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 P(G)=P(H>60, z=0.875, P(H>60)=1−0.809, N(53,82)
0.190786
P(G)=0.191 A1 N2
[3 marks]
finding P(H>70)=0.01679 (seen anywhere) (A1)
recognizing conditional probability (R1)
eg P(A|B), P(H>70|H>60)
correct working (A1)
eg 0.016790.191
0.0880209
P(X>70|G)=0.0880 A1 N3
[6 marks]
attempt to square their P(G) (M1)
eg 0.1912
0.0363996
P(both G)=0.0364 A1 N2
[2 marks]
correct substitution into formula for E(X) (A1)
eg 100(0.191)
E(G)=19.1 [19.0, 19.1] A1 N2
[3 marks]
recognizing binomial probability (may be seen in part (c)(i)) (R1)
eg X∼B(n, p)
valid approach (seen anywhere) (M1)
eg P(X⩾25)=1−P(X⩽24), 1−P(X<a)
correct working (A1)
eg P(X⩽24)=0.913…, 1−0.913…
0.0869002
P(X⩾25)=0.0869 A1 N2
[3 marks]