June 2019 Paper 3 Q5
5. A machine puts liquid into bottles of perfume. The amount of liquid put into each bottle, \(D\,\text{ml}\), follows a normal distribution with mean \(25\,\text{ml}\)
Given that 15% of bottles contain less than \(24.63\,\text{ml}\)
A random sample of 200 bottles is taken.
The machine is adjusted so that the standard deviation of the liquid put in the bottles is now \(0.16\,\text{ml}\)
Following the adjustments, Hannah believes that the mean amount of liquid put in each bottle is less than \(25\,\text{ml}\)
She takes a random sample of 20 bottles and finds the mean amount of liquid to be \(24.94\,\text{ml}\)
You should state your hypotheses clearly. (5)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{24.63 - 25}{\text{‘}\sigma\text{’}} = -1.0364\) | M1 | 3.1b |
| \([\sigma =]\ 0.357\) (must come from compatible signs) | A1 | 1.1b |
| \(\mathrm{P}(D \gt k) = 0.4\) or \(\mathrm{P}(D \lt k) = 0.6\) | B1 | 1.1b |
| \(\dfrac{k - 25}{\text{‘}0.357\text{’}} = 0.2533\) | M1 | 3.4 |
| \(k = \text{awrt } \underline{\mathbf{25.09}}\) | A1 | 1.1b |
| (5) |
Notes
M1: for standardising 24.63, 25 and ‘\(\sigma\)’ (ignore label) and setting \(=\) to \(z\) where \(1 \lt |z| \lt 2\)
A1: \([\sigma =]\) awrt 0.36. Do not award this mark if signs are not compatible.
B1: for either correct probability statement (may be implied by correct answer)
this mark may be scored for a correct region shown on a diagram
M1: for a correct expression with \(z = \text{awrt } 0.253\) (may be implied by correct answer)
A1: awrt 25.09 (Correct answer with no incorrect working scores 5 out of 5)
| Scheme | Marks | AO |
|---|---|---|
| \([Y \sim \mathrm{B}(200,\ 0.45) \to]\ W \sim \mathrm{N}(90,\ 49.5)\) | B1 | 3.3 |
| \(\mathrm{P}(Y \lt 100) \approx \mathrm{P}(W \lt 99.5)\ \left[= \mathrm{P}\left(Z \lt \dfrac{99.5 - 90}{\sqrt{49.5}}\right)\right]\) | M1 | 3.4 |
| \(= 0.9115\ldots\) awrt 0.912 | A1 | 1.1b |
| (3) |
Notes
B1: setting up normal distribution approximation of binomial \(\mathrm{N}(90,\ 49.5)\) (may be implied by a correct answer) Look out for e.g. \(\sigma = \frac{3\sqrt{22}}{2}\) or \(\sigma = \text{awrt } 7.04\)
M1: attempting a probability using a continuity correction i.e. \(\mathrm{P}(W \lt 100.5)\), \(\mathrm{P}(W \lt 99.5)\) or \(\mathrm{P}(W \lt 98.5)\) condone \(\leqslant\) (The continuity correction may be seen in a standardisation).
A1: awrt 0.912 [Note: 0.911299… from binomial scores 0 out of 3]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0 : \mu = 25 \qquad \mathrm{H}_1 : \mu \lt 25\) | B1 | 2.5 |
| \([\bar{D} \sim]\ \mathrm{N}\left(25,\ \dfrac{0.16^2}{20}\right)\) | M1 | 3.3 |
| \(\mathrm{P}(\bar{D} \lt 24.94)\ [= \mathrm{P}(Z \lt -1.677\ldots)] = 0.046766\ldots\) | A1 | 3.4 |
| \(p = 0.047 \lt 0.05\) or \(z = -1.677\ldots \lt -1.6449\) or \(24.94 \lt 24.94115\ldots\) or reject \(\mathrm{H}_0\)/in the critical region/significant | M1 | 1.1b |
| There is sufficient evidence to support Hannah’s belief. | A1 | 2.2b |
| (5) | ||
| (13 marks) |
Notes
B1: for both hypotheses in terms of \(\mu\)
M1: selecting suitable model must see N(ormal), mean 25, sd \(= \frac{0.16}{\sqrt{20}}\) (o.e.) or var \(= \frac{4}{3125}\) (o.e.)
Condone \(\mathrm{N}\left(25,\ \frac{0.16}{\sqrt{20}}\right)\) if \(\frac{0.16}{\sqrt{20}}\) then used as s.d.
A1: \(p\) value \(= \text{awrt } 0.047\) or test statistic awrt \(-1.68\) or CV awrt 24.941
(any of these values imply the M1 provided they do not come from Normal mean = 24.94)
M1: a correct comparison (including compatible signs) or correct non-contextual conclusion (f.t. their \(p\) value, test statistic or critical value in the comparison)
M1 may be implied by a correct contextual statement
NB Any contradictory non contextual statements/comparisons score M0A0 e.g. ‘\(p \lt 0.05\), not significant’
A1: correct conclusion in context mentioning Hannah’s belief
or the mean amount/liquid in each bottle is now less than 25ml (dep on M1A1M1)