| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Frequency distribution and Poisson fit |
| Difficulty | Standard +0.3 This is a standard goodness-of-fit test for Poisson distribution with straightforward calculations. Part (a) requires basic mean/variance comparison (mean ≈ variance property of Poisson), and part (b) involves routine chi-squared test calculations with some values already provided. The question is slightly easier than average as it's highly structured with given summary statistics and a partially completed table, requiring mainly procedural application rather than problem-solving insight. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.06b Fit prescribed distribution: chi-squared test |
| No. of defective ornaments in a week, \(r\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| No. of weeks with \(r\) defective ornaments, \(f\) | 2 | 14 | 13 | 5 | 3 | 1 | 2 | 0 |
| No. of defective ornaments in a week, \(r\) | Observed frequency | Probability | Expected frequency | Chi-squared contribution |
| 0 | 2 | 0.13534 | 5.4134 | 2.15232 |
| 1 | 14 | |||
| 2 | 13 | 0.27067 | 0.43620 | |
| 3 | 5 | 7.2179 | ||
| \(\geqslant 4\) | 6 | 0.14288 | 0.01421 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | mean = 84/40 = 2.1 |
| Answer | Marks |
|---|---|
| suggestion.) | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 3.3 | FT their mean |
| Answer | Marks | Guidance |
|---|---|---|
| both round to 2. | Allow 2.1 ≈ 2.04 or mean ≈ variance. | |
| 3 | (b) | (i) |
| Answer | Marks |
|---|---|
| 0.14288 5.7151 0.01421 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | At least 1 EF correct to at least |
| Answer | Marks |
|---|---|
| least 3d.p. | If M0M0 then SC1 for at least one |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | (ii) |
| Answer | Marks |
|---|---|
| data. | B1 |
| Answer | Marks |
|---|---|
| [6] | 3.3 |
| Answer | Marks |
|---|---|
| 3.5a | or H : The Poisson model is |
| Answer | Marks |
|---|---|
| assertive “post-conclusion” is A0. | Condone “the Poisson model is a good |
| Answer | Marks | Guidance |
|---|---|---|
| Prob | EF | χ2 cont |
| 0.13534 | 5.4134 | 2.15232 |
| 0.27067 | 10.8268 | 0.93001 |
| 0.27067 | 10.8268 | 0.43620 |
| 0.18045 | 7.2179 | 0.68150 |
| 0.14288 | 5.7151 | 0.01421 |
| 3 | (c) | These defects did not occur independently of each |
| other | B1 | |
| [1] | 3.5b | or these defects did not occur |
Question 3:
3 | (a) | mean = 84/40 = 2.1
variance:
2 = (256 – 402.12)/39 = awrt 2.04
n–1
The mean and variance have a similar value. (In a
Poisson distribution the mean and variance are
equal so this supports the Poisson model
suggestion.) | B1
B1FT
B1
[3] | 1.1
1.1
3.3 | FT their mean
Provided their mean and variance
both round to 2. | Allow 2.1 ≈ 2.04 or mean ≈ variance.
3 | (b) | (i) | Prob EF χ2 cont
0.13534 5.4134 2.15232
0.27067 10.8268 0.93001
0.27067 10.8268 0.43620
0.18045 7.2179 0.68150
0.14288 5.7151 0.01421 | M1
A1
M1
A1
[4] | 1.1
1.1
1.1
1.1 | At least 1 EF correct to at least
2d.p.
EF column all correct to at least
2d.p.
At least 1 contribution correct to at
least 3d.p.
Contributions both correct to at
least 3d.p. | If M0M0 then SC1 for at least one
probability correct to 3 s.f. shown in
the table.
3 | (b) | (ii) | H : The Poisson model fits the data
0
H : The Poisson model does not fit the data
1
χ2 = awrt 4.21
ν = 4
(χ2 ) = 7.779
4 10%
4.21 < 7.779 so the result is not significant
we do not reject H oe
0
There is insufficient evidence at the 10% level to
suggest that the Poisson model does not fit the
data. | B1
B1FT
M1
A1
M1
A1
[6] | 3.3
3.4
1.1
1.1
2.2b
3.5a | or H : The Poisson model is
0
suitable
H : The Poisson model is not
1
suitable
FT for 2.60273 + their contributions
Can be implied by 7.779
For correctly comparing their test
statistic with their critical value and
then making a consistent
conclusion.
From correct values and hypotheses
only.
Contextual and non-assertive. An
assertive “post-conclusion” is A0. | Condone “the Poisson model is a good
fit” or “the given model …”
Can be implied by p-value = 0.378
A1 for p-value = 0.378
or p-value = awrt 0.378 > 0.1
Allow “It is reasonable to believe that
the Poisson model is suitable”
Prob | EF | χ2 cont
0.13534 | 5.4134 | 2.15232
0.27067 | 10.8268 | 0.93001
0.27067 | 10.8268 | 0.43620
0.18045 | 7.2179 | 0.68150
0.14288 | 5.7151 | 0.01421
3 | (c) | These defects did not occur independently of each
other | B1
[1] | 3.5b | or these defects did not occur
singly.3 A glassware factory produces a large number of ornaments each week. Just before they leave the factory, all the ornaments are checked and some may be found to be defective. The Quality Assurance Manager of the factory wishes to model the number of defective ornaments that are found each week using a Poisson distribution.
The numbers of defective ornaments found each week in a period of 40 weeks are shown in Table 3.1.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 3.1}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
No. of defective ornaments in a week, $r$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
No. of weeks with $r$ defective ornaments, $f$ & 2 & 14 & 13 & 5 & 3 & 1 & 2 & 0 \\
\hline
\end{tabular}
\end{center}
\end{table}
You are given that summary statistics for the data are $\sum f = 40 , \sum \mathrm { rf } = 84$ and $\sum \mathrm { r } ^ { 2 } \mathrm { f } = 256$.
\begin{enumerate}[label=(\alph*)]
\item By using the summary statistics to determine estimates for the mean and variance of the number of defective ornaments produced by the factory each week, explain how the data support the suggestion that the number of defective ornaments produced each week can be modelled using a Poisson distribution.
The Quality Assurance Manager is asked by the head office to carry out a chi-squared hypothesis test for goodness of fit based on a $\operatorname { Po } ( 2 )$ distribution.
\item Table 3.2, which is incomplete, gives observed frequency, probability, expected frequency and chi-squared contribution.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 3.2}
\begin{tabular}{|l|l|l|l|l|}
\hline
No. of defective ornaments in a week, $r$ & Observed frequency & Probability & Expected frequency & Chi-squared contribution \\
\hline
0 & 2 & 0.13534 & 5.4134 & 2.15232 \\
\hline
1 & 14 & & & \\
\hline
2 & 13 & 0.27067 & & 0.43620 \\
\hline
3 & 5 & & 7.2179 & \\
\hline
$\geqslant 4$ & 6 & 0.14288 & & 0.01421 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\roman*)]
\item Complete the copy of the table in the Printed Answer Booklet.
\item Carry out the test at the $10 \%$ significance level.
\end{enumerate}\item On one occasion a fork-lift truck in the factory drops a crate containing eight ornaments and all of them are subsequently found to be defective.
Explain why the Poisson model cannot model defects occurring in this manner.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2024 Q3 [14]}}