| No. of errors | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of pages | 37 | 65 | 60 | 49 | 27 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{0 \times 37 + 1 \times 65 + 2 \times 60 + \ldots + 5 \times 12}{37 + 65 + 60 + \ldots + 12} = \frac{500}{250} = 2\) | M1 A1 cso | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{var} = \frac{\sum x^2}{250} - 2^2 = \frac{1478}{250} - 4 = 1.912\) (or \(s^2 = 1.9196\ldots\)) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| For a Poisson distribution the mean must equal the variance; parts (a) and (b) are very close, so a Poisson might be a suitable model. | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| This is less than 5% so a significant result and there is evidence that the secretary has improved. | B1 B1 M1 M1 A1 A1 ft | (6 marks) |
## Part (a)
$\bar{x} = \frac{0 \times 37 + 1 \times 65 + 2 \times 60 + \ldots + 5 \times 12}{37 + 65 + 60 + \ldots + 12} = \frac{500}{250} = 2$ | M1 A1 cso | (2 marks)
## Part (b)
$\text{var} = \frac{\sum x^2}{250} - 2^2 = \frac{1478}{250} - 4 = 1.912$ (or $s^2 = 1.9196\ldots$) | M1 A1 | (2 marks)
## Part (c)
For a Poisson distribution the mean must equal the variance; parts (a) and (b) are very close, so a Poisson might be a suitable model. | B1 | (1 mark)
## Part (d)
$H_0: \mu = 2; H_1: \mu < 2$
$X =$ number of errors over 4 pages. Under $H_0 \sim P_0(8)$
$P(X \leq 3) = 0.0424$
This is less than 5% so a significant result and there is evidence that the secretary has improved. | B1 B1 M1 M1 A1 A1 ft | (6 marks)
**Total: 11 marks**
---A company director monitored the number of errors on each page of typing done by her new secretary and obtained the following results:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
No. of errors & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
No. of pages & 37 & 65 & 60 & 49 & 27 & 12 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that the mean number of errors per page in this sample of pages is 2. [2]
\item Find the variance of the number of errors per page in this sample. [2]
\item Explain how your answers to parts (a) and (b) might support the director's belief that the number of errors per page could be modelled by a Poisson distribution. [1]
\end{enumerate}
Some time later the director notices that a 4-page report which the secretary has just typed contains only 3 errors. The director wishes to test whether or not this represents evidence that the number of errors per page made by the secretary is now less than 2.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Assuming a Poisson distribution and stating your hypothesis clearly, carry out this test. Use a 5\% level of significance. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q4 [11]}}