| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question with standard bookwork parts (a-c) requiring basic recall of properties and probability calculations, followed by a routine hypothesis test (d-e) at a specified significance level. The scaling of λ for different time periods is standard, and identifying error types is textbook knowledge. Slightly above average only due to the multi-part nature and hypothesis testing component. |
| Spec | 2.05b Hypothesis test for binomial proportion5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sqrt{2.1}\) or 1.45 (3 sf) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 4.2\) | B1 | |
| \(1 - e^{-4.2}(1 + 4.2)\) | M1 | \(1 - P(X \leqslant 1)\) any \(\lambda\), allow one end error |
| \(= 0.922\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 6.3\) | M1 | \(P(X = 5, 6, 7)\) any \(\lambda\), allow one end error |
| \(e^{-6.3}\left(\frac{6.3^5}{5!} + \frac{6.3^6}{6!} + \frac{6.3^7}{7!}\right)\) | ||
| \(= 0.455\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \lambda = 6.3\); \(H_1: \lambda < 6.3\) | B1 | Accept \(\mu\), accept 2.1 (per week) |
| \(P(X \leqslant 2) = e^{-6.3}\left(1 + 6.3 + \frac{6.3^2}{2!}\right)\) | M1 | |
| \(= 0.0498\) or \(0.0499\) | A1 | Accept 0.0499 |
| \(\text{'0.0498'} < 0.1\) | M1 | For valid comparison. For CV method the comparison can be '2 lies in CR of \(X \leqslant 2\)' |
| There is evidence that mean number of absences has decreased | A1 FT | In context, not definite. No contradictions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\) rejected | *B1 FT | OE |
| Hence Type I error possible | DB1 FT |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{2.1}$ or 1.45 (3 sf) | B1 | |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 4.2$ | B1 | |
| $1 - e^{-4.2}(1 + 4.2)$ | M1 | $1 - P(X \leqslant 1)$ any $\lambda$, allow one end error |
| $= 0.922$ (3 sf) | A1 | |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 6.3$ | M1 | $P(X = 5, 6, 7)$ any $\lambda$, allow one end error |
| $e^{-6.3}\left(\frac{6.3^5}{5!} + \frac{6.3^6}{6!} + \frac{6.3^7}{7!}\right)$ | | |
| $= 0.455$ (3 sf) | A1 | |
## Question 5(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \lambda = 6.3$; $H_1: \lambda < 6.3$ | B1 | Accept $\mu$, accept 2.1 (per week) |
| $P(X \leqslant 2) = e^{-6.3}\left(1 + 6.3 + \frac{6.3^2}{2!}\right)$ | M1 | |
| $= 0.0498$ or $0.0499$ | A1 | Accept 0.0499 |
| $\text{'0.0498'} < 0.1$ | M1 | For valid comparison. For CV method the comparison can be '2 lies in CR of $X \leqslant 2$' |
| There is evidence that mean number of absences has decreased | A1 FT | In context, not definite. No contradictions |
## Question 5(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$ rejected | *B1 FT | OE |
| Hence Type I error possible | DB1 FT | |
---5 The number of absences per week by workers at a factory has the distribution $\operatorname { Po } ( 2.1 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the standard deviation of the number of absences per week.
\item Find the probability that the number of absences in a 2-week period is at least 2 .
\item Find the probability that the number of absences in a 3-week period is more than 4 and less than 8 .\\
Following a change in working conditions, the management wished to test whether the mean number of absences has decreased. They found that, in a randomly chosen 3-week period, there were exactly 2 absences.
\item Carry out the test at the $10 \%$ significance level.
\item State, with a reason, which of the errors, Type I or Type II, might have been made in carrying out the test in part (d).
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q5 [13]}}