| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Frequency distribution and Poisson fit |
| Difficulty | Standard +0.3 This is a standard A-level statistics question on Poisson distribution fitting and chi-squared goodness of fit test. It requires routine calculations (mean, variance, Poisson probabilities) and a standard hypothesis test procedure. All steps are textbook applications with no novel problem-solving required, making it slightly easier than average for Further Maths statistics content. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of visitors in a 10 -minute period | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | \(\geqslant 9\) |
| Number of 10 -minute periods | 2 | 2 | 12 | 8 | 11 | 13 | 4 | 7 | 1 | 0 |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | \(\geqslant 9\) | ||
| 1.10 | 8.79 | 11.72 | 9.38 | 6.25 | 3.57 | 1.79 | 1.28 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = 240/60 = 4\) | B1 | Find mean of sample data |
| \(\sigma^2 = 1174/60 - 4^2 = 3.57\) | B1 | Find variance of sample data |
| \(4 \approx 3.57\) (no \(\sqrt{}\) on \(\bar{x},\ \sigma^2\)) | B1 | State valid reason why Poisson distribution suitable (AEF); allow unsuitable since \(4 \neq 3.57\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4.40;\ 11.7_{[2]}\) (to 1 d.p.) | B1; B1 | Find expected values \(60\lambda^r e^{-\lambda}/r!\) with \(\lambda=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): [Poisson] distribution fits data | B1 | State (at least) null hypothesis (AEF) |
| \(O_i: 4\ 12\ \ldots\ 6\ 8\) | ||
| \(E_i: \underline{5.50}\ 8.79\ \ldots\ 6.25\ \underline{6.64}\) | *M1 A1 | Combine cells so that all exp. value \(\geq 5\) |
| \(\chi^2 = 0.409 + 1.172 + 1.181 + 0.044 + 1.397 + 0.81 + 0.279 = 5.29\) | M1 DA1 | Calculate value of \(\chi^2\) to 2 d.p.; A1 dep *M1 |
| 7 cells: \(\chi^2_{5,\ 0.9} = 9.236\); 8 cells: \(\chi^2_{6,\ 0.9} = 10.64\); 9 cells: \(\chi^2_{7,\ 0.9} = 12.02\); 10 cells: \(\chi^2_{8,\ 0.9} = 13.36\) | \(\text{B1}\sqrt{}\) | State or use consistent tabular value to 2 d.p. |
| Accept \(H_0\) if \(\chi^2 <\) tabular value (AEF) | M1 | State or imply valid method for conclusion |
| \(5.29 < 9.24\) so distribution fits | A1 | Conclusion (AEF, requires both values correct) |
# Question 9(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 240/60 = 4$ | B1 | Find mean of sample data |
| $\sigma^2 = 1174/60 - 4^2 = 3.57$ | B1 | Find variance of sample data |
| $4 \approx 3.57$ (no $\sqrt{}$ on $\bar{x},\ \sigma^2$) | B1 | State valid reason why Poisson distribution suitable (AEF); allow unsuitable since $4 \neq 3.57$ |
---
# Question 9(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4.40;\ 11.7_{[2]}$ (to 1 d.p.) | B1; B1 | Find expected values $60\lambda^r e^{-\lambda}/r!$ with $\lambda=4$ |
---
# Question 9(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: [Poisson] distribution fits data | B1 | State (at least) null hypothesis (AEF) |
| $O_i: 4\ 12\ \ldots\ 6\ 8$ | | |
| $E_i: \underline{5.50}\ 8.79\ \ldots\ 6.25\ \underline{6.64}$ | *M1 A1 | Combine cells so that all exp. value $\geq 5$ |
| $\chi^2 = 0.409 + 1.172 + 1.181 + 0.044 + 1.397 + 0.81 + 0.279 = 5.29$ | M1 DA1 | Calculate value of $\chi^2$ to 2 d.p.; A1 dep *M1 |
| 7 cells: $\chi^2_{5,\ 0.9} = 9.236$; 8 cells: $\chi^2_{6,\ 0.9} = 10.64$; 9 cells: $\chi^2_{7,\ 0.9} = 12.02$; 10 cells: $\chi^2_{8,\ 0.9} = 13.36$ | $\text{B1}\sqrt{}$ | State or use consistent tabular value to 2 d.p. |
| Accept $H_0$ if $\chi^2 <$ tabular value (AEF) | M1 | State or imply valid method for conclusion |
| $5.29 < 9.24$ so distribution fits | A1 | Conclusion (AEF, requires both values correct) |
---9 The number of visitors arriving at an art exhibition is recorded for each 10 -minute period of time during the ten hours that it is open on a particular day. The results are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
Number of visitors in a 10 -minute period & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & $\geqslant 9$ \\
\hline
Number of 10 -minute periods & 2 & 2 & 12 & 8 & 11 & 13 & 4 & 7 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
(i) Calculate the mean and variance for this sample and explain whether your answers support a suggestion that a Poisson distribution might be a suitable model for the number of visitors in a 10-minute period.\\
(ii) Use an appropriate Poisson distribution to find the two expected frequencies missing from the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Number of visitors in \\
a 10-minute period \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & $\geqslant 9$ \\
\hline
\begin{tabular}{ l }
Expected number of \\
10 -minute periods \\
\end{tabular} & 1.10 & & 8.79 & & 11.72 & 9.38 & 6.25 & 3.57 & 1.79 & 1.28 \\
\hline
\end{tabular}
\end{center}
(iii) Test, at the $10 \%$ significance level, the goodness of fit of this Poisson distribution to the data.
\hfill \mbox{\textit{CAIE FP2 2016 Q9 [13]}}