Find the probability that in a certain week (Monday to Friday only)

(i)     there are fewer than 12 accidents;

(ii)     there are more than 8 accidents, given that there are fewer than 12 accidents.

Due to the increased usage, it is found that the probability of more than 3 accidents in a day at the weekend (Saturday and Sunday) is 0.24.

Assuming a Poisson model,

(i)     calculate the mean number of accidents per day at the weekend (Saturday and Sunday);

(ii)     calculate the probability that, in the four weekends in February, there will be more than 5 accidents during at least two of the weekends.

It is found that 20 % of skiers having accidents are at least 25 years of age and 40 % are under 18 years of age.

Assuming that the ages of skiers having accidents are normally distributed, find the mean age of skiers having accidents.

(i)     \(X \sim {\text{Po}}(11)\)     (M1)

\({\text{P}}(X \leqslant 11) = 0.579\)     (M1)A1

(ii)     \({\text{P}}(X > 8\left| {x < 12) = } \right.\)     (M1)     

\( = \frac{{{\text{P}}(8 < X < 12)}}{{{\text{P}}(X < 12)}}{\text{ }}\left( {{\text{or }}\frac{{{\text{P}}(X \leqslant 11) - {\text{P}}(X \leqslant 8)}}{{{\text{P}}(X \leqslant 11)}}{\text{ or }}\frac{{0.3472...}}{{0.5792...}}} \right)\)     A1

\( = 0.600\)     A1     N2

[6 marks]

(i)     \(Y \sim {\text{Po}}(m)\)

\({\text{P}}(Y > 3) = 0.24\)     (M1)

\({\text{P}}(Y \leqslant 3) = 0.76\)     (M1)

\({{\text{e}}^{ - m}}\left( {1 + m + \frac{1}{2}{m^2} + \frac{1}{6}{m^3}} \right) = 0.76\)     (A1) 

Note: At most two of the above lines can be implied.

Attempt to solve equation with GDC     (M1)

\(m = 2.49\)     A1

(ii)     \(A \sim {\text{Po}}(4.98)\)

\({\text{P}}(A > 5) = 1 - {\text{P}}(A \leqslant 5) = 0.380...\)     M1A1

\(W \sim {\text{B}}(4,\,0.380...)\)     (M1)

\({\text{P}}(W \geqslant 2) = 1 - {\text{P}}(W \leqslant 1) = 0.490\)     M1A1

 [10 marks]

\({\text{P}}(A < 25) = 0.8,{\text{ P}}(A < 18) = 0.4\)

\(\frac{{25 - \mu }}{\sigma } = 0.8416...\)     (M1)(A1)

\(\frac{{18 - \mu }}{\sigma } = -0.2533...{\text{ (or}} -0.2534{\text{ from tables)}}\)     (M1)(A1)

solving these equations     (M1) 

\(\mu  = 19.6\)     A1 

Note: Accept just 19.6, 19 or 20; award A0 to any other final answer.

[6 marks]

Generally, candidates had difficulties with this question, mainly in applying conditional probability and interpreting the expressions ‘more than’, ‘at least’ and ‘under’ to obtain correct expressions. Although many candidates identified the binomial distribution in part (b) (ii), very few succeeded in answering this question due to incorrect interpretation of the question or due to accuracy errors.

Generally, candidates had difficulties with this question, mainly in applying conditional probability and interpreting the expressions ‘more than’, ‘at least’ and ‘under’ to obtain correct expressions. Although many candidates identified the binomial distribution in part (b) (ii), very few succeeded in answering this question due to incorrect interpretation of the question or due to accuracy errors.

Generally, candidates had difficulties with this question, mainly in applying conditional probability and interpreting the expressions ‘more than’, ‘at least’ and ‘under’ to obtain correct expressions. Although many candidates identified the binomial distribution in part (b) (ii), very few succeeded in answering this question due to incorrect interpretation of the question or due to accuracy errors.