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Date May 2014 Marks available 3 Reference code 14M.2.sl.TZ2.10
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 10 Adapted from N/A

Question

A forest has a large number of tall trees. The heights of the trees are normally distributed with a mean of \(53\) metres and a standard deviation of \(8\) metres. Trees are classified as giant trees if they are more than \(60\) metres tall.

A tree is selected at random from the forest.

Find the probability that this tree is a giant.

[3]
a(i).

A tree is selected at random from the forest.

Given that this tree is a giant, find the probability that it is taller than \(70\) metres.

[3]
a(ii).

Two trees are selected at random. Find the probability that they are both giants.

[2]
b.

\(100\) trees are selected at random.

Find the expected number of these trees that are giants.

[3]
c(i).

\(100\) trees are selected at random.

Find the probability that at least \(25\) of these trees are giants.

[3]
c(ii).

Markscheme

valid approach     (M1)

eg     \({\text{P}}(G) = {\text{P}}(H > 60,{\text{ }}z = 0.875,{\text{ P}}(H > 60) = 1 - 0.809,{\text{ N}}\left( {53, {8^2}} \right)\)

\(0.190786\)

\({\text{P}}(G) = 0.191\)     A1     N2

[3 marks]

a(i).

finding \({\text{P}}(H > 70) = 0.01679\)  (seen anywhere)     (A1)

recognizing conditional probability     (R1)

eg     \({\text{P}}(A\left| {B),{\text{ P}}(H > 70\left| {H > 60)} \right.} \right.\)

correct working     (A1)

eg     \(\frac{{0.01679}}{{0.191}}\)

\(0.0880209\)

\({\text{P}}(X > 70\left| {G) = 0.0880} \right.\)     A1     N3

[6 marks]

a(ii).

attempt to square their \({\text{P}}(G)\)     (M1)

eg     \({0.191^2}\)

\(0.0363996\)

\({\text{P}}({\text{both }}G) = 0.0364\)     A1     N2

[2 marks]

b.

correct substitution into formula for \({\text{E}}(X)\)     (A1)

eg     \(100(0.191)\)

\({\text{E}}(G) = 19.1{\text{ }}[19.0,{\text{ }}19.1]\)     A1     N2

[3 marks]

c(i).

recognizing binomial probability (may be seen in part (c)(i))     (R1)

eg     \(X \sim {\text{B}}(n,{\text{ }}p)\)

valid approach (seen anywhere)     (M1)

eg     \({\text{P}}(X \geqslant 25) = 1 - {\text{P}}(X \leqslant 24),{\text{ }}1 - {\text{P}}(X < a)\)

correct working     (A1)

eg     \({\text{P}}(X \leqslant 24) = 0.913 \ldots ,{\text{ }}1 - 0.913 \ldots \)

\(0.0869002\)

\({\text{P}}(X \geqslant 25) = 0.0869\)     A1     N2

[3 marks]

c(ii).

Examiners report

[N/A]
a(i).
[N/A]
a(ii).
[N/A]
b.
[N/A]
c(i).
[N/A]
c(ii).

Syllabus sections

Topic 5 - Statistics and probability » 5.8 » Binomial distribution.
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