S1 June 2013 Q6
6. The weight, in grams, of beans in a tin is normally distributed with mean \(\mu\) and standard deviation 7.8
Given that 10% of tins contain less than 200 g, find
The machine settings are adjusted so that the weight, in grams, of beans in a tin is normally distributed with mean 205 and standard deviation \(\sigma\).
| Scheme | Marks |
|---|---|
| [Let \(X\) be the amount of beans in a tin. \(\mathrm{P}(X \lt 200) = 0.1\)] | |
| \(\dfrac{200 - \mu}{7.8} = -1.2816\) [ calc gives 1.28155156…] | M1 B1 |
| \(\mu = 209.996\ldots\) awrt 210 | A1 |
| (3) |
Notes
Condone poor handling of notation if answers are correct but A marks must have correct working.
M1 for an attempt to standardise (allow \(\pm\)) with 200 and 7.8 and set \(= \pm\) any \(z\) value (\(|z| \gt 1\))
B1 for \(z = \pm 1.2816\) (or better used as a \(z\))[May be implied by 209.996(102…) or better seen]
A1 for awrt 210 (can be scored for using 1.28 but then they get M1B0A1). The 210 must follow from correct working− sign scores A0
If answer is awrt 210 and 209.996… or better seen then award M1B1A1
\(z = 1.28\) gives 209.984 and \(z = 1.282\) gives 209.9996 and both score M1B0A1
If answer is awrt 210 or awrt 209.996 then award M1B0A1 (unless of course \(z = 1.2816\) is seen)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \gt 225) = \mathrm{P}\left(Z \gt \dfrac{225 - \text{"}210\text{"}}{7.8}\right)\) | M1 |
| \(= \mathrm{P}(Z \gt 1.92)\) or \(1 - \mathrm{P}(Z \lt 1.92)\) (allow 1.93) | A1 |
| \(= 1 - 0.9726 = 0.0274\) (or better) [calc gives 0.0272037…] \(= 0.0274\) = awrt 2.7% allow 0.027 | A1 |
| (3) |
Notes
M1 for attempting to standardise with 225, their mean and 7.8 . Allow \(\pm\)
1st A1 for \(Z \gt\) awrt 1.92/3. Allow a diagram but must have 1.92/3 and correct area indicated. Must have the \(Z\) so \(\mathrm{P}(X \gt 225)\) with or without a diagram is not sufficient. Award for 1 – 0.9726 or 1 – 0.9732
2nd A1 for 2.7 % or better (calculator gives 2.72…) Allow awrt 0.027. Correct ans scores 3/3
| Scheme | Marks |
|---|---|
| [Let \(Y\) be the new amount of beans in a tin] | |
| \(\dfrac{210 - 205}{\sigma} = 2.3263\) or \(\dfrac{200 - 205}{\sigma} = -2.3263\) [ calc gives 2.3263478…] | M1 B1 |
| \(\sigma = \dfrac{5}{2.3263}\) | dM1 |
| \(\sigma = 2.15\) (2.14933…) | A1 |
| (4) | |
| (10 marks) |
Notes
1st M1 for an attempt to standardise with 200 or 210, 205 and \(\sigma\) and set \(= \pm\) any \(z\) value (\(|z| \gt 2\))
B1 for \(z = 2.3263\) (or better) and compatible signs. If B0 in (a) for using a value in [1.28, 1.29] but not using 1.2816: allow awrt 2.33 here
2nd dM1 Dependent on the first M1 for correctly rearranging to make \(\sigma = \ldots\) May be implied e.g. \(\frac{5}{\sigma} = 2.32 \to \sigma = 2.16\) (M1A0) BUT must have \(\sigma \gt 0\)
A1 for awrt 2.15 . Must follow from correct working but a range of possible \(z\) values will do. NB \(2.320 \lt z \leqslant 2.331\) will give an answer of awrt 2.15