June 2018 Paper 3 Q5
5. The lifetime, \(L\) hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours.
Alice’s calculator requires 4 batteries and will stop working when any one battery reaches the end of its lifetime.
At the start of her exams Alice put 4 new batteries in her calculator.
She has used her calculator for 16 hours, but has another 4 hours of exams to sit.
Alice only has 2 new batteries so, after the first 16 hours of her exams, although her calculator is still working, she randomly selects 2 of the batteries from her calculator and replaces these with the 2 new batteries.
After her exams, Alice believed that the lifetime of the batteries was more than 18 hours. She took a random sample of 20 of these batteries and found that their mean lifetime was 19.2 hours.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(L \gt 16) = 0.69146\ldots\) awrt 0.691 | B1 | 1.1b |
| (1) |
Notes
B1 for evaluating probability using their calculator (awrt 0.691) Accept 0.6915
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(L \gt 20 \mid L \gt 16) = \dfrac{\mathrm{P}(L \gt 20)}{\mathrm{P}(L \gt 16)}\) | M1 | 3.1b |
| \(= \dfrac{0.308537\ldots}{\text{(a)}}\) or \(\dfrac{1 - \text{(a)}}{\text{(a)}},\ = 0.44621\ldots\) | A1ft, A1 | 1.1b 1.1b |
| For calc to work require \((0.44621\ldots)^4 = 0.03964\ldots\) awrt 0.0396 | dM1 A1 | 2.1 1.1b |
| (5) |
Notes
1st M1 for a first step of identifying a suitable conditional probability (either form)
1st A1ft for a ratio of probabilities with numerator = awrt 0.309 or 1 – (a) and denom = their (a)
2nd A1 for awrt 0.446 (o.e.) Accept 0.4465 (from \(\frac{0.3085}{0.691} = 0.44645\ldots\))
NB \(\dfrac{\mathrm{P}(16 \lt L \lt 20)}{\mathrm{P}(L \gt 16)} = 0.5538\ldots\) scores M1A1A1 when they do \(1 - 0.5538 = 0.4462\ldots\)
2nd M1 (dep on 1st M1) for 2nd correct step i.e. (their \(0.446\ldots)^4\) or \(X \sim \mathrm{B}(4, \text{``}0.446\text{''})\) and \(\mathrm{P}(X = 4)\)
3rd A1 for awrt 0.0396
| Scheme | Marks | AO |
|---|---|---|
| Require: \([\mathrm{P}(L \gt 4)]^2 \times [\mathrm{P}(L \gt 20 \mid L \gt 16)]^2\) | M1 | 1.1a |
| \(= (0.99976\ldots)^2 \times (\text{``}0.44621\ldots\text{''})^2\) | A1ft | 1.1b |
| \(= 0.19901\ldots\) awrt 0.199 (*) | A1cso* | 1.1b |
| (3) |
Notes
1st M1 for a correct approach to solving the problem (May be implied by A1ft)
1st A1ft for \(\mathrm{P}(L \gt 4) =\) awrt 0.9998 used and ft their 0.44621 in correct expression
If use \(\mathrm{P}(L \gt 20) = 0.3085\ldots\) as \(0.446\ldots\) in (b) then M1 for \((0.3085\ldots)^2 \times [\mathrm{P}(L \gt 4)]^2\); A1ft as above
2nd A1cso for 0.199 or better with clear evidence of M1 [NB \((0.4462\ldots)^2 = 0.199\ldots\) is M0A0A0]
Must see M1 scored by correct expression in symbols or values (M1A1ft)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0: \mu = 18 \qquad \mathrm{H}_1: \mu \gt 18\) | B1 | 2.5 |
| \(\bar{L} \sim \mathrm{N}\left(18, \left(\dfrac{4}{\sqrt{20}}\right)^2\right)\) | M1 | 3.3 |
| \(\mathrm{P}(\bar{L} \gt 19.2) = \mathrm{P}(Z \gt 1.3416\ldots) = 0.089856\ldots\) | A1 | 3.4 |
| \((0.0899 \gt 5\%)\) or \((19.2 \lt 19.5)\) or \(1.34 \lt 1.6449\) so not significant | A1 | 1.1b |
| Insufficient evidence to support Alice’s claim (or belief) | A1 | 3.5a |
| (5) | ||
| (14 marks) |
Notes
B1 for both hypotheses in terms of \(\mu\).
M1 for selecting a suitable model. Sight of normal, mean 18, sd \(\frac{4}{\sqrt{20}}\) (o.e.) or variance = 0.8
1st A1 for using the model correctly. Allow awrt 0.0899 or 0.09 from correct prob. statement
ALT CR \((\bar{L}) \gt 19.471\ldots\) (accept awrt 19.5) or CV of 1.6449 (or better: calc 1.6448536..)
2nd A1 for correct non-contextual conclusion. Wrong comparison or contradictions A0
Error giving 2nd A0 implies 3rd A0 but just a correct contextual conclusion can score A1A1
3rd A1 dep on M1 and 1st A1 for a correct contextual conclusion mentioning Alice’s claim /belief or there is insufficient evidence that the mean lifetime is more than 18 hours