CAIE S2 2020 March — Question 4 7 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2020
SessionMarch
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeFinding maximum n for P(X=0) threshold
DifficultyStandard +0.3 Part (a) is a straightforward Poisson calculation requiring scaling the mean from 50 days to 365 days, then computing P(X<3). Part (b) requires solving P(X=0)=e^{-λn}>0.95 for n, involving logarithms and rate conversion—slightly more challenging but still a standard textbook exercise with clear methodology.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!

4 The number of accidents on a certain road has a Poisson distribution with mean 0.4 per 50-day period.
  1. Find the probability that there will be fewer than 3 accidents during a year (365 days).
  2. The probability that there will be no accidents during a period of \(n\) days is greater than 0.95 . Find the largest possible value of \(n\).

Question 4(a):
AnswerMarks Guidance
AnswerMark Guidance
\(\lambda = 0.4 \times 365 \div 50 = 2.92\)B1
\(e^{-2.92}(1 + 2.92 + \frac{2.92^2}{2})\)M1 Any \(\lambda\). Allow one end error
\(= 0.441\) (3 sf)A1
Question 4(b):
AnswerMarks Guidance
AnswerMark Guidance
\(e^{-\lambda} > 0.95\)M1 Allow '\(=\)' throughout
\(-\lambda > \ln 0.95\) or \(\lambda < 0.051293\) OEM1 Attempt ln both sides
\('0.051293' \times 50 \div 0.4 \; (= 6.411)\)M1
Largest \(n\) is 6 (3 sf). Allow \(n = 6\) or \(n \leqslant 6\) (NOT \(n < 6\) or \(n \geqslant 6\) as final answer)A1 SC Trial and Improvement: M1 for \(e^{-\lambda} > 0.95\) SOI; M1 for \(\lambda = n \times \frac{0.4}{50}\); M1 for use of both; \(n=6\) giving 0.9531 and \(n=7\) giving 0.9455; A1 \(n=6\)
## Question 4(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 0.4 \times 365 \div 50 = 2.92$ | B1 | |
| $e^{-2.92}(1 + 2.92 + \frac{2.92^2}{2})$ | M1 | Any $\lambda$. Allow one end error |
| $= 0.441$ (3 sf) | A1 | |

## Question 4(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-\lambda} > 0.95$ | M1 | Allow '$=$' throughout |
| $-\lambda > \ln 0.95$ or $\lambda < 0.051293$ OE | M1 | Attempt ln both sides |
| $'0.051293' \times 50 \div 0.4 \; (= 6.411)$ | M1 | |
| Largest $n$ is 6 (3 sf). Allow $n = 6$ or $n \leqslant 6$ (NOT $n < 6$ or $n \geqslant 6$ as final answer) | A1 | SC Trial and Improvement: M1 for $e^{-\lambda} > 0.95$ SOI; M1 for $\lambda = n \times \frac{0.4}{50}$; M1 for use of both; $n=6$ giving 0.9531 and $n=7$ giving 0.9455; A1 $n=6$ |
4 The number of accidents on a certain road has a Poisson distribution with mean 0.4 per 50-day period.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that there will be fewer than 3 accidents during a year (365 days).
\item The probability that there will be no accidents during a period of $n$ days is greater than 0.95 . Find the largest possible value of $n$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2020 Q4 [7]}}