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.
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.
Two trees are selected at random. Find the probability that they are both giants.
\(100\) trees are selected at random.
Find the expected number of these trees that are giants.
\(100\) trees are selected at random.
Find the probability that at least \(25\) of these trees are giants.
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]
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]
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]
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]
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]