S1 January 2013 Q4
4. The length of time, \(L\) hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours.
Alice is about to go on a 6 hour journey.
Given that it is 127 hours since Alice last charged her phone,
| Scheme | Marks |
|---|---|
| \(\dfrac{127 - 100}{15}\) | M1 |
| So \(\mathrm{P}(L \gt 127) = \mathrm{P}(Z \gt 1.8)\) or \(1 - \mathrm{P}(Z \lt 1.8)\) o.e. | A1 |
| \(= 1 - 0.9641 = \underline{\mathbf{0.0359}}\) (awrt 0.0359) | A1 |
| (3) |
Notes
M1 for attempting to standardise with 127, 100 and 15. Allow \(\pm\)
1st A1 for \(Z \gt 1.8\). Allow a diagram but must have 1.8 and correct area indicated. Must have the \(Z\) so \(\mathrm{P}(L \gt 127)\) with or without a diagram is insufficient. May be implied by 0.0359
2nd A1 for awrt 0.0359 (calc. gives 0.035930266…). Correct ans only 3/3. M1A0A1 not poss.
| Scheme | Marks |
|---|---|
| \(\dfrac{d - 100}{15} = -1.2816\) (Calculator gives \(-1.2815515\ldots\)) | M1, B1 |
| \(d = 80.776\) (awrt 80.8) | A1 |
| (3) |
Notes
M1 for an attempt to standardise with 100 and 15 and set \(= \pm\) any \(z\) value (\(|z| \gt 1\))
B1 for \(z = \pm 1.2816\) (or better) seen anywhere [May be implied by 80.776(72…) or better seen]
A1 for awrt 80.8 (can be scored for using 1.28 but then they get M1B0A1). The 80.8 must follow from correct working.
Calc If answer is awrt 80.8 and awrt 80.777 or 80.776… or better seen then award M1B1A1
If answer is awrt 80.8 or 80.77 then award M1B0A1 (unless of course \(z = 1.2816\) is seen)
| Scheme | Marks |
|---|---|
| Require \(\mathrm{P}(L \gt 133 \mid L \gt 127)\) | M1 |
| \(= \left[\dfrac{\mathrm{P}(L \gt 133)}{\mathrm{P}(L \gt 127)}\right] = \dfrac{\mathrm{P}(Z \gt 2.2)}{\mathrm{P}(L \gt 127)}\) | dM1 |
| \(= \left[\dfrac{1 - 0.9861}{1 - 0.9641}\right] = \dfrac{0.0139}{[0.0359]}\) | A1 |
| \(= 0.3871\ldots\) = awrt 0.39 | A1 |
| (4) | |
| (10 marks) |
Notes
S.C. An attempt at \(\mathrm{P}(L \lt 133 \mid L \gt 127)\) that leads to awrt 0.61 (M0M1A0A0)
1st M1 for clear indication of correct conditional probability or attempt at correct ratio. So clear attempt at \(\dfrac{\mathrm{P}(L \gt 133)}{\mathrm{P}(L \gt 127)}\) is sufficient for the 1st M1
2nd dM1 dependent on 1st M1 for \(\mathrm{P}(L \gt 133)\) leading to \(\mathrm{P}(Z \gt 2.2)\).
1st A1 for 0.0139 or better seen coming from \(\mathrm{P}(Z \gt 2.20)\). Dependent on both Ms
2nd A1 for awrt 0.39. Both Ms required
ALT If they assume Alice did not check that the phone was working you may see:
\([\mathrm{P}(L \lt 127).0] + \mathrm{P}(L \gt 127).\underline{\mathrm{P}(L \gt 133 \mid L \gt 127)}\) Provided the conditional probability is seen as part of this calculation the 1st M1 can be scored and their final answer will be 0.0139(4/4)
An answer of 0.0139 without sight of the conditional probability is 0/4.