AS June 2019 Q3
3. Andreia’s secretary makes random errors in his work at an average rate of 1.7 errors every 100 words.
Andreia asks the secretary to produce a 250-word article for a magazine.
Andreia offers the secretary a choice of one of two bonus schemes, based on a random sample of 40 pieces of work each consisting of 100 words.
In scheme A the secretary will receive the bonus if more than 10 of the 40 pieces of work contain no errors.
In scheme B the bonus is awarded if the total number of errors in all 40 pieces of work is fewer than 56
Following the bonus scheme, Andreia randomly selects a single 500-word piece of work from the secretary to test if there is any evidence that the secretary’s rate of errors has decreased.
| Scheme | Marks | AO |
|---|---|---|
| [\(X =\) number of errors in 100-word piece] \(X \sim \mathrm{Po}(1.7)\) | M1 | 3.3 |
| \(\mathrm{P}(X \lt 2) = \mathrm{P}(X \leqslant 1) = 0.49324\ldots\) awrt 0.493 | A1 | 1.1b |
| (2) |
Notes
M1 for selecting the correct Poisson distribution
A1 for awrt 0.493
| Scheme | Marks | AO |
|---|---|---|
| [\(R =\) number of errors in the article] \(R \sim \mathrm{Po}(4.25)\) | M1 | 3.3 |
| \(\mathrm{P}(R = 5) = 0.16482\ldots\) awrt 0.165 | A1 | 1.1b |
| (2) |
Notes
M1 for selecting the correct Poisson distribution
A1 for awrt 0.165
| Scheme | Marks | AO |
|---|---|---|
| Scheme A: Let \(A \sim \mathrm{B}(40, \mathrm{e}^{-1.7})\) or \(\mathrm{B}(40, 0.18268\ldots)\) | M1 | 3.3 |
| \(\mathrm{P}(A \gt 10) = 1 - \mathrm{P}(A \leqslant 10)\) | M1 | 1.1b |
| \(= 0.0995591\ldots\) awrt 0.0996 | A1 | 1.1b |
| Scheme B: Let \(B \sim \mathrm{Po}(40 \times 1.7)\) or \(\mathrm{Po}(68)\) \(\mathrm{P}(B \lt 56) = \mathrm{P}(B \leqslant 55) = 0.061133\ldots\) | M1 | 3.3 |
| So choose scheme A (since the probability of a bonus is greater) | A1 | 2.4 |
| (5) |
Notes
1st M1 for choosing a correct model for scheme A i.e. \(\mathrm{B}(40, \mathrm{P}(X = 0))\), where \(X \sim \mathrm{Po}(1.7)\)
Allow use of awrt 0.183 for \(\mathrm{P}(X = 0)\) … 0.183 gives answer awrt 0.101
Condone \(\mathrm{B}(0.183, 40)\) (o.e.) if it leads to a prob rounding to range (0.09~0.1) otherwise M0
2nd M1 for \(1 - \mathrm{P}(A \leqslant 10)\)
1st A1 for awrt 0.0996 [NB use of 0.183 will give awrt 0.101 and scores M1M1A0]
3rd M1 for selecting a correct Poisson model for scheme B i.e. \(\mathrm{Po}(40 \times 1.7)\) or better
2nd A1 for a correct conclusion based on comparing two probs: awrt 0.1 vs 0.061 or better
So can allow 0.1 > 0.061 leading to choosing A [Probably scores M1M1A0M1A1]
NB [ Normal approx.(not on spec) leading to \(0.06477\ldots\) might score 3rd M1 if \(\mathrm{Po}(68)\) seen but 2nd A0]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{H}_0: \lambda = 1.7\) (or \(\mu = 8.5\)) \(\mathrm{H}_1: \lambda \lt 1.7\) (or \(\mu \lt 8.5\)) | B1 | 2.5 |
| [\(E =\) no. of errors in the piece of work] \(E \sim \mathrm{Po}(8.5)\) | M1 | 3.3 |
| \(\mathrm{P}(E \leqslant 3) = 0.0301\) or \(\mathrm{P}(E \leqslant 4) = 0.0744\) | A1 | 1.1b |
| So critical region is \(E \leqslant 3\) | A1 | 2.2a |
| (4) | ||
| (13 marks) |
Notes
B1 for both hypotheses in terms of \(\lambda\) or \(\mu\) (can be interchanged)
M1 for selecting \(\mathrm{Po}(8.5)\) (sight of or use of e.g. may be implied by 1st A1)
1st A1 for some evidence of correct use of \(\mathrm{Po}(8.5)\) i.e. either of these probs (2dp or better)
May be implied by a correct critical region
2nd A1 for a correct critical region. Allow \(E \lt 4\) and allow any letter for \(E\).
Two different regions (e.g. from 2 tail test) is 2nd A0
SC Use of binomial throughout: (with hypotheses \(\mathrm{H}_0: p = 0.017\) and \(\mathrm{H}_1: p \lt 0.017\) in (d))
Scores 0 in (a) 0 in (b) possibly just 2nd M1 in (c) But allow all 4 marks in (d): B1 hypotheses, M1 for \(Y \sim \mathrm{B}(500, 0.017)\), 1st A1 for \(\mathrm{P}(Y \leqslant 3) = 0.02913\ldots\) or \(\mathrm{P}(Y \leqslant 4) = 0.07266\ldots\) 2nd A1 \(Y \leqslant 3\)
Allow probs to be to 2dp or better so 0.03 and 0.07 as in main scheme.