 6.1 kg and variance 0.52 kg2.
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 4.5 kg and standard deviation 0.45 kg.
Ten female cats are chosen at random.
Sketch a diagram showing the above information.
Find the proportion of male Persian cats weighing between 5.5 kg and 6.5 kg.
Determine the expected number of cats in this group that have a weight of less than 5.3 kg.
Find the probability that exactly one of them weighs over 4.62 kg.
Let N be the number of cats weighing over 4.62 kg.
Find the variance of N.
A cat is selected at random from all 160 cats.
Find the probability that the cat was female, given that its weight was over 4.7 kg.
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]
X~N(6.1, 0.52)
P(5.5<X<6.5) OR labelled sketch of region (M1)
=0.673 (0.673074…) A1
[2 marks]
(P(X<5.3)=) 0.0547992… (A1)
0.0547992…×80 (M1)
=4.38 (4.38393…) A1
[3 marks]
Y~N(4.5, 0.452),
(P(Y>4.62)=) 0.394862… (A1)
use of binomial seen or implied (M1)
using B(10, 0.394862…) (M1)
0.0430 (0.0429664…) A1
[4 marks]
np(1-p)=2.39 (2.38946…) A1
[1 mark]
P(F∩(W>4.7))=0.5×0.3284 (=0.1642) (A1)
attempt use of tree diagram OR use of P(F W>4.7)=P(F∩(W>4.7))P(W>4.7) (M1)
0.5×0.32840.5×0.9974+0.5×0.3284 (A1)
=0.248 (0.247669…) A1
[4 marks]