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Date May 2016 Marks available 2 Reference code 16M.2.sl.TZ2.6
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 6 Adapted from N/A

Question

A competition consists of two independent events, shooting at 100 targets and running for one hour.

The number of targets a contestant hits is the \(S\) score. The \(S\) scores are normally distributed with mean 65 and standard deviation 10.

The distance in km that a contestant runs in one hour is the \(R\) score. The \(R\) scores are normally distributed with mean 12 and standard deviation 2.5. The \(R\) score is independent of the \(S\) score.

Contestants are disqualified if their \(S\) score is less than 50 and their \(R\) score is less than \(x\) km.

A contestant is chosen at random. Find the probability that their \(S\) score is less than 50.

[2]
a.

Given that 1% of the contestants are disqualified, find the value of \(x\).

[4]
b.

Markscheme

0.0668072

\({\text{P}}(S < 50) = 0.0668{\text{ }}({\text{accept P}}(S \leqslant 49) = 0.0548)\)     A2     N2

[2 marks]

a.

valid approach     (M1)

Eg\(\,\,\,\,\,\)\({\text{P}}(S < 50) \times {\text{P}}(R < x)\)

correct equation (accept any variable)     A1

eg\(\,\,\,\,\,\)\({\text{P}}(S < 50) \times {\text{P}}(R < x) = 1\% ,{\text{ }}0.0668072 \times p = 0.01,{\text{ P}}(R < x) = \frac{{0.01}}{{0.0668}}\)

finding the value of \({\text{P}}(R < x)\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{0.01}}{{0.0668}},{\text{ }}0.149684\)

9.40553

\(x = 9.41{\text{ }}({\text{accept }}x = 9.74{\text{ from }}0.0548)\)     A1     N3

[4 marks]

b.

Examiners report

The first part of this question was a direct application of the normal distribution and most candidates who attempted the question obtained the correct value. In some cases, candidates gave the answer to 2 or 1 sf, losing a mark and taking the risk of obtaining an incorrect answer in the following question.

a.

Part b) proved challenging for various reasons. Many did not recognize that 0.01 was the probability of an intersection. Others did not know how to find that probability using the fact that the events were independent. Some candidates thought that the independence formula was \({\text{P}}(A) + {\text{P}}(B) = 0.01\) instead of \({\text{P}}(A) \times {\text{P}}(B) = 0.01\).

Of those that were able to find the correct value of \({\text{P}}(R < x)\), only some continued to find the value of \(x\).

Premature rounding in the answer to (a) sometimes caused the final mark in (b) to be lost unnecessarily.

b.

Syllabus sections

Topic 5 - Statistics and probability » 5.9 » Normal distributions and curves.
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