S2 June 2015 Q5
5. Liftsforall claims that the lift they maintain in a block of flats breaks down at random at a mean rate of 4 times per month. To test this, the number of times the lift breaks down in a month is recorded.
Over a randomly selected 1 month period the lift broke down 3 times.
The residents in the block of flats have a maintenance contract with Liftsforall. The residents pay Liftsforall £500 for every quarter (3 months) in which there are at most 3 breakdowns. If there are 4 or more breakdowns in a quarter then the residents do not pay for that quarter.
Liftsforall installs a new lift in the block of flats.
Given that the new lift breaks down at a mean rate of 2 times per month,
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{Po}(4)\) \(\mathrm{P}(X = 0) = 0.0183 \qquad \mathrm{P}(X \geqslant 8) = 0.0511\) \(\mathrm{P}(X \leqslant 1) = 0.0916 \qquad \mathrm{P}(X \geqslant 9) = 0.0214\) | M1 |
| CR \(X = 0\) \(\phantom{\text{CR }}X \geqslant 9\) | A1 A1 |
Notes
M1: using Po(4), need to see a probability from Po(4), need not be one of the 4 given here. May be implied by a single correct CR
A1: \(X = 0\) or \(X \leqslant 0\) or \(X \lt 1\)
A1: \(X \geqslant 9\) or \(X \gt 8\)
Any letter(s) may be used instead of \(X\) eg CR or \(Y\) or in words
SC candidates who write \(\mathrm{P}(X = 0)\) and \(\mathrm{P}(X \geqslant 9)\) award M1A1 A0
NB Candidates who write \(8 \lt x \leqslant 0\) oe get M1A0A0
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \lambda = 4 \quad \mathrm{H}_1 : \lambda \ne 4\) | B1 |
| There is evidence that Liftsforall’s claim is true or There is insufficient evidence to doubt Liftforall’s claim | B1ft |
Notes
B1: both hypotheses correct, labelled \(\mathrm{H}_0\) or NH or \(\mathrm{H}_n\) and \(\mathrm{H}_1\) or AH or \(\mathrm{H}_a\) may use \(\lambda\) or \(\mu\). These must be seen in part (b)
B1: ft their CR only, Do not ft hypotheses. Needs to include the word Liftsforall. If no Critical region stated in part (a) award B0 or \(\mathrm{P}(X \leqslant 3)\) = awrt 0.434 and a correct conclusion.
| Scheme | Marks |
|---|---|
| \(0.0183 + 0.0214 = 0.0397\) | B1 |
Notes
B1: Awrt 0.0397
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(B \leqslant 3 \mid B \sim \mathrm{Po}(6)) = 0.1512\) | M1 A1 |
| \(X \sim \mathrm{B}(4, 0.1512)\) | dB1ft |
| If \(0 \lt p \lt 0.5\) | |
| \(\mathrm{P}(X \leqslant 1) = \mathrm{P}(X = 0) + \mathrm{P}(X = 1)\) | M1 |
| \((1 - 0.1512)^4 + 4 \times (1 - 0.1512)^3 \times 0.1512\) | dM1 |
| \(= 0.889\) | A1 |
| If \(0.5 \lt p \lt 1\) | |
| \(\mathrm{P}(Y \geqslant 3) = \mathrm{P}(Y = 3) + \mathrm{P}(Y = 4)\) | M1 |
| \(4 \times (0.8488)^3 \times 0.1512 + (0.8488)^4\) | dM1 |
| \(= 0.889\) | A1 |
Notes
M1: using Po(6) and writing or using \(\mathrm{P}(B \leqslant 3)\) oe. A1: awrt 0.151
B1ft: dep on M1 being awarded. Using or writing B(4, “their 0.151”) for use they need \((1 - p)^4\) or \(p(1 - p)^3\) or \(p^2(1 - p)^2\)
Alternative method for first 3 marks
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(B \geqslant 4 \mid B \sim \mathrm{Po}(6)) = 0.8488\) | M1 A1 |
| \(Y \sim \mathrm{B}(4, 0.8488)\) | dB1ft |
M1: using Po(6) and writing or using \(\mathrm{P}(B \geqslant 4)\) oe A1: awrt 0.849
B1ft: dep on M1 being awarded. Using or writing B(4, “their 0.849”) for use they need \((p)^4\) or \(p^3(1 - p)\) or \(p^2(1 - p)^2\)
If \(0 \lt p \lt 0.5\): M1: using or writing \(\mathrm{P}(X = 0) + \mathrm{P}(X = 1)\) oe; M1: \((1 - p)^4 + 4 \times (1 - p)^3 \times p\) oe; A1: awrt 0.889
If \(0.5 \lt p \lt 1\): M1: using or writing \(\mathrm{P}(X = 3) + \mathrm{P}(X = 4)\) oe; M1: \((p)^4 + 4 \times (p)^3 \times (1 - p)\) oe; A1: awrt 0.889
NB: a correct answer implies full marks, lose the final A mark if got awrt 0.889 and go on to do more work