S1 January 2011 Q8
8. The weight, \(X\) grams, of soup put in a tin by machine \(A\) is normally distributed with a mean of 160 g and a standard deviation of 5 g.
A tin is selected at random.
The weight stated on the tin is \(w\) grams.
The weight, \(Y\) grams, of soup put into a carton by machine \(B\) is normally distributed with mean \(\mu\) grams and standard deviation \(\sigma\) grams.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \gt 168) = \mathrm{P}\left(Z \gt \dfrac{168 - 160}{5}\right)\) | M1 |
| \(= \mathrm{P}(Z \gt 1.6)\) | A1 |
| \(= 0.0548\) awrt 0.0548 | A1 |
| (3) |
Notes
M1 for an attempt to standardize 168 with 160 and 5 i.e. \(\pm\left(\dfrac{168 - 160}{5}\right)\) or implied by 1.6
1st A1 for P(\(Z\) > 1.6) or P(\(Z\) < -1.6) ie \(z\) = 1.6 and a correct inequality or 1.6 on a shaded diagram
Correct answer to (a) implies all 3 marks
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \lt w) = \mathrm{P}\left(Z \lt \dfrac{w - 160}{5}\right)\) | |
| \(\dfrac{w - 160}{5} = -2.3263\) | M1 B1 |
| \(w = 148.37\) awrt 148 | A1 |
| (3) |
Notes
M1 for attempting \(\pm\left(\dfrac{w - 160}{5}\right)\) = recognizable \(z\) value (\(|z| \gt 1\))
B1 for \(z = \pm\) 2.3263 or better. Should be \(z\) =… or implied so: \(1 - 2.3263 = \dfrac{w - 160}{5}\) is M0B0
A1 for awrt 148. This may be scored for other \(z\) values so M1B0A1 is possible
For awrt 148 only with no working seen award M1B0A1
| Scheme | Marks |
|---|---|
| \(\dfrac{160 - \mu}{\sigma} = 2.3263\) | M1 B1 |
| \(\dfrac{152 - \mu}{\sigma} = -1.2816\) | B1 |
| \(160 - \mu = 2.3263\sigma\) \(152 - \mu = -1.2816\sigma\) | |
| \(8 = 3.6079\sigma\) | M1 |
| \(\sigma = 2.21\ldots.\) awrt 2.22 | A1 |
| \(\mu = 154.84\ldots\) awrt 155 | A1 |
| (6) | |
| (12 marks) |
Notes
M1 for attempting to standardize 160 or 152 with \(\mu\) and \(\sigma\) (allow \(\pm\)) and equate to \(z\) value ( \(|z| \gt 1\))
1st B1 for awrt \(\pm\) 2.33 or \(\pm\) 2.32 seen
2nd B1 for awrt \(\pm\) 1.28 seen
2nd M1 for attempt to solve their two linear equations in \(\mu\) and \(\sigma\) leading to equation in just one variable
1st A1 for \(\sigma\) = awrt 2.22 . Award when 1st seen
2nd A1 for \(\mu\) = awrt 155. Correct answer only for part (c) can score all 6 marks.
NB \(\sigma\) = 2.21 commonly comes from \(z\) = 2.34 and usually scores M1B0B1M1A0A1
The A marks in (c) require both M marks to have been earned