S2 June 2009 Q5
5. An administrator makes errors in her typing randomly at a rate of 3 errors every 1000 words.
The administrator is given an 8000 word report to type and she is told that the report will only be accepted if there are 20 or fewer errors.
| Scheme | Marks |
|---|---|
| \(X\) = the number of errors in 2000 words so \(X \sim \mathrm{Po}(6)\) | B1 |
| \(\mathrm{P}(X \geqslant 4) = 1 - \mathrm{P}(X \leqslant 3)\) | M1 |
| \(= 1 - 0.1512 \quad = 0.8488\) awrt 0.849 | A1 |
| (3) |
Notes
B1 for seeing or using Po(6)
M1 for \(1 - \mathrm{P}(X \leqslant 3)\) or \(1 - [\mathrm{P}(X = 0) + \mathrm{P}(X = 1) + \mathrm{P}(X = 2) + \mathrm{P}(X = 3)]\)
A1 awrt 0.849
SC If B(2000, 0.003) is used and leads to awrt 0.849 allow B0 M1 A1
If no distribution indicated awrt 0.8488 scores B1M1A1 but any other awrt 0.849 scores B0M1A1
| Scheme | Marks |
|---|---|
| \(Y\) = the number of errors in 8000 words. \(Y \sim \mathrm{Po}(24)\) so use a Normal approx | M1 |
| \(Y \approx\sim \mathrm{N}\left(24, \sqrt{24}^{\,2}\right)\) | A1 |
| Require \(\mathrm{P}(Y \leqslant 20) = \mathrm{P}\left(Z \lt \dfrac{20.5 - 24}{\sqrt{24}}\right)\) | M1 M1 |
| \(= \mathrm{P}(Z \lt -0.714\ldots)\) | A1 |
| \(= 1 - 0.7611\) | M1 |
| \(= 0.2389\) awrt (0.237~0.239) | A1 |
| [N.B. Exact Po gives 0.242 and no \(\pm\) 0.5 gives 0.207] | |
| (7) | |
| (10 marks) |
Notes
1st M1 for identifying the normal approximation
1st A1 for [mean = 24] and [sd = \(\sqrt{24}\) or var = 24]
These first two marks may be given if the following are seen in the standardisation formula : 24
\(\sqrt{24}\) or awrt 4.90
2nd M1 for attempting a continuity correction (20/ 28 \(\pm\) 0.5 is acceptable)
3rd M1 for standardising using their mean and their standard deviation.
2nd A1 correct z value awrt \(\pm\)0.71 or this may be awarded if see \(\dfrac{20.5 - 24}{\sqrt{24}}\) or \(\dfrac{27.5 - 24}{\sqrt{24}}\)
4th M1 for 1 - a probability from tables (must have an answer of < 0.5)
3rd A1 answer awrt 3 sig fig in range 0.237 – 0.239