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

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

0.0668072

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

[2 marks]

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]

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.

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.