AQA S2 2013 January — Question 3 11 marks

Exam BoardAQA
ModuleS2 (Statistics 2)
Year2013
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSum of Poisson processes
TypeMinimum time or stock level
DifficultyStandard +0.3 This is a straightforward application of Poisson distribution properties with standard calculations: single Poisson probabilities, sum of independent Poisson variables (weekday + weekend totaling λ=9.5 for the week), and finding a threshold using cumulative probabilities. All parts follow textbook procedures with no novel insight required, though part (b) requires slightly more careful work with tables/calculator. Slightly easier than average due to clear structure and routine methods.
Spec5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson

3 A large office block is busy during the five weekdays, Monday to Friday, and less busy during the two weekend days, Saturday and Sunday. The block is illuminated by fluorescent light tubes which frequently fail and must be replaced with new tubes by John, the caretaker. The number of fluorescent tubes that fail on a particular weekday can be modelled by a Poisson distribution with mean 1.5. The number of fluorescent tubes that fail on a particular weekend day can be modelled by a Poisson distribution with mean 0.5 .
  1. Find the probability that:
    1. on one particular Monday, exactly 3 fluorescent light tubes fail;
    2. during the two days of a weekend, more than 1 fluorescent light tube fails;
    3. during a complete seven-day week, fewer than 10 fluorescent light tubes fail.
  2. John keeps a supply of new fluorescent light tubes. More new tubes are delivered every Monday morning to replace those that he has used during the previous week. John wants the probability that he runs out of new tubes before the next Monday morning to be less than 1 per cent. Find the minimum number of new tubes that he should have available on a Monday morning.
  3. Give a reason why a Poisson distribution with mean 0.375 is unlikely to provide a satisfactory model for the number of fluorescent light tubes that fail between 1 am and 7 am on a weekday.

3 A large office block is busy during the five weekdays, Monday to Friday, and less busy during the two weekend days, Saturday and Sunday. The block is illuminated by fluorescent light tubes which frequently fail and must be replaced with new tubes by John, the caretaker.

The number of fluorescent tubes that fail on a particular weekday can be modelled by a Poisson distribution with mean 1.5.

The number of fluorescent tubes that fail on a particular weekend day can be modelled by a Poisson distribution with mean 0.5 .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that:
\begin{enumerate}[label=(\roman*)]
\item on one particular Monday, exactly 3 fluorescent light tubes fail;
\item during the two days of a weekend, more than 1 fluorescent light tube fails;
\item during a complete seven-day week, fewer than 10 fluorescent light tubes fail.
\end{enumerate}\item John keeps a supply of new fluorescent light tubes. More new tubes are delivered every Monday morning to replace those that he has used during the previous week. John wants the probability that he runs out of new tubes before the next Monday morning to be less than 1 per cent. Find the minimum number of new tubes that he should have available on a Monday morning.
\item Give a reason why a Poisson distribution with mean 0.375 is unlikely to provide a satisfactory model for the number of fluorescent light tubes that fail between 1 am and 7 am on a weekday.
\end{enumerate}

\hfill \mbox{\textit{AQA S2 2013 Q3 [11]}}