| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Assess model suitability before testing |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with a Poisson distribution. Part (a) requires recognizing that for Poisson distributions, mean equals variance (here 2.4 ≈ 1.5² = 2.25). Part (b) follows a routine procedure: calculate expected frequencies using Poisson(2.5), combine cells if needed, compute chi-squared statistic, find degrees of freedom, and compare to critical value. While it requires careful calculation across multiple steps, it's a textbook application with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 5.02i Poisson distribution: random events model5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.06b Fit prescribed distribution: chi-squared test |
| Number of breakages | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 or more |
| Frequency | 21 | 33 | 46 | 31 | 23 | 16 | 10 | 0 |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | (Mean of distribution = 2.4) | |
| Variance = 2.25 | M1 | Allow 1.52. |
| Mean approx. equal to variance (so Poisson might be suitable) | A1 | Must have ‘approximately’ oe, not equality. |
| Answer | Marks | Guidance |
|---|---|---|
| 5(b) | Po(2.5) leads to frequencies: | M1 |
| 14.77[5], 36.938, 46.173, 38.477, 24.048, 12.024, 5.010, 2.554 | A1 | All correct to 4sf. |
| Combine last two values: 7.564 | M1 | FT their table values if final figure is less than 5. |
| Answer | Marks | Guidance |
|---|---|---|
| +1.3148+0.7845 | M1 | Apply correct formula to their frequencies. |
| 6.64 | A1 | Correct to 3sf. |
| 5(b) | H : distribution fits the data | |
| 0 | B1 | Must mention data and distribution . |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | M1 | Compare with correct tabular value and conclusion. |
| Answer | Marks | Guidance |
|---|---|---|
| distribution | A1 | Correct conclusion, in context, following correct work. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | (Mean of distribution = 2.4)
Variance = 2.25 | M1 | Allow 1.52.
Mean approx. equal to variance (so Poisson might be suitable) | A1 | Must have ‘approximately’ oe, not equality.
Must have 2.25.
2
--- 5(b) ---
5(b) | Po(2.5) leads to frequencies: | M1 | At least 3 correct to 3sf.
14.77[5], 36.938, 46.173, 38.477, 24.048, 12.024, 5.010, 2.554 | A1 | All correct to 4sf.
Combine last two values: 7.564 | M1 | FT their table values if final figure is less than 5.
( O−E )2
=2.6227+0.4198+0.00065+1.4529+0.04570
E
+1.3148+0.7845 | M1 | Apply correct formula to their frequencies.
6.64 | A1 | Correct to 3sf.
5(b) | H : distribution fits the data
0 | B1 | Must mention data and distribution .
Tabular value, 5 degrees of freedom, is 9.236
‘6.64’ < 9.236, so accept H
0 | M1 | Compare with correct tabular value and conclusion.
Insufficient evidence to show that data does not follow a Poisson
distribution | A1 | Correct conclusion, in context, following correct work.
Level of uncertainty in language is used.
Allow ±1 difference in third significant figure.
8
Question | Answer | Marks | GuidanceChai packs china mugs into cardboard boxes. Chai's manager suspects that breakages occur at random times and that the number of breakages may follow a Poisson distribution. He takes a small sample of observations and finds that the number of breakages in a one-hour period has a mean of 2.4 and a standard deviation of 1.5.
\begin{enumerate}[label=(\alph*)]
\item Explain how this information tends to support the manager's suspicion. [2]
\end{enumerate}
The manager now takes a larger sample and claims that the numbers of breakages in a one-hour period follow a Poisson distribution. The numbers of breakages in a random sample of 180 one-hour periods are summarised in the following table.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Number of breakages & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 or more \\
\hline
Frequency & 21 & 33 & 46 & 31 & 23 & 16 & 10 & 0 \\
\hline
\end{tabular}
The mean number of breakages calculated from this sample is 2.5.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use the data from this larger sample to carry out a goodness of fit test, at the 10% significance level, to test the claim. [8]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q5 [10]}}