CAIE S2 2003 November — Question 4 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2003
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeFinding maximum n for P(X=0) threshold
DifficultyStandard +0.3 This is a straightforward application of the Poisson distribution with two standard parts: (i) calculating P(X≥3) = 1 - P(X≤2) with mean λ=20/80=0.25, and (ii) solving e^(-k/80) = 0.9 for k using logarithms. Both parts require only direct substitution into formulas with minimal algebraic manipulation, making this slightly easier than average for A-level.
Spec5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities

4 The number of emergency telephone calls to the electricity board office in a certain area in \(t\) minutes is known to follow a Poisson distribution with mean \(\frac { 1 } { 80 } t\).
  1. Find the probability that there will be at least 3 emergency telephone calls to the office in any 20-minute period.
  2. The probability that no emergency telephone call is made to the office in a period of \(k\) minutes is 0.9 . Find \(k\).

(i) \(\lambda = \frac{20}{80} = 0.25\)
AnswerMarks Guidance
\(P(X \geq 3) = 1 - P(X \leq 2) = 1 - e^{-0.25}\left(1 + 0.25 + \frac{0.25^2}{2}\right) = 0.00216\)B1, M1, M1, A1 For \(\lambda = 0.25\); For calculating a relevant Poisson prob( any \(\lambda\)); For calculating expression for P(\(X \geq 3\)) their \(\lambda\); For correct answer
[4]
(ii) \(e^{-\frac{k}{80}} = 0.9\)
\(\frac{-k}{80} = -0.10536\)
AnswerMarks Guidance
\(k = 8.43\)M1, M1, M1, A1 For using \(\lambda = -t/80\) in an expression for P(0); For equating their expression to 0.9; For solving the associated equation; For correct answer cwo
[4]
**(i)** $\lambda = \frac{20}{80} = 0.25$

$P(X \geq 3) = 1 - P(X \leq 2) = 1 - e^{-0.25}\left(1 + 0.25 + \frac{0.25^2}{2}\right) = 0.00216$ | B1, M1, M1, A1 | For $\lambda = 0.25$; For calculating a relevant Poisson prob( any $\lambda$); For calculating expression for P($X \geq 3$) their $\lambda$; For correct answer

| [4]

**(ii)** $e^{-\frac{k}{80}} = 0.9$

$\frac{-k}{80} = -0.10536$

$k = 8.43$ | M1, M1, M1, A1 | For using $\lambda = -t/80$ in an expression for P(0); For equating their expression to 0.9; For solving the associated equation; For correct answer cwo

| [4]
4 The number of emergency telephone calls to the electricity board office in a certain area in $t$ minutes is known to follow a Poisson distribution with mean $\frac { 1 } { 80 } t$.\\
(i) Find the probability that there will be at least 3 emergency telephone calls to the office in any 20-minute period.\\
(ii) The probability that no emergency telephone call is made to the office in a period of $k$ minutes is 0.9 . Find $k$.

\hfill \mbox{\textit{CAIE S2 2003 Q4 [8]}}