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Date November Example question Marks available 1 Reference code EXN.3.AHL.TZ0.1
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term State Question number 1 Adapted from N/A

Question

An estate manager is responsible for stocking a small lake with fish. He begins by introducing 1000 fish into the lake and monitors their population growth to determine the likely carrying capacity of the lake.

After one year an accurate assessment of the number of fish in the lake is taken and it is found to be 1200.

Let N be the number of fish t years after the fish have been introduced to the lake.

Initially it is assumed that the rate of increase of N will be constant.

When t=8 the estate manager again decides to estimate the number of fish in the lake. To do this he first catches 300 fish and marks them, so they can be recognized if caught again. These fish are then released back into the lake. A few days later he catches another 300 fish, releasing each fish after it has been checked, and finds 45 of them are marked.

Let X be the number of marked fish caught in the second sample, where X is considered to be distributed as Bn,p. Assume the number of fish in the lake is 2000.

The estate manager decides that he needs bounds for the total number of fish in the lake.

The estate manager feels confident that the proportion of marked fish in the lake will be within 1.5 standard deviations of the proportion of marked fish in the sample and decides these will form the upper and lower bounds of his estimate.

The estate manager now believes the population of fish will follow the logistic model Nt=L1+Ce-kt where L is the carrying capacity and C,k>0.

The estate manager would like to know if the population of fish in the lake will eventually reach 5000.

Use this model to predict the number of fish in the lake when t=8.

[2]
a.

Assuming the proportion of marked fish in the second sample is equal to the proportion of marked fish in the lake, show that the estate manager will estimate there are now 2000 fish in the lake.

[2]
b.

Write down the value of n and the value of p.

[2]
c.i.

State an assumption that is being made for X to be considered as following a binomial distribution.

[1]
c.ii.

Show that an estimate for Var(X) is 38.25.

[2]
d.i.

Hence show that the variance of the proportion of marked fish in the sample, VarX300, is 0.000425.

[2]
d.ii.

Taking the value for the variance given in (d) (ii) as a good approximation for the true variance, find the upper and lower bounds for the proportion of marked fish in the lake.

[2]
e.i.

Hence find upper and lower bounds for the number of fish in the lake when t=8.

[2]
e.ii.

Given this result, comment on the validity of the linear model used in part (a).

[2]
f.

Assuming a carrying capacity of 5000 use the given values of N0 and N1 to calculate the parameters C and k.

[5]
g.i.

Use these parameters to calculate the value of N8 predicted by this model.

[2]
g.ii.

Comment on the likelihood of the fish population reaching 5000.

[2]
h.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

N8=1000+200×8        M1

=2600        A1

 

[2 marks]

a.

45300=300N        M1A1

N=2000        AG

 

[2 marks]

b.

n=300, p=3002000=0.15     A1A1

 

[2 marks]

c.i.

Any valid reason for example:          R1

Marked fish are randomly distributed, so p constant.

Each fish caught is independent of previous fish caught

 

[1 mark]

c.ii.

Var(X)=np1-p          M1

=300×3002000×17002000          A1

=38.25          AG

 

[2 marks]

d.i.

VarX300=Var(X)3002          M1A1

=0.000425             AG

 

[2 marks]

d.ii.

0.15±1.50.000425          (M1)

0.181 and 0.119             A1

 

[2 marks]

e.i.

300N=0.181,  300N=0.119            M1

Lower bound 1658 upper bound 2519             A1

 

[2 marks]

e.ii.

Linear model prediction falls outside this range so unlikely to be a good model            R1A1

 

[2 marks]

f.

1000=50001+C            M1

C=4            A1

1200=50001+4e-k            M1

e-k=38004×1200            (M1)

k=-ln0.7916=0.2336         A1

 

[5 marks]

g.i.

N8=50001+4e-0.2336×8=3090         M1A1

 

Note: Accept any answer that rounds to 3000.

 

[2 marks]

g.ii.

This is much higher than the calculated upper bound for N(8) so the rate of growth of the fish is unlikely to be sufficient to reach a carrying capacity of 5000.        M1R1

 

[2 marks]

h.

Examiners report

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

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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