June 2022 Paper 3 Q2
2. A manufacturer uses a machine to make metal rods.
The length of a metal rod, \(L\) cm, is normally distributed with
- a mean of 8 cm
- a standard deviation of \(x\) cm
Given that the proportion of metal rods less than 7.902 cm in length is 2.5%
The cost of producing a single metal rod is 20p
A metal rod
- where \(L \lt 7.94\) is sold for scrap for 5p
- where \(7.94 \leqslant L \leqslant 8.09\) is sold for 50p
- where \(L \gt 8.09\) is shortened for an extra cost of 10p and then sold for 50p
Give your answer to the nearest pound. (5)
The same manufacturer makes metal hinges in large batches.
The hinges each have a probability of 0.015 of having a fault.
A random sample of 200 hinges is taken from each batch and the batch is accepted if fewer than 6 hinges are faulty.
The manufacturer’s aim is for 95% of batches to be accepted.
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\mathrm{P}(L \lt 7.902) = 0.025 \Rightarrow\right]\ \dfrac{7.902 - 8}{x} = -1.96\) oe | M1 | 3.4 |
| \([x =]\ 0.05\,*\) | A1cso* | 1.1b |
| SC B1 (mark as M0A1) for \(\dfrac{7.902 - 8}{0.05} = -1.96 \Rightarrow 0.024998\) | ||
| (2) |
Notes
M1: Using the normal distribution to set up equation. Allow \(\sigma\) for \(x\) and awrt \(\pm 1.96\)
A1*: cso For a correct expression for \(x\) followed by 0.05 or 0.05000… No incorrect working seen
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(7.94 \leqslant L \leqslant 8.09) = 0.8490\ldots\) awrt 0.849 | B1 | 1.1b |
| (1) |
Notes
B1: awrt 0.849
| Scheme | Marks | AO |
|---|---|---|
| \([\mathrm{P}(L \lt 7.94) =]\ 0.115069\ldots\) (awrt 0.115) or \([\mathrm{P}(L \gt 8.09) =]\ 0.03593\ldots\) (awrt 0.036) | B1 | 1.1b |
| \([\mathrm{P}(L \lt 7.94) =]\ 0.115069\ldots\) (awrt 0.115) & \([\mathrm{P}(L \gt 8.09) =]\ 0.03593\ldots\) (awrt 0.036) | B1 | 1.1b |
| Expected income per 500 rods = \(\sum(\text{Income} \times \text{probability} \times 500)\) \((500 \times \text{``}0.849\text{''} \times 0.5) + (500 \times \text{``}0.1150\ldots\text{''} \times 0.05) + (500 \times \text{``}0.03593\ldots\text{''} \times 0.4)\) or Expected profit per rod = \(\sum(\text{Profit} \times \text{probability})\) \(0.30 \times \text{``}0.849\text{''} + {-0.15} \times \text{``}0.1150\ldots\text{''} + 0.20 \times \text{``}0.03593\ldots\text{''}\ [= 0.2446\ldots]\) | M1 | 3.4 |
| Expected profit per 500 rods \(500 \times \sum(\text{Profit} \times \text{probability})\) or \(\sum(\text{Income} \times \text{probability} \times 500) - 500 \times 0.2\) \(= 500 \times \text{``}0.2446\ldots\text{''}\) or \(= \text{``}222.3\text{''} - 500 \times 0.2\) | M1d | 3.1b |
| = [£]\(122.3\ldots\) awrt [£]122 | A1 | 1.1b |
| (5) |
Notes
B1: awrt 0.115 (Implied by awrt 57.5 for number of rods) or awrt 0.036 (Implied by awrt 18 for number of rods)
B1: awrt 0.115 (Implied by awrt 57.5 for number of rods) and awrt 0.036 (Implied by awrt 18 for number of rods)
M1: Correct method to find the total income of 500 rods. Attempt at all 3 with at least two correct and no extras
or Correct method to find sum of all three profits with at least two of 30, −15 or 20 correct. May work in pence but need to be consistent. Allow awrt 24.5 or 0.245
M1d: Dep on previous method for finding profit for 500 rods. May work in pence but need to be consistent. Allow “\(0.2446\ldots\)” \(\times\, 500\) or “their income” for 500 rods \(-\ 500 \times 0.2\) (accept 499 or 501)
A1: All previous marks must be awarded for awrt 122 awrt 12200p
NB if uses any integer values for numbers of rods then it is A0 other than for 18 for \(L \gt 8.09\)
| Scheme | Marks | AO |
|---|---|---|
| Let \(X \sim \mathrm{B}(200,\ 0.015)\) | M1 | 3.3 |
| \(\mathrm{P}(X \leqslant 5) =\) or \(\mathrm{P}(X \geqslant 6) =\) | M1 | 1.1b |
| \(0.9176\ldots\) or \(0.0824\) | A1 | 1.1b |
| Manufacturer is unlikely to achieve their aim since \(0.9176 \lt 0.95\) or Manufacturer is unlikely to achieve their aim since \(0.0824 \gt 0.05\) | A1ft | 2.4 |
| (4) | ||
| (12 marks) |
Notes
M1: Selecting the appropriate model. May be seen or used. Allow B(200, 0.985) or Po(3)
Condone B(0.015, 200) or B(0.985, 200).
M1: Writing or using \(\mathrm{P}(X \leqslant 5)\) Do not accept \(\mathrm{P}(X \lt 6)\) unless found \(\mathrm{P}(X \leqslant 5)\)
or Writing or using \(\mathrm{P}(X \geqslant 6)\) Do not accept \(\mathrm{P}(X \gt 5)\) unless found \(\mathrm{P}(X \geqslant 6)\)
A1: 0.92 (Poisson 0.916…) or 0.08 or better
A1ft: Need at least one of the method marks to be awarded. Correct conclusion with the comparison (may be in words). Ft “their \(p = 0.9176\ldots\)” as long as \(p \gt 0.9\) If “their 0.9176…” < 0.95 must … be unlikely… If “their 0.9176…” > 0.95 they must say … be likely… To ft the alternative then \(p \lt 0.1\)