Notes

B1: One contextual correct assumption relating to independence/constant rate/singly
Must mention goals or match, or equivalent context.

B1: Two correct contextual assumptions with goals/matches mentioned at least once.
Assumptions relating to probability or number of goals are B0, but do not penalise if two correct assumptions also given.

Notes

B1: Stating or using both correct Poisson distributions [0.164… or 0.365… implies \(\mathrm{Po}(0.9)\), 0.160… implies \(\mathrm{Po}(1.8)\). Both are required for this mark]
or writing or using \(\mathrm{B}(3, 0.5)\)

M1: Correct use of conditional probability formula. Implied by \(\dfrac{0.164\ldots \times 0.365\ldots}{0.160\ldots}\)
Or writing or using any binomial distribution and \(\mathrm{P}(X = 2)\)

M1: Correct expression with probabilities   or   writing or using \(\mathrm{B}(3, 0.5)\) and \(\mathrm{P}(X = 2)\)

A1: Correct exact answer. Do not award if there is clear evidence of rounding.

Notes

(i) B1: Stating correct binomial distribution

(ii) M1: Correct method to find the mean or variance of \(X\) (a mean of 2.2 from Poisson is M1)
ft their \(n\) and \(p\)

M1: Correct approximate normal distribution [condone using \(X\) not \(\overline{X}\)]. Allow variance awrt 0.0248

A1: awrt 0.898
NB: awrt 0.898 implies the M1M1A1

Notes

1^st B1 for awrt 0.209

2^nd B1 for awrt 0.308

Notes

B1 for both hypotheses correct in terms of \(\lambda\) or \(\mu\) (allow \(\lambda = 6\) etc)

1^st M1 for selecting the correct model. Sight or use of \(\mathrm{Po}(6)\)

2^nd M1 for use of the correct model with two probs correct to 2.s.f. (accept \(\mathrm{P}(E \geqslant 12) = 0.02\))
Must see attempt at lower and upper limit. Probabilities may be seen in (c).

A1 for correct critical region (both parts). Allow \(E \leqslant 1\) and \(E \geqslant 12\) or \(E \leqslant 1, E \geqslant 12\) etc
Writing CR as probability statements is A0
NB: Completely correct CR implies M1M1A1

SC: 1-tailed test
B0 as hypotheses are incorrect
M1 for sight or use of \(\mathrm{Po}(6)\)
M1 (dep on \(\mathrm{H}_1\)) for sight of \(\mathrm{P}(E \leqslant 1) = 0.0174\) or \(\mathrm{P}(E \geqslant 11) = 0.0426\), in line with their \(\mathrm{H}_1\)
A1 for CR: \(E \leqslant 1\) or CR: \(E \geqslant 11\), in line with their hypotheses

Notes

B1ft for 0.0375 or 0.0374 or summing their two appropriate probs (ft their CR)
NB: If candidate uses a 1-tailed test, this mark cannot be gained

Notes

B1 for both hypotheses correct which must be attached to \(\mathrm{H}_0\) and \(\mathrm{H}_1\)
must be in terms of \(\lambda\) or \(\mu\)    allow either 1.7 or 17

M1 for stating or using the correct Poisson model.
may be implied by sight of awrt 0.0406/7 or awrt 0.959 or 0.9747… or better

1^st A1 for correct prob of awrt 0.04
or for correct CR found \(X \geqslant 27\) (\(X \gt 26\)) (ignore lower tail CR if found) allow CV \(X = 27\)

2^nd A1 (dep on M1A1) for a correct conclusion in context mentioning “rate of calls” o.e.
Allow e.g. ‘The rate of calls is 1.7 per minute/17 per 10 minutes’
Must be rate o.e. not “number”
A0 if inconsistent comments are seen e.g. “reject \(\mathrm{H}_0\), no change in rate of calls”

Notes

1^st M1 for sight or use of the correct binomial model.
may be implied by sight of awrt: 0.0523 or 0.948 or 0.795 or 0.205

2^nd M1 for correct interpretation of more than 2 (allow 1 – 0.95 or better)

A1 for awrt 0.0523 (correct answer only scores 3 out of 3)

SC: Use of \(\mathrm{Po}(70 \times 0.012)\) leading to an answer of 0.0533(45…) and scores M1M1A0

Notes

1^st M1 for sight or use of \(\mathrm{Po}(900p)\) (as a suitable approx. to \(\mathrm{B}(900, p)\))
(may be implied by correct answer awrt 0.00333)

2^nd M1 for a correct equation in \(p\) or correct use of \(\mathrm{P}(C = 0)\) from Po e.g. \(\mathrm{e}^{-\lambda} = 0.05\)

3^rd M1 for a correct method to solve for \(p\) (allow \(p = \pm\ln(0.05)/900\))
or to solve for \(\lambda\), i.e. \(\lambda =\) awrt 3(.00)

A1 for \(p\) = awrt 0.00333 Must see Po used    condone \(\dfrac{1}{300}\) o.e.
Allow standard form (awrt \(3.33 \times 10^{-3}\)) or percentage (awrt 0.333%)

SC: Use of Binomial gives 0.003323… awrt 0.00332 scores M0M0M0A1

Notes

M1 For using the model \(\mathrm{Po}(11.2)\) implied by sight of: 0.02077… or 0.9889.. or 0.9792..

A1 awrt 0.021

Notes

B1 awrt 0.0616

1^st M1 Setting up a new model \(\mathrm{B}(250, \text{“}0.0616\text{”})\) [condone \(\mathrm{B}(\text{“}0.0616\text{”}, 250)\)]

2^nd M1 Seeing the model \(\mathrm{Po}(\text{their } np)\) implied by sight of: 0.1475.. or 0.89975 or 0.8524…

A1 awrt 0.148

SC if no approximation used (and 1^st M1 not seen) an answer of awrt 0.140 could get B1M1M0A0

Notes

1^st B1 Both hypotheses correct using \(\lambda\) or \(\mu\) and 16.8 or 0.4   [Accept their ans to \(0.4 \times 42\)]

2^nd B1 Realising \(\mathrm{Po}(16.8)\) needs to be used.   Sight or use of, implied by correct prob or CR

M1 For 0.014 or better (0.0141..) or CR \(X \leqslant 9\) oe must be CR and not probability.
[Allow CR \(X \leqslant 10\) with probability \(\mathrm{P}(X \leqslant 10) = 0.054\) or better]

A1 Indep of 1^st B1 (must see 2^nd B1 and M1 scored) for a correct inference in context

Notes

1^st M1 Selecting a suitable model.   Sight of \(\mathrm{Po}(0.6n)\) or \(\mathrm{B}(n, \mathrm{e}^{-0.6})\) or implied by 2^nd M1

2^nd M1 For a correct inequality or equality involving \(n\)   [Condone slips in solving]
Allow MR i.e. misread of 0.01 for 0.001 (or similar) to score M1M1A0

A1 \(n = 12\) cao   [Correct answer with no incorrect working seen scores 3/3]

Notes

B1 Allow equivalent statements.   Underlined words required.

Notes

M1: For using the given information to find \(\mathrm{E}(N)\)
ALT \(a + b + c = 0.5\) oe

M1: For use of \(\sum n\mathrm{P}(N = n) = \text{“}3.2\text{”}\)   At least 3 terms correct
ALT \(\sum(4n + 2)\mathrm{P}(N = n) = 14.8 \Rightarrow 2a + 1.2 + 0.5 + 3.5 + 18b + 22c = 14.8\)   At least 3 terms correct

M1: Forming an equation in \(b\) and \(c\) using conditional probability

M1: For using \(\sum n^2\mathrm{P}(N = n)\)   Allow with the letters \(b\) and \(c\)

dM1: Dependent on previous method mark. Correct method to find \(\mathrm{Var}(N)\)

A1*: All previous marks must be awarded and 2.76 stated

Notes

M1: Setting up a new model with the correct fees. At least 3 terms correct. Allow 0.5, 0.7, 0.9, 1

M1: Correct method for calculating E(fee) Allow with the letters \(b\) and \(c\)

A1: 81[p]   No units needed. Allow 0.81 if fees are in pounds

Notes

B1: A correct limitation.

Notes

M1: For calculating the mean and setting up the correct model. Poisson may be implied by 0.8883 or better or 1 – awrt 0.1117 but must see 3.75 or \(1.25 \times 3\)

A1*cso: \(\mathrm{P}(C \geqslant 2) =\) awrt 0.8883 or 1 – awrt 0.1117 = 0.888 Must see \(\mathrm{P}(C \geqslant 2)\) oe

Notes

M1: Setting up a new model using their answer to (a) Implied by correct answer

A1: awrt 0.0216 or awrt 0.0215

Notes

B1: awrt 0.0228

M1: Setting up a new model \(\mathrm{B}(150, \text{“}0.0228\text{”})\) and using \(np\) (working seen if incorrect)

M1: Using the model \(\mathrm{Po}(\text{their } np)\) Must be clearly stated and \(\mathrm{P}(E \geqslant 3)\) oe seen

A1*cso: Only award if the previous 3 marks have been awarded and 0.664 is stated.
NB Use of \(\mathrm{B}(150\ 0.02281)\) gives 0.668

Notes

B1: Idea that \(n = 150\) (number of periods selected) is large and \(p\) is 0.022… (exactly 8 calls in the time period) is small.

Notes

B1: Both hypotheses correct using \(\lambda\) or \(\mu\)   allow 1.25 or 3.75

Notes

B1: Realising \(\mathrm{Po}(30)\) needs to be used. NB Implied by correct answer or \(\mathrm{P}(X = 40) = 0.0139\ldots\)

M1: Writing or using \(1 - \mathrm{P}(X \leqslant 39)\) or if CR method for \(\mathrm{P}(X \geqslant 42) = 0.0221\ldots\)

A1: 0.04… or awrt 0.05 or CR \(X \geqslant 42\) oe must be CR and not probability

A1: A fully correct solution and correct inference in context. Calls required
If put this prob but then give Cr X >= 40 M1A1A0

Notes

B1: Correct reason why the model would not be appropriate and correct conclusion. Condone e.g. ‘\(p\) is close to 0.5’ for \(p\) is not small.
Mean is not equal to variance on its own in B0.

Notes

B1: Correct explanation mentioning independence (oe).
Ignore extraneous comments.

Notes

M1: Using \(\mathrm{Po}(7)\) with 2.4

A1: awrt 0.0296

Notes

(i) B1: For realising the distribution is \(\mathrm{Po}(24)\) (May be seen or implied in part (ii))

B1: awrt 0.0719

(ii) M1: Writing or using \(1 - \mathrm{P}(X \leqslant 20)\)

A1: awrt 0.757

Notes

B1: Both hypotheses correct (must use \(\mu\) or \(\lambda\))

B1: awrt 0.0774   Allow awrt 0.08 from a correct probability statement.
allow CR: \(X \leqslant 9\)

M1: Correct non-contextual conclusion (may be implied by correct contextual conclusion). Allow a f.t. comparison of ‘their \(p\)’ with 0.05
(Ignore any contradictory contextual comments for this mark)

A1: A fully correct solution drawing a correct inference in context with all previous marks in (b) scored.

Notes

M1: Use of \(\mathrm{Po}(8)\) to attempt critical region [\(\mathrm{P}(Y \leqslant 3) = 0.0423..\ \ \mathrm{P}(Y \leqslant 4) = 0.0996..\)]

A1: Finding critical region for the test \(Y \leqslant 3\) which must come from \(\mathrm{Po}(8)\).

B1: Identifying the need to use \(\mathrm{Po}(4)\) as the true distribution.
Allow \(\mathrm{Po}(4)\) seen or used for this mark.

M1: Writing or using \(\mathrm{P}(W \geqslant \text{‘}4\text{’})\) or \(1 - \mathrm{P}(W \leqslant \text{‘}3\text{’})\) from \(\mathrm{Po}(4)\). Allow f.t. on their identified CR but must be using \(\mathrm{Po}(4)\)

A1: awrt 0.567

Notes

B1 for both hypotheses in terms of \(\lambda\) or \(\mu\)  (either way around)

1^st M1 for selecting the correct Po model. Sight or use of \(\mathrm{Po}(7.5)\) may be implied by 2^nd M1

2^nd M1 for using the correct model to find one of these probs with correct label (2sf or better)

1^st A1 for one end correct

2^nd A1 for a fully correct CR
Allow any letter, even CR \(\leqslant 2\) or set notation but not \(\mathrm{P}(X \leqslant 2)\)
Can have \(X \lt 3\) and \(X \gt 13\) etc

Notes

B1ft for awrt 0.0419 or awrt 0.0418
or ft addition of their two probs provided both are \(0 \lt \text{prob} \lt 0.025\)  (awrt 3sf)

Notes

1^st M1 for selecting a correct binomial model, ft their answer to part (b)

2^nd M1 for a correct probability statement of \(1 - \mathrm{P}(M \leqslant 1)\) dep on a binomial selected

A1cso for answer in range [0.04139, 0.04154] dep on use of \(\mathrm{B}(8, \text{“}0.0419\text{”})\) or better

Notes

1^st M1 for selecting a \(\mathrm{Po}(6.3)\) model

2^nd M1 for a correct probability statement using their Poisson model and their CR in (a) which may have just one tail.

A1 for awrt 0.945

Notes

1^st M1 for selecting a Poisson model – written or used.  May be implied by 2^nd M1 or a correct Answer.

2^nd M1 for the correct Poisson \(\mathrm{Po}\left(\frac{20}{3}\right)\) or \(\mathrm{Po}(6.67)\) or better seen and writing or using \(1 - \mathrm{P}(C \leqslant 4)\)

A1 for awrt 0.794  (correct ans with no incorrect working scores 3/3)

Notes

M1 for selecting a correct model \(\mathrm{B}(4, 0.189)\) or better  (calc: \(0.188875\ldots\))

A1 for using the model to get awrt 0.0219 (correct ans with no incorrect working scores 2/2)

Notes

1^st M1 for a correct prob of 1 call (expressions in e or values)
(allow \(0.31479\ldots\) or awrt 0.315 or \(0.033689\ldots\) or awrt 0.0337)

2^nd M1 for a correct probability statement or expression.
E.g. \(\mathrm{P}\left(S = 1 \mid S \sim \mathrm{Po}\left(\frac{5}{3}\right)\right) \times \mathrm{P}(T = 1 \mid T \sim \mathrm{Po}(5))\)

SC e.g. \(F \sim \mathrm{Po}(\lambda)\) used in (b) to find \(\mathrm{P}(F = 0)\)
Then if we see \(Y \sim \mathrm{Po}(3\lambda)\) and statement \(\mathrm{P}(F = 1) \times \mathrm{P}(Y = 1)\) award M0M1

A1 for awrt 0.0106  (correct ans with no incorrect working scores 3/3)

Notes

(i) M1 for using or writing \(1 - \mathrm{P}(X \leqslant 5)\) or \(1 - \mathrm{P}(X \lt 6)\) may be implied by awrt 0.884

(ii) M1 for using or writing \(\mathrm{P}(X \leqslant 9) - \mathrm{P}(X \leqslant 3)\) oe may be implied by awrt 0.566

Notes

M1 Using normal approximation with mean = variance = 378 or sd = \(\sqrt{378}\) (awrt 19.4) or writing N(378,378) May be seen in standardisation.

M1 \(\pm\left(\dfrac{(369 \text{ or } 370 \text{ or } 369.5 \text{ or } 370.5) - \text{their mean}}{\text{their sd}}\right)\) If they have not given a mean and variance they must be correct in here. (allow 1 – standardisation)

M1d dep on previous method mark being awarded. Using a continuity correction \(370 \pm 0.5\)

A1ft standardisation with correct CC ie \(\pm\dfrac{369.5 - \text{"their 378"}}{\sqrt{\text{"their 378"}}}\) or awrt \(\pm\)0.44 or implied by 0.330 or 0.331 (allow 1 – standardisation) (0.33 must be from correct standardisation) NB 0.33 with no working gains NO marks. 0.330 or 0.331 with no working gains full marks.

Notes

M1 writing \(\mathrm{B}(5, \text{"}0.33\text{"})\) or \(\mathrm{B}(5, 1 - \text{"}0.33\text{"})\) or seeing \({}^5C_n(\text{"}0.33\text{"})^n(1 - \text{"}0.33\text{"})^{5-n}\) where \(1 \leqslant n \leqslant 4\) Allow if \({}^nC_r\) calculated or in factorial form

M1 \(1 - (1 - \text{"}0.33\text{"})^5 - 5(\text{"}0.33\text{"})^1(1 - \text{"}0.33\text{"})^4 - 10(\text{"}0.33\text{"})^2(1 - \text{"}0.33\text{"})^3 - 10(\text{"}0.33\text{"})^3(1 - \text{"}0.33\text{"})^2\) oe Allow if using \({}^nC_r\) form or factorial form

NB awrt 0.044 with no incorrect working gains M1M1A1

Notes

B1 both hypotheses, allow use of \(\mu\)

M1 for seeing \(\left[\mathrm{P}(X \geqslant 10) =\right] 0.0839\) or \(\left[\mathrm{P}(X \geqslant 11) =\right] 0.0426\) or \(\left[\mathrm{P}(X \leqslant 9) =\right] 0.9161\) or \(\left[\mathrm{P}(X \leqslant 10) =\right] 0.9574\) oe allow a sideways slip of 1. ie 6.5/5.5

A1 for seeing \(\mathrm{P}(X \leqslant 10) = 0.9574\) or \(\mathrm{P}(X \geqslant 11) = 0.0426\) or CR \(X \geqslant 11\)

A1 0.0426

NB An answer of 0.0426 implies will get M1A1A1

Notes

M1 must have 9/ value oe is not in CR allow \(0.153 \gt 0.05\)

A1ft correct statement in context – need accidents or Jonty

Notes

M1 \(\mathrm{P}(X \leqslant c - 1 \mid \lambda = 8)\) with \(c - 1\) being correct or using their \(c\). Allow if a CR is stated in the form \(X \leqslant c\) for \(1 - \mathrm{P}(X \leqslant c \mid \lambda = 8)\)

A1 awrt 0.816

Notes

(i) M1 writing or using \(1 - \mathrm{P}(X \leqslant 3)\) implied by awrt 0.242
A1 awrt 0.242

(ii) B1 Using Po(0.625)
M1 finding \(\mathrm{P}(X = 3)\) with any \(\lambda\) e.g \(\dfrac{\mathrm{e}^{-\lambda}\lambda^3}{3!}\) or \(\mathrm{P}(X \leqslant 3) - \mathrm{P}(X \leqslant 2)\) – may be implied by awrt 0.0218
A1 awrt 0.0218

Notes

1st M1 for writing or using \(1 - \mathrm{P}(X = 0) \lt 0.2\) or \(\mathrm{P}(X = 0) \gt 0.8\) oe allow use of = instead of > or <. May be implied by \(\mathrm{e}^{-\lambda} = 0.8\) or \(\mathrm{e}^{-\lambda} \gt 0.8\) or by awrt 5.36 or 0.089

2nd M1 writing an inequality of the form \(\mathrm{e}^{-\lambda} \gt 0.8\) using any \(\lambda\). May be implied by or by awrt 5.36 or 0.089 Do not allow \(\mathrm{e}^{-\lambda} = 0.8\)

A1cso both the method marks must be awarded. Accept 5 or \(t = 5\) or \(t \lt 5\)

Notes

B1 both hypotheses using \(\lambda\) or \(\mu\) - allow 5 or 2.5 and it must be clear which is \(\mathrm{H}_0\) and which is \(\mathrm{H}_1\)

1st M1 writing or using Po(5) and \(1 - \mathrm{P}(X \leqslant 9)\) May be implied by a correct CR. Do not allow for writing \(\mathrm{P}(X \geqslant 10)\)

1st A1 awrt 0.0318. Allow CR \(X \geqslant 10\) or \(X \gt 9\)

NB allow M1A1 if not using CR route for \(\mathrm{P}(X \leqslant 9)\) = awrt 0.968

2nd M1 dependent on previous M being awarded. A correct statement (do not allow if there are contradicting non-contextual statements). ft their Prob/CR compared with 0.05/10 (0.95 if using 0.968)

2nd A1 A correct contextual statement must include the word calls and the idea the rate has increased. (do not allow “it has changed” on its own oe). All previous marks must be awarded for this mark to be awarded.
M1A1 is awarded for a correct contextual statement on its own provided previous marks have been awarded

Notes

B1: Cao Allow 141/100

M1: using \(\dfrac{\sum fx^2}{100} - (\text{their mean})^2\) or \(\dfrac{100}{99}\left(\dfrac{\sum fx^2}{100} - (\text{their mean})^2\right)\) oe
NB Allow the square root of this for the M mark.
If no working shown for \(\sum fx^2\) then you must see 343, 3.43 or a correct answer

A1: awrt 1.44 or 1.46 for \(s^2\)

Notes

B1: Cao - allow alternative wording
Allow mean equals variance

Notes

(i) M1: writing or using \(\dfrac{\mathrm{e}^{-\lambda}\lambda^2}{2!}\) or \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1)\)
A1: awrt 0.251

(ii) M1: writing or using \(1 - \mathrm{P}(X = 0)\) oe
A1: awrt 0.777

Notes

B1: Writing Po(7.5)

M1: writing \(\mathrm{P}(Y \geqslant 11)\) or \(1 - \mathrm{P}(Y \leqslant 10)\) oe

A1: Seeing 1 – 0.8622 leading to 0.1378 cso (both B1 and M1 awarded)

Notes

M1: using \((p)^n(1 - p)^{12 - n}\) where \(p = 0.1378\) or 0.138 condone missing \(n\mathrm{C}r\)

M1: \(\dbinom{12}{3}(p)^3(1 - p)^9\), with \(0 \lt p \lt 1\) Allow 220 or 12 C 3 instead of \(\dbinom{12}{3}\)

A1: awrt 0.152

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

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.

Notes

B1: Awrt 0.0397

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

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

Notes

M1: using or writing \(1 - \mathrm{P}(N \leqslant 9)\) or \(1 - \mathrm{P}(N \lt 10)\)

A1: awrt 0.413

Notes

B1: using or writing Po(1.8)

M1: for a single term of the form \(\dfrac{\mathrm{e}^{-\lambda}\lambda^2}{2!}\) with any value for \(\lambda\) or \(\mathrm{P}(X \leqslant 2) - \mathrm{P}(X \leqslant 1)\)

A1: awrt 0.268

Notes

M1: Using or writing, normal approximation with mean = 450

M1: Using or writing the mean = variance. Does not need to be 450. May be seen in the standardisation calculation.

M1: \(\pm\left(\dfrac{(470 \text{ or } 469.5 \text{ or } 470.5) - \text{their mean}}{\text{their sd}}\right)\) May be implied by a correct answer or \(z\) = awrt 0.92

M1: dep on previous method mark being awarded. Using a continuity correction \(470 \pm 0.5\) May be implied by a correct answer or \(z\) = awrt 0.92

A1: correct standardisation no need to subtract from 1. Award for \(\dfrac{469.5 - 450}{\sqrt{450}}\) or awrt 0.92 or a correct answer

A1: awrt 0.179

Notes

B1 for suitable hypotheses

M1 for correct use of Po(5). Award if one relevant probability is seen or a correct CR. Allow if a correct CR written as a Probability statement

1st A1 for \(X \leqslant 1\) or \(X \lt 2\) or \(0 \lt X \lt 2\) or \(0 \leqslant X \lt 2\) or \(0 \lt X \leqslant 1\) oe. Allow any letter

2nd A1 for \(X \geqslant 10\) or \(X \gt 9\) or \(10 \leqslant x \leqslant 40\) or \(9 \lt x \leqslant 40\) oe. Allow any letter

Ignore any \(\cup\) or \(\cap\) signs

Do not allow CR written as probability statements

Notes

M1 for adding their probabilities of ‘their’ critical regions if sum gives a probability less than 1 or award if a correct answer given

A1 for awrt 0.0722 (o.e)

Notes

B1 for suitable hypotheses

1st M1 for normal approximation

1st A1 for mean = 25 and variance = 25 or sd = 5 may be seen in the standardisation formula or implied by a correct answer

2nd M1 for attempting a continuity correction (Method 1: \(19 \pm 0.5\) / Method 2: \(x \pm 0.5\))

3rd M1 for standardising using their mean and their standard deviation and using either Method 1 [19.5, 19, 18.5 accept \(\pm z\).] Method 2 [\((x \pm 0.5)\) and equal to a \(\pm z\) value]

2nd A1 for awrt 0.136 or 35.3 or \(-1.1 \gt -1.96\)

3rd A1 for a correct contextualised conclusion. cao for a one tailed test, must come from correct working. Condone incorrect hypotheses.

NB if finding \(\mathrm{P}(X = 19)\) ie \(\mathrm{P}(X \leqslant 19.5) - \mathrm{P}(X \leqslant 18.5)\) they can get B1 M1 A1 M1 M1 A0 A0

(Note: the printed critical-value method uses 1.96, 35.3 and −1.96, which are as printed in the mark scheme. For this one-tailed 5% test the critical value is \(z = -1.6449\), which gives \(\dfrac{x + 0.5 - 25}{5} = -1.6449\), \(x = 16.3\); 19 is not in the critical region, so the conclusion is the same.)

Notes

M1 for Poisson (accept Po). Condone P(9)
A1 for mean of 9

Notes

M1 for writing \(1 - \mathrm{P}(X \leqslant 7)\). This may be implied by \(1 - 0.3239\) or a correct answer
A1 for awrt 0.676

Notes

B1 Po(1.5) written or used
M1 writing or using \(1 - \mathrm{P}(Y = 0)\) or \(1 - \mathrm{P}(Y \leqslant 0)\) or \(1 - \mathrm{e}^{-\lambda}\) [may not be \(Y\)]
A1 for at least (1 – 0.223) or better. No need for final comment. * answer given so 0.777 does not imply all three marks

Notes

1st M1 for identifying binomial with \(n = 6\) and \(p = 0.777\) or better. Condone use of \(p = 0.223\). May be implied by \((p)^4(1 - p)^2\), \(p\) = awrt 0.777 or awrt 0.223

2nd M1 Must have \({}^6\mathrm{C}_4\,(0.777)^4(1 - 0.777)^2\)

A1 for awrt 0.272

Notes

1st M1 for a correct expression/selection of probabilities

2nd M1 for use of Po(6) and at least one correct prob. seen
May see \(\mathrm{P}(X = 9) = \dfrac{\mathrm{e}^{-6}6^9}{9!} = 0.06883\ldots\) or \(\mathrm{P}(X = 10) = \dfrac{\mathrm{e}^{-6}6^{10}}{10!} = 0.04130\ldots\)

1st A1 for a fully correct expression
2nd A1 for awrt 0.0473

Notes

B1 for evidence of use of Po(8)
M1 for an expression of the correct form with at least one correct prob.
A1 for awrt 0.225

Notes

B1 any correct statement with context of emails in

B1d Dependent on previous B1. Any correct statement, need not have context

SC for 2 correct statements without context B1 B0

Notes

B1 \(X = 0\) or \(X \leqslant 0\) Allow any letter.

B1 \(X \geqslant 9\) or \(X \gt 8\) Allow any letter.

SC if write correct CR’s as probability statements award B1 B0

For these 2 marks ignore any union sign (\(\cup\)) or intersection sign (\(\cap\))

Notes

M1 adding their probabilities of ‘their’ critical regions if sum gives a probability less than 1 or award if a correct answer given

A1 awrt 0.0397

Notes

M1 correct reason ft their CR. Do not allow non-contextual contradictions.

A1 correct conclusion for their CR. Allow conclusion in context of emails are received at a rate of 2 every 5 mins

Notes

(corrected from the printed mark scheme: the alternative form of \(\mathrm{H}_1\) is printed as “(or \(\lambda = 2\))”; it should be \(\lambda \lt 2\))

B1 both hypotheses correct, must have \(\lambda\) or \(\mu\) and either 2 or 6.

M1 using Po(6) may be implied by correct answer.

A1 0.062 or \(X \leqslant 2\)

M1 dependent on previous method being awarded. Do not allow conflicting non-contextual statements. Follow through their hypotheses.

Notes

M1 use of Po(6)

A1 correct CR. May be implied by correct probability. Alow if written as part of a 2 tailed CR
A1 awrt 0.062

Notes

M1 using Po(4) and \(1 - \mathrm{P}(X \leqslant 2)\), ft their CR in (b) if one tail
A1 awrt 0.762

Notes

B1 Po(9) written or used in either (i) or (ii)

(i) M1 writing \(\mathrm{P}(X \leqslant 7) - \mathrm{P}(X \leqslant 6)\) or \(\dfrac{\mathrm{e}^{-\lambda}\lambda^7}{7!}\)

This may be implied by 0.3239 – 0.2068

A1 awrt 0.117

(ii) M1 writing \(1 - \mathrm{P}(X \leqslant 9)\)

This may be implied by 1 – 0.5874.

A1 awrt 0.413

Notes

B1 Po(1.5) written or used

M1 writing or using 1 – P(0) or \(1 - \mathrm{e}^{-\lambda}\).

This may be implied by 1 – 0.2231

A1 awrt 0.777

Notes

B1 for Po(8) seen or implied by use

M1 for clear evidence of use of Po(8), may be implied by a correct CR (allow written as a probability statement) or a probability seen in part(b). If they give 3 and 14

1st A1 for \(X \leqslant 3\) or \(0 \leqslant X \leqslant 3\) or 0,1,2,3 or [0,3] Allow any letter

2nd A1 for \(X \geqslant 14\) or \([14, \infty)\) condone \([14, \infty]\) Allow any letter

These A marks must be for statements with \(X\) only – not in prob statements

Notes

M1 for showing they are adding together the two probabilities that correspond to their CR or allow M1 A1for correct answer

Notes

B1 for both hypotheses. Must be in terms of \(\lambda\) or \(\mu\), 8 or 80 can be swapped

1st M1 for normal approx

1st A1 E(Y) = 80 and Var(Y) = 80 (or correct st. dev seen somewhere)

2nd M1 for use of a continuity correction 94.5 or 95.5

3rd M1 Standardising using their mean and their sd, If they have not written down a mean and sd then these need to be correct here to award the mark. They must also use 94.5, 95.5 or 95 and find the correct area ie using \(1 - \mathrm{P}(Z \leqslant\) “their 1.62”)

2nd A1 for awrt 0.053 or awrt 0.947

4th M1 for a correct statement based on their probability and 0.05

3rd A1 cso for a correct contextualised statement and a fully correct solution with no errors seen. Need either bakery’s claim or Raisins and muffin

NB If Found \(\mathrm{P}(X = 95)\) they can get B1 M1 A1 M0M0A0M0A0

Notes

M1 for a correct method. May use incorrect \(\lambda\)

A1 for awrt 0.168

Notes

B1 for writing or using Po(6)

M1 for writing or using \(1 - \mathrm{P}(Y \leqslant 10)\)

A1 for awrt 0.0426

Notes

1st M1 for identifying that \(C \sim \mathrm{B}(5, 0.3)\). Follow through their 0.3. May be implied

2nd M1 for writing or using \(1 - \mathrm{P}(C \leqslant 2)\)

A1 for awrt 0.163

SC if they use normal distribution they may get M0 M1 A0 if they find \(\mathrm{P}(C \geqslant 2.5)\)

Notes

B1 for 0.8 for P(Bethan is served in the next 4 minutes)

M1 for identifying Po(1.2)

A1 for \(\mathrm{e}^{-1.2}\) or awrt 0.301...

A1 for awrt 0.241

Notes

M1 for correct expression for size using Po(2)

Notes

1st M1 for a correct expression in terms of probabilities. Allow \(1 - \mathrm{P}(X \leqslant 2)\) or \(1 - \mathrm{P}(X \lt 3)\)

1st A1 for correct equation in \(\lambda\)

2nd A1 cso

Notes

SC if both correct but not to 2dp award B1B0

Notes

1st B1ft points

2nd B1ft curve (or straight lines) through points

Notes

M1 for correct expression for size using Po(2)

Notes

M1 for correct expression using Po(5)

Notes

1^st M1 for a correct expression in terms of probabilities
Alternate answer \(1 - [\mathrm{P}(X = 0) + \mathrm{P}(X = 1) \times \mathrm{P}(X \leqslant 1)]\)
2^nd M1 for an attempt at a correct equation in \(\lambda\)
1^st A1 for a correct expression in \(\lambda\)

Notes

1^st B1 points
2^nd B1 curve (or straight lines)

Notes

1^st M1 for an attempt to calculate expected time
Alternate method \(15 + 15 \times \mathrm{P}(X = 1)\)
1^st A1 for a correct expression in terms of \(\lambda\)
2^nd M1 for attempt at correct inequality

Notes

1^st B1 for a comment about power & timings
2^nd B1 for selecting second test

Notes

B1 stating or using Po(7)

M1 stating or using \(1 - \mathrm{P}(X \leqslant 10)\)

Notes

M1 using or writing \(\mathrm{P}(X \gt d) \lt 0.05\) or \(\mathrm{P}(X \lt d) \gt 0.95\) (condone \(\geqslant\) instead of \(\gt\) and \(\leqslant\) instead of \(\lt\)) May be implied by correct answer. Different letters may be used.

1st A1 \(\mathrm{P}(X \leqslant 12)\)/ \(\mathrm{P}(X \lt 13) =\) awrt 0.973 or \(\mathrm{P}(X \leqslant 11)\) / \(\mathrm{P}(X \lt 12) =\) awrt 0.947 May be implied by a correct answer

2nd A1 12 or 13

NB An answer of 12/13 on its own with no working gains M1A1A1

Notes

1st B1 both hypotheses correct using \(\lambda\) or \(\mu\), and 1 or 28

2nd B1 for writing or using a normal approximation with correct mean and Var (may be given if sd correct in standardisation formula)

1st M1 for use of a continuity correction 35.5 or 36.5 or \(x \pm 0.5\)

2nd M1 Standardising using their mean and their sd. If they have not written down a mean and sd then these need to be correct here to award the mark. They must use [35.5, 36.5, 36, \(x\) or \(x \pm 0.5\)] For CR must have = awrt 1.64 or 1.65

1st A1 awrt 0.0778 or 0.9222 or the statement \(1.42 \lt\) awrt 1.65/1.64 or CR \(X \geqslant 37.2\)/ \(X \gt 37.2\)

3rd M1 a correct conclusion for their probability. May be implied by a correct contextual conclusion. NB Non contextual contradicting statements gets M0

2nd A1 a correct contextual conclusion for their hypotheses and a fully correct solution with no errors seen. Need the words “rate/average number”, “sales” and “increased”oe

NB If found \(\mathrm{P}(X = 36)\) they can get B1B10M0A0M0A0

Notes

M1 \(0.25\mathrm{e}^{-0.25}\) o.e

Notes

B1 stating or using Po(1.5)

M1 stating or using \(1 - \mathrm{P}(X \leqslant 2)\)

Notes

1st B1 for normal approximation and correct mean

2nd B1 Var \((X) = 75\) or sd \(= \sqrt{75}\) or awrt 8.66 (may be given if correct in standardisation formula)

1st M1 using either 89.5 or 88.5

2nd M1 Standardising using their mean and their sd, using [89.5, 88.5 or 89] and for finding correct area

NB use of Poisson gives an answer of 0.9498 and gains no marks

Notes

B1 Writing or using Po(3) in either (i) or (ii)

(i) M1 writing or using \(\mathrm{P}(X \leqslant 7) - \mathrm{P}(X \leqslant 6)\) or \(\dfrac{\mathrm{e}^{-\lambda}\lambda^7}{7!}\)

(ii) M1 writing or using \(1 - \mathrm{P}(X \leqslant 3)\). (Do not accept writing \(1 - \mathrm{P}(X \lt 4)\) unless they have used \(1 - \mathrm{P}(X \leqslant 3)\)).

Notes

1st M1 for writing or using a normal approximation

1st A1 for correct mean and sd (may be given if correct in standardisation formula)

2nd M1 Standardising using their mean and their sd and using [18.5, 19, 19.5, 20 or 20.5] and for finding correct area by doing 1 – P(\(Z \leqslant\) “their 1.92”) If they have not written down a mean and sd then these need to be correct here to award the mark

3rd M1 for attempting a continuity correction (\(19 \pm 0.5\)) i.e. 18.5 or 19.5 only.

2nd A1 for \(\pm\dfrac{19.5 - 30}{\sqrt{30}}\) or \(\pm\) awrt 1.9 or better.

3rd A1 awrt 0.0274, 0.0275 or 0.0276

SC using \(\mathrm{P}(X \lt 20.5/19.5) - \mathrm{P}(X \lt 19.5/18.5)\) can get M1A1 M0M1A0A0

Notes

B1 Accept \(n\) (the number of trials) large / high / big / \(n \gt 50\) (accept any number larger than 50)

B1 Accept \(p\) (the probability) small / close to 0 / \(p \lt 0.2\) ( accept any number less than 0.2). Do not accept low.

These must appear in part (a).

Notes

B1 writing or using Po(10)

M1 using a Poisson (\(\lambda\) need not equal 10) and for writing or using \(1 - \mathrm{P}(X \leqslant 3)\). (Do not accept writing \(1 - \mathrm{P}(X \lt 4)\) unless they have used \(1 - \mathrm{P}(X \leqslant 3)\)).

A1 0.9897 cao must be 4 dp

NB

An awrt 0.990 on its own gains B0M0A0 unless there is evidence that Po(10) is used. In which case it gets B1M1A0

Using B(1000,0.01) gives 0.989927…. and gains B0M0A0

Notes

1st B1 for writing or using Po(8) in either (i) or (ii)

(i) M1 writing or using \(\mathrm{P}(X \leqslant 3) - \mathrm{P}(X \leqslant 2)\) or \(\dfrac{e^{-8}8^3}{3!}\)

(ii) M1 writing or using \(1 - \mathrm{P}(X \leqslant 5)\)

Notes

M1 writing or attempting to use B(12,their (a(ii))) NB ft their a(ii) to at least 2sf

M1 \(\dfrac{12!}{9!3!}\)(a(ii))\(^9\)(1- a(ii))\(^3\)   allow \({}^{12}\mathrm{C}_3\) or \({}^{12}\mathrm{C}_9\) or 220 instead of \(\dfrac{12!}{9!3!}\) NB ft their a(ii) to at least 1sf but an expression must be seen (No use of tables)

Notes

1st M1 for writing or using a normal approximation

1st A1 for correct mean and sd (may be given if correct in standardisation formula)

2nd M1 Standardising using their mean and their sd and using [24.5, 25, 25.5, 26 or 26.5] and for finding correct area by doing \(1 - \mathrm{P}(Z \leqslant\) “their 1.23”)

NB if they have not written down a mean and sd then they need to be correct in the standardisation to gain this mark.

3rd M1 for attempting a continuity correction (\(26 \pm 0.5\))

2nd A1 for \(\pm\dfrac{25.5 - 20}{\sqrt{20}}\) or \(\pm\) awrt 1.2 or better.

SC using \(\mathrm{P}(X \lt 26.5/25.5) - \mathrm{P}(X \lt 25.5/24.5)\) can get M1A1 M0M1A0A0

Notes

1st B1 for \(\mathrm{H}_0 : p = 0.03\)

2nd B1 for \(\mathrm{H}_1 : p \gt 0.03\)

SC If both hypotheses are correct but a different letter to \(p\) is used they get B1 B0

Also allow B1 B0 for \(\mathrm{H}_0 : \lambda = 6\) and \(\mathrm{H}_1 : \lambda \gt 6\)

B1 writing or using Po(6)

One tail

1st M1 for writing or using \(1 - \mathrm{P}(X \leqslant 11)\) or giving \(\mathrm{P}(X \leqslant 10) = 0.9574\) or giving \(\mathrm{P}(X \geqslant 11) = 0.0426\). May be implied by correct CR or probability = 0.0201

1st A1 for 0.0201 or CR \(X \geqslant 11\)/ \(X \gt 10\). NB \(\mathrm{P}(X \leqslant 11) = 0.9799\) on its own scores M1A1

2nd M1 dependent on the 1st M1 being awarded. For a correct statement based on the table below. Do not allow non-contextual conflicting statements eg “significant” and “accept \(\mathrm{H}_0\)”. Ignore comparisons.

2nd A1 ft for a correct contextualised statement. NB A correct contextual statement on its own scores M1A1.

Two tail

1st M1 for writing or using \(1 - \mathrm{P}(X \leqslant 11)\) or giving \(\mathrm{P}(X \geqslant 12) = 0.0201\) or giving \(\mathrm{P}(X \leqslant 11) = 0.9799\). May be implied by correct CR or probability = 0.0201

1st A1 for 0.0201 or CR \(X \geqslant 12\)/ \(X \gt 11\). NB \(\mathrm{P}(X \leqslant 11) = 0.9799\) on its own scores M1A1

2nd M1 dependent on the 1st M1 being awarded. For a correct statement based on the table below. Do not allow non-contextual conflicting statements eg “significant” and “accept \(\mathrm{H}_0\)”. Ignore comparisons.

2nd A1 ft for a correct contextualised statement. NB A correct contextual statement on its own scores M1A1.

Use of N(6,5.82) May get B1 B1 B0 M1 (must use 11.5)A0 M1dep A1 ft

Notes

(i) Allow accept H₁ instead of reject H₀. It must be clear which hypothesis gets rejected/accepted.
(ii) Allow equivalent wording.

Notes

M1 Writing or using Po(8). May be implied by correct critical region.
A1 allow \(0 \leqslant X \leqslant 3\) or CR \(\leqslant 3\) or \(X \leqslant 3\). Any letter may be used but not \(\mathrm{P}(X \leqslant 3)\). This must be on its own.

Notes

B1 both hypotheses correct. Must use \(\lambda\) or \(\mu\).

One tail
1st M1 for writing or using \(1 - \mathrm{P}(X \leqslant 12)\) or writing \(\mathrm{P}(X \leqslant 13) = 0.9658\) or \(\mathrm{P}(X \geqslant 14) = 0.0342\). May be implied by correct CR.or probability = 0.0638
A1 for 0.0638 or \(X \geqslant 14\). Allow \(X\) >13. NB \(\mathrm{P}(X \leqslant 12) = 0.9362\) on its own scores M1A1
2nd M1 dependent on the 1st M1 being awarded. For a correct statement based on the table below. Do not allow non-contextual conflicting statements eg “not significant” and “reject H₀”. Ignore comparisons.
2nd A1 for a correct contextualised statement. NB A correct contextual statement on its own scores M1A1.

Two tail
1st M1 for writing or using \(1 - \mathrm{P}(X \leqslant 12)\) or writing \(\mathrm{P}(X \leqslant 14) = 0.9827\) or \(\mathrm{P}(X \geqslant 15) = 0.0173\). May be implied by correct CR.or probability = 0.0638
A1 for 0.0638 or \(X \geqslant 15\). Allow \(X\) >14. NB \(\mathrm{P}(X \leqslant 12) = 0.9362\) on its own scores M1A1
2nd M1 dependent on the 1st M1 being awarded . For a correct statement based on the table below. Do not allow non-contextual conflicting statements eg “not significant” and “reject H₀”.Ignore comparisons.
2nd A1 for a correct contextualised statement. NB A correct contextual statement on its own scores M1A1.

Notes

B1 Writing or use of B(120,0.075) may be implied by using Po(9) or N(9,8.325)
1st M1 writing or use of Poisson
1st A1 writing or use of Po(9)
2nd M1 for writing or using \(1 - \mathrm{P}(X \leqslant 3)\) or this may be implied by an awrt 0.972 using normal approximation.

Notes

M1 ((their (a))² or \(0.979^2\) or \(0.9788^2\) or \(0.98^2\)

Notes

1st B1 Any one of the 3 statements - no context required. NB It must be a constant (mean) rate and not a constant probability or a constant mean.
2nd B1 A different statement with context of hits. NB random and independent are the same statement.
If only one mark awarded give the 1st B1. Never award B0 B1

Notes

B1 writing or using Po(5)
M1 writing or using \(\mathrm{P}(X \leqslant 10) - \mathrm{P}(X \leqslant 9)\) or \(\dfrac{e^{-5}5^{10}}{10!}\)

Notes

B1 writing or using Po(10)
M1 writing or using \(1 - \mathrm{P}(X \leqslant 14)\)

Notes

1st B1 for a normal approximation 2nd B1 for correct mean and sd (may be seen in standardisation formula
1st M1 for attempting a continuity correction (71 \(\pm\) 0.5)
2nd M1 Standardising using their mean and their sd and using [69.5, 70, 70.5, 71 or 71.5] allow \(\pm\) z
NB if they have not written down a mean and sd then they need to be correct in the standardisation to gain this mark.
1st A1 for \(z = \pm\) awrt 2.9 or better. May be awarded for \(\pm\dfrac{70.5 - 50}{\sqrt{50}}\)
3rd M1 for 1 - tables value

SC using P(\(X\)< 70.5/71.5) – P(\(X\)<69.5/70.5) can get B1B1 M0M1A0 M0A0

Notes

M for at least 2 terms on numerator. 359/80 only award M0A0

Notes

M for 80xPoisson probability with 4.4875 and either 2 or 4.

1st A1 for awrt 9.06 and 2^nd A1 for awrt 15.20 or 15.21