| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Goodness-of-fit test for Poisson |
| Difficulty | Challenging +1.2 This is a goodness-of-fit test for a geometric distribution with a given parameter (p=1/6 for a fair die). While it requires calculating expected frequencies, computing chi-squared statistic, and comparing to critical values, it follows a standard procedure taught in Further Statistics. The main challenge is careful arithmetic and possibly pooling cells, but no novel insight is required—making it moderately above average difficulty for A-level. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of throws | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| Frequency | 126 | 43 | 22 | 3 | 5 | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(\text{Geo}(0.6)\) to calculate expected values | 1 | M1 |
| 120, 48, 19.2, 7.68, 3.072, 1.2288, 0.8192 | 1 | A1 |
| Last 3 cells combined | 1 | M1 |
| \(\frac{(126-120)^2}{120} + \frac{(43-48)^2}{48} + \frac{(22-19.2)^2}{19.2} + \frac{(3-7.68)^2}{7.68} + \frac{(6-5.12)^2}{5.12}\) | 1 | M1 |
| \(\chi^2 = 4.23\) | 1 | A1 |
| Use appropriate tabular value for their test | 1 | M1 |
| \(4.23 < 9.49\) so the distribution \(\text{Geo}(0.6)\) is a satisfactory fit | 1 | A1 |
| Total | 7 |
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $\text{Geo}(0.6)$ to calculate expected values | 1 | M1 | Need at least 3 values correct |
| 120, 48, 19.2, 7.68, 3.072, 1.2288, 0.8192 | 1 | A1 | All 7 correct, stated or implied |
| Last 3 cells combined | 1 | M1 | Observed frequency $= 6$, expected frequency $= 5.12$ |
| $\frac{(126-120)^2}{120} + \frac{(43-48)^2}{48} + \frac{(22-19.2)^2}{19.2} + \frac{(3-7.68)^2}{7.68} + \frac{(6-5.12)^2}{5.12}$ | 1 | M1 | $0.3 + 0.5208 + 0.4083 + 2.8519 + 0.15125$ |
| $\chi^2 = 4.23$ | 1 | A1 | Correct value to (at least) 3 sf |
| Use appropriate tabular value for their test | 1 | M1 | |
| $4.23 < 9.49$ so the distribution $\text{Geo}(0.6)$ is a satisfactory fit | 1 | A1 | Correct conclusion from correct values; conclusion must refer to the distribution explicitly or via a statement of the null hypothesis for the test |
| **Total** | **7** | | |
---2 Each of 200 identically biased dice is thrown repeatedly until an even number is obtained. The number of throws needed is recorded and the results are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of throws & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Frequency & 126 & 43 & 22 & 3 & 5 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
Carry out a goodness of fit test, at the $5\%$ significance level, to test whether $\operatorname{Geo}(0.6)$ is a satisfactory model for the data.
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q2 [7]}}