Notes

M1: for selecting \(\mathrm{B}(36, 0.015)\) sight of or use of.
Must be \(\mathrm{B}(\ldots)\) [Implied by sight of \(\mathrm{Po}(0.54)\)]
May be implied by sight of: 0.0848 or awrt 0.085 or 0.00203 or 0.998 or 0.002

(i) A1: for awrt 0.0848

(ii) A1: for awrt 0.00203 [Allow \(2.03 \times 10^{-3}\)]

Use of Poisson \(\mathrm{Po}(0.54)\) (i) 0.08496 scores A0 (ii) 0.002309… awrt 0.00231 scores A1

Notation: We do not accept “calculator speak” e.g. \(N = 36\), \(p = 0.015\) instead of \(\mathrm{B}(36, 0.015)\)

Notes

Notation: We do not accept “calculator speak” e.g. \(N = 36\), \(p = 0.015\) instead of \(\mathrm{B}(36, 0.015)\)

M1: for sight of \(\mathrm{B}(260, 0.015)\)
may be implied by probability of awrt 0.0441 or awrt 0.0264 or awrt 0.956
or correct CR \(X = 0\) [ or \(\leqslant 0\)] and \(X \geqslant 9\) or \(\mathrm{P}(X \geqslant 8) = 0.04\) or better

A1: for probability of awrt 0.0441 (selected if more probabilities seen)
Condone \(\mathrm{P}(X = 8) = 0.0441\ldots\)
If \(\mathrm{P}(X \geqslant 8)\) is explicitly seen allow 0.04 or better
or for the correct CR \(X = 0\) [ or \(\leqslant 0\)] and \(X \geqslant 9\). Allow acceptance region [1, 8] (o.e.)

A1: dep on M1A1 for a correct conclusion in context.
Allow e.g:
proportion/probability of defective components has not changed
proportion/probability of defective components is still 0.015
Allow “defectives” instead of “components”
If a comparison is seen (other than \(0.956 < 0.975\)) it must be with 0.025
Incorrect comparisons score A0 (condone missing comparison)
Do not accept contradictory statements e.g. “significant so insufficient evidence of change in proportion” would score A0
“Insufficient evidence that \(p \neq 0.015\)” is A0, we need the words.

SC: Poisson Approximation (If you see \(\mathrm{Po}(3.9)\) or probability 0.0454…)
B2 and mark as M0A1A1

B1: for probability of awrt 0.0454 or 0.04… if \(\mathrm{P}(X \geqslant 8)\) seen or CR as above

B1: for correct conclusion in context (same rules as above)

Notes

Mark (a) and (b) together for sight of the probabilities.

B1 for both hypotheses in terms of \(p\) or \(\pi\). Must be attached to \(\mathrm{H_0}\) and \(\mathrm{H_1}\)

M1 for use of the correct model. Implied by sight of at least one probability truncated or rounded to at least 2sf from: 0.0155, 0.0148, 0.0805, 0.9845, 0.9581, 0.9949
Implied by sight of fully correct CR (with no probs) so e.g. \(X = 0, X \gt 8\) scores M1A0A0

1^st A1 for at least one correct probability (to at least 3sf) with its probability statement
i.e. for \(\mathrm{P}(X = 0) =\) awrt 0.0148 or \(\mathrm{P}(X \geqslant 9) =\) awrt 0.0155

2^nd A1 (dep on M1 and 1^st A1 but not on B1)
for both correct probs (to at least 3 sf) and the correct critical region
Do not need set notation. Allow \(X \lt 1\) and \(X \gt 8\) or words e.g. “0 or greater than 8” etc
Allow “,” or “and” or “or” or “\(\cap\)” between \(X \leqslant 0\) and \(X \geqslant 9\)
\(\mathrm{P}(X = 0)\) and \(\mathrm{P}(X \geqslant 9)\) is 2^nd A0

Notes

B1ft for awrt 3.03% or correct sum of their two probabilities (provided each is less than 0.5)
Their probabilities must be to at least 2sf and relate to their CR

Notes

To score in (c) they must have a CR of the form (\(X = 0\) or \(X \lt a\)) and \(X \gt b\), where \(b\) is \(\geqslant 7\)
May be implied by \(\mathrm{P}(X \lt a)\) and \(\mathrm{P}(X \gt b)\) i.e. 2^nd A0 in (a) but correct form. Need \(b \gt a\)

B1 for a suitable comment in context that suggests no support for Freya’s belief/claim
or e.g. insufficient evidence of change in proportion/percentage of left-handed adults
or e.g. proportion/percentage of left-handed adults is not different from 10% (or …is 10%)
or e.g. 10% of adults in the country are left-handed
Do not allow contradictory comments e.g. “in CR so no support for Freya’s belief” is B0

NB A correct contextual answer in (c) using an acceptance region please send to review.

Notes

If you see any attempt using an \(n\)-sided die with \(n\) not equal to 6 please send to review.

(a) M1 for sight or use of the correct distribution. Must have B, or Bin or Bpd or Bcd and the correct value for \(n\) and \(p\), just \(n = 10, p = \tfrac{1}{6}\) is M0
Implied by one answer correct to 2dp or by sight of \(\dbinom{10}{3}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^7\) or one of:
[\(\mathrm{P}(X = 0) =\)] 0.16(1…), [\(\mathrm{P}(X = 1) =\)] 0.32(3…), [\(\mathrm{P}(X = 2) =\)] 0.29(0…), [\(\mathrm{P}(X \leqslant 3) =\)] 0.93(0…)

(i) 1^st A1 for awrt 0.155

(ii) 2^nd A1 for awrt 0.775

Notes

1^st M1 for selecting a correct model. Sight or use of correct binomial, ft their (a)(i)
May be implied by sight of [\(\mathrm{P}(D \leqslant 11) =\)] 0.78… or 0.79 or [\(\mathrm{P}(D \leqslant 12) =\)] 0.87…

2^nd M1 for correct interpretation of “at least 12” and writing or using \(1 - \mathrm{P}(D \leqslant 11)\)
We are not attempting to ft their incorrect 0.155 on our calculators here.

A1 for awrt 0.212   [Answer only 3/3]

Notes

B1 for 100 but must be seen in part (c) i.e. between (b) and (d)

Notes

1^st M1 for attempting normal with mean = 100 or ft their answer to (c)
May be implied by the correct mean and a correctly labelled s.d. (\(\sigma\)) or var (\(\sigma^2\))

1^st A1 for correctly labelled standard deviation allow \(\sqrt{\tfrac{250}{3}} = \sqrt{83.3\ldots} = 9.1(28\ldots)\) or correctly labelled variance. Implied by \(\mathrm{N}\left(\mu, \tfrac{250}{3}\right)\) or correct answer.

2^nd M1 for attempt at continuity correction i.e. sight of \(95 \pm 0.5\)

2^nd A1 for awrt 0.689   [Answer only 4/4]

NB If they don’t state the model for 1^st M1 but just give probabilities with probability statements (\(Y\) is any letter):

\(\sigma = \tfrac{250}{3}\): 1^st M1 implied by: \(\mathrm{P}(Y \gt 94.5) = 0.52(63\ldots)\), \(\mathrm{P}(Y \gt 95) = 0.52(39\ldots)\), \(\mathrm{P}(Y \gt 95.5) = 0.52(15..)\)

No cc: 1^st M1 1^st A1 implied by: \(\mathrm{P}(T \gt 95) = 0.70(805\ldots)\)
1^st M1 1^st A1 2^nd M1 implied by: \(\mathrm{P}(T \gt 94.5) = 0.72(657\ldots)\)

Exact binomial gives 0.68567… and will likely score 0/4

Typical solution

Mean = 3

Typical solution

Variance = 1.5

Typical solution

The mean is the same but the variance is different, so this is not a suitable model.

Typical solution

(d)(i)

Opportunity sampling.

(d)(ii)

Easy to do.

Typical solution

(a)(i)

0.00378

(a)(ii)

0.617

(a)(iii)

Typical solution

The event of a diner is vegan is independent from one diner to the next.

Typical solution

(i)

(ii)

Critical region is \(x \leqslant 16\), \(x \geqslant 24\)

(iii)

There is insufficient evidence to suggest that proportion of diesel cars registered in 2022 with CO emissions less than 0.3 g/km has changed

Typical solution

No diesel cars in the LDS have CO emissions more than 0.5 g/km

The LDS only uses data from cars in some regions in England but not all

Typical solution

8.4

Typical solution

Variance = 48 × 0.175 × 0.825

= 6.93

Typical solution

0.6742

Typical solution

Typical solution

Typical solution

Model is appropriate because 48 seats and 17.5% passengers travelling for business match the values of \(n\) and \(p\).

However the plane may not be full so \(n\) may not be 48.

Typical solution

Under null hypothesis

Do not reject \(\mathrm{H}_0\)

There is insufficient evidence of an increase in the proportion of local businesses that made a profit in their first year.

Typical solution

The probability of passing the driving test is constant

Typical solution

0.0157

Typical solution

0.908

Typical solution

Typical solution

12.8

Typical solution

Typical solution

\(X\) = number of customers with an internet banking account

Under null hypothesis

Do not reject \(H_0\)

There is insufficient evidence to suggest an increase in the proportion of bank customers registered for an internet banking account.

Typical solution

The probability of getting a complaint call is fixed

Calls occur independently of each other

Typical solution

(i)

0.00684

(ii)

(iii)

Typical solution

Notes

B1B1: Allow probability or percentage. Allow use of any letter as long as the parameter is defined correctly e.g. allow ‘\(p = 95\%\) where \(p\) is the % of services that arrive on time’ or ‘\(p = \mathrm{P}\)(service arrives on time)’
Deduct one B mark for each error as follows:

• 2-tail B1B0
• undefined \(p\) B1B0
• missing 0.95 or 95% B0B0.
• eg \(\mathrm{H}_0 = 0.95\) etc B0B0

M1: Correct distribution and value of \(X\), both may be stated or implied by an appropriate probability value e.g. by [\(\mathrm{P}(X \leqslant 44) =\) or \(\mathrm{P}(X \lt 45) =\)] 0.0378 or [\(\mathrm{P}(X \leqslant 43) =\) or \(\mathrm{P}(X \lt 44) =\)] 0.0118 or 0.962 or 0.988. Allow M1 for correct distribution and use of \(X = 44\) even if within an incorrect statement (e.g. \(\mathrm{P}(X = 44) = 0.026\))

A1: This whole statement must be correct (i.e. do not accept \(\mathrm{P}(X \lt 44)\))
Accept 0.038 (2sf) or 0.0378 to 3sf or 0.0377 ... rot
Accept \(\mathrm{P}(X \gt 44) = 0.96(22)\) or \(\mathrm{P}(X \lt 45) = 0.0378\) etc.

A1FT: FT their probability - correct comparison for their value e.g. \(0.962 \gt 0.95\) FT hypotheses (e.g. 2-tail 0.025)

M1: Dependent on \(\mathrm{P}(X \leqslant 44)\) or \(\mathrm{P}(X \lt 44)\) soi (i.e. 0.0378 or 0.0118 or 0.962 or 0.988). Condone ‘Accept H₁’

A1: Conclusion must be in context, not definite and consistent with their comparison.
Dependent on all previous M and A marks, except the first A1 (i.e. this mark can be given if 0.0378 value is given incorrectly)

• Allow percentage or probability instead of proportion
• Allow 95% instead of 0.95
• Allow ‘sufficient evidence to suggest that Riley’s claim is correct’
• Condone ‘significant’ for ‘sufficient’ and ‘it is likely that…’
• Do not accept ‘Show’ or ‘Prove’ A0

Notes

B1: Must refer to one of the assumptions for using a binomial distribution i.e.

• The probability/proportion may not be constant
• The events may not be independent
• There may not be a fixed number of trials
• The trials may not have only 2 outcomes

And explain correctly in context by referring to e.g.

• Trains
• Service
• On time

But candidates need not suggest a ‘mechanism’ for how this could come about (i.e. ‘leaves on the line’ or ‘staff shortage’ etc.).
So accept e.g.:

• “the proportion/probability of services arriving on time may not be constant”
• “the event that a service arrives on time may not be independent”
• “the service may be delayed for related reasons on consecutive [or connected] days so the events are not independent”
• “the train service may not run on all 50 days sampled”
• “the train service may be cancelled [meaning it is neither on time nor delayed] and so there is a third outcome”

Do not accept statements about successive stations or successive trains (the question is about the same train service on successive days).
Do not accept just ‘the trains are not independent’ – must refer to ‘train times’ or ‘arrival on time’ oe.
Do not accept ‘equal chance’ without a reference to probability or proportion.

Notes

M1: May be implied by A1 or 0.066 or 0.067 (2sf).
Allow \(1 - \mathrm{P}(W \leqslant 7)\,[= 1 - 0.9812\ldots]\) or 0.0188 for M1

A1: Accept awrt 0.0664 or awrt 0.0665 \([0.066447\ldots]\)

Notes

B1: awrt 0.106

Notes

M1: Attempt to standardise and equate to \(\Phi^{-1}(0.75)\,[= 0.67449]\)
May be implied by A1.

A1: awrt 7.41 or 7.42

Notes

M1: May see N(40, 32)

A1: awrt 49.3

Notes

M1: Attempt \(\mathrm{P}(X \geqslant \text{‘50’})\) using \(\mathrm{B}\left(200, \tfrac{1}{5}\right)\) for their ‘49’+1
Allow this mark for \(\mathrm{P}(X \lt 50) = 0.95\ldots\)
The value must be linked to their correct threshold value of ‘49’+1 (= 50) (or ‘49’), if an inequality is given it must be correct.

M1: Attempt \(\mathrm{P}(X \geqslant \text{‘49’})\) using \(\mathrm{B}\left(200, \tfrac{1}{5}\right)\) for their ‘49’
Allow this mark for \(\mathrm{P}(X \lt 49) = 0.93\ldots\)
The value must be linked to their correct threshold value of ‘49’ (or ‘49’ − 1 = 48), if an inequality is given it must be correct.

A1: Condone \(X \geqslant 50\) (dependent on both method marks)
SCB1 for correct answer with no/insufficient working (max. 1/3)

Further marking guidance (Appendix):
For the M1 marks, require two distinct probabilities from the correct binomial distribution \(\mathrm{B}\left(200, \tfrac{1}{5}\right)\), which correspond to two values centred around \(49 \pm 1\) or their answer from (a) \(\pm 1\).
Candidates need not state an inequality, but each probability must be associated with an appropriate value of \(X\) and if an inequality is given it must be correct.

Additional clarifications:
• The above table shows likely values for \(x = 49 \pm 1\)
• FT their ‘49’ from (a) for M marks only (check values)
• NB for candidates whose ‘49’ is ~65 or greater, their probabilities may appear as ‘1’ and ‘0’ – allow M1M1 if this follows correct setup.
• Condone 2 probabilities for the same \(x\) provided inequalities are given correctly
• Probability values may be given to any degree of accuracy, truncated or rounded.
Do not accept any single probabilities i.e. \(\mathrm{P}(X = x)\)

Notes

(a)(i) B1: awrt 0.219 (Condone 21.9%)

(a)(ii)

M1: Allow M1 for \(1 - \mathrm{P}(X \leqslant 4)\) or \(1 - 0.515\) or 0.484
May be implied by correct answer.

A1: awrt 0.703 (Condone 70.3%)

Notes

(b)(i) B1: Allow \((1 - 0.3)\) for 0.7
Accept \({}^{15}\mathrm{C}_r\) etc. but must be in terms of \(r\)

(b)(ii)

B1: Must be stated (not implied)

B1: Allow \(\mathrm{P}(X = r)\) is a term of the expansion
Condone “appears in the expansion” but not “it is the expansion” and not “the coefficient of a term”
Condone \(\mathrm{P}(X = r)\) is the \(r\)th term of the expansion

Notes

M1: For indication of B(100, 1/6) used and an attempt at any \(\mathrm{P}(X \leqslant a)\) in the right region, i.e. finding \(\mathrm{P}(X \leqslant a)\) for any \(a \leqslant 16\) BC

M1: For finding both \(\mathrm{P}(X \leqslant 10)\) and \(\mathrm{P}(X \leqslant 11)\) BC (allow numerical slip if clear indication that the right distribution is used)
Allow \(\lt\) in both cases if values are correct.

A1: Dep on both M marks so must have found both \(\mathrm{P}(X \leqslant 11)\) and \(\mathrm{P}(X \leqslant 10)\) correctly (condone values correct to 2sf throughout but do not accept answers that are not correct to 2sf, e.g. 0.077 or 0.042 unless there is a clear indication of truncation such as ‘…’)

Notes

B1FT: FT their \(\mathrm{P}(X \leqslant 10)\) (allow 2sf)
Accept 4.27%

Notes

B1: Allow ‘\(n = 20, p = \frac{1}{3}\)’ or just \(\left(20, \frac{1}{3}\right)\). Accept \(p = 0.\dot{3}\) but not 0.333

Notes

(b)(i)

B1FT: FT their values of \(n\) and \(p\) and allow \(\dbinom{20}{x}\) as alternative to \({}^{20}\mathrm{C}_x\)
(If decimals used both must be correct to 3sf)

(b)(ii)

B1FT: FT their ‘20’ only
Any acceptable alternative in set notation
Indication that \(x\) is an integer must be present (but condone ‘integer’ or ‘whole number’ or \(x \in \mathbb{N}\))

Notes

M1: Condone \(\mathrm{P}(X \leqslant 9) - \mathrm{P}(X \leqslant 5)\), or \(0.90810 - 0.29721\) or 0.611
May be implied by correct working or answer.
Allow sum of \(\mathrm{P}(X = x)\) for \(x\) from 5 to 9.

A1: BC cao

Notes

B1: For an explanation (which must be in context) showing clear recognition that a random sample (of this size) would not be representative across the year groups.
Acceptable similar answers in context include:

• ‘should include roughly equal numbers from each year’
• ‘not representative of each year group’
• ‘may not include students from every year group’

Do not accept ‘may end up with more students from Y7-11 than from Y12-13’ (because a representative sample would have this)
Must be in context – so do not accept generic answers without context such as just: ‘It might be biased’ or ‘It’s not representative’ or ‘it should be a stratified sample’

Do not accept answers solely referring to the suitability of the individual students within the sample to represent the school

Notes

M1: soi, eg \(\mathrm{H}_0: p = 0.02\) and \(\mathrm{B}(600, p)\). Allow \(n = 600\), \(p = 0.02\)

M1: May be implied by 0.0991 or 0.0202 or 0.9798 or 0.9009 or correct values

A1: or \((\mathrm{P}(X \leqslant 17) =)\ 0.939\) or 0.94 (2 sf)

A1: or \((\mathrm{P}(X \leqslant 18) =)\ 0.964\) or 0.96 (2 sf)
These two probabilities seen imply M1M1A1A1
Condone errors such as \(\mathrm{P}(X \gt 18) = 0.0610\)

A1: Ignore hypotheses and/or “Reject \(\mathrm{H}_0\)” or similar
Unsupported answers:
0.0359: M1M1A1A0A0
Critical region is \(X \geqslant 19\): M1M1A0A0A0

Alternative method (normal with no cc)

M1: soi. Can be scored either for \(\mathrm{N}(12, 11.76)\) or \(\mathrm{B}(600, 0.02)\)

M1: \(\mathrm{P}(x \gt a) = 0.05 \Rightarrow a = 17.64\) only gets M1 if a probability is calculated

Alternative method (normal with cc)

M1: soi. Can be scored either for \(\mathrm{N}(12, 11.76)\) or \(\mathrm{B}(600, 0.02)\)

Notes

B1: or 18 is under the significance level
Allow You have to do a proper hypothesis test
No other answers acceptable

Notes

B1: allow equivalent in words; do not allow percentages

B1: or \(p\) is the proportion of people (aged 16 or over from England or Wales selected at random) who are married or in a civil partnership; must mention either married or in a civil partnership

B1B1 for other symbol instead of \(p\) used if correctly defined

Notes

M1: for calculation of a value for \(\mathrm{P}(X \leqslant k)\), ignore slip in notation
M0 for attempt at Normal approximation

M1: may see invBinom(0.01,78,0.494) = 28 for M1M1;

A1: allow critical region \(\lt 28\) oe; if M0M0 allow SC1 for correct critical region with no working or insufficient working

Notes

M1: allow eg 26 is in critical region

A1: from correct critical region; condone from \(0 \lt X \leqslant 27\) oe

A1: dependent on award of previous two marks and correct critical region; allow percentage/probability but not “number”
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest;
if M0 allow SC1 for \(\mathbf{P}(\boldsymbol{X} \leqslant \mathbf{26}) = \mathbf{0.003} \lt \mathbf{0.01}\), so reject \(\mathrm{H_0}\) and conclude there is sufficient evidence at the 1% level to suggest that the proportion of people aged 16 or over who are married or in a civil partnership is lower in 2024 than in 2022

Notes

B1: allow \([\mathrm{P}(X \leqslant 27) =]\ 0.005\,946\ldots\)

Notes

B1: allow
eg because he/she/they used more data
eg the sample is bigger

Notes

B1: mark the final answer

Notes

B1: FT \(\dfrac{\text{their } 0.76}{4}\)

Notes

B1: for 1 correct frequency to 2 d.p.

B1: for 3 correct frequencies to 2 d.p.

B1: for all 5 correct to 2 d.p.

FT use of B(4, their 0.19);
allow SC2 for all 5 frequencies awrt 22, 20, 7, 1, 0 or better
allow SC1 for 3 of the above frequencies awrt nearest whole number or better
if B0B0B0 allow SC1 for probabilities: 0.430467…, 0.403895…, 0.142111…, 0.022223…, 0.001303 given to 3 dp or more; FT use of B(4, their 0.19) with \(0 \lt p \lt 1\)

Notes

B1: allow comparison of one pair of frequencies or one pair of probabilities;

Notes

B1: allow equivalent in words; do not allow percentages
allow other variable only if correctly defined

B1: or \(p\) is the proportion of adults that are vegetarian
B1B1 if other symbol instead of \(p\) used if correctly defined

M1*: may be implied by
\((\mathrm{P}(X \geqslant 18)) = 0.0295 - 0.030\) or
\((\mathrm{P}(X \geqslant 19)) = 0.015 - 0.015331\) or
\((\mathrm{P}(X \geqslant 20)) = 0.0075 - 0.00754\)
NB M0 for \(\mathrm{P}(X = 19) = 0.00779\)
NB \(\mathrm{P}(X \leqslant 17) = 0.97049\ldots\), \(\mathrm{P}(X \leqslant 18) = 0.984669\ldots\) and \(\mathrm{P}(X \leqslant 19) = 0.99246\ldots\) imply M1

A1: or \([\mathrm{P}(X \leqslant 18) =]\ 0.984669 - 0.985\) or 0.98

A1FT: A0 if their \(0.015 \gt 0.025\) or their \(0.985 \lt 0.975\)

A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest

Alternatively, using critical region

B1: allow equivalent in words; do not allow percentages
allow other variable only if correctly defined

B1: or \(p\) is the proportion of adults that are vegetarian
B1B1 if other symbol instead of \(p\) used if correctly defined

M1*: allow calculation of upper tail only for M1

A1: from \(\mathrm{P}(X \geqslant 19) = 0.015 - 0.015331\) and \(\mathrm{P}(X \leqslant 4) = 0.010\)
must see both tails for A1

A1FT: A0 if \(19 \lt\) their critical value

A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest

Alternatively, using Normal approximation

B1: allow equivalent in words; do not allow percentages
allow other variable only if correctly defined

B1: or \(p\) is the proportion of adults that are vegetarian
B1B1 if other symbol instead of \(p\) used if correctly defined

M1*: NB M0 for \(\mathrm{P}(X = 19) = 0.006145\) (from normPdf\((19, 11.2, \sqrt{10.08})\)) or \(\mathrm{P}(X = 19) = 0.00627\) (from using continuity correction, may see normCdf\((18.5, 19.5, 11.2, \sqrt{10.08})\))
NB \(\mathrm{P}(X \leqslant 19.5) = 0.99553\ldots\) and \(\mathrm{P}(X \leqslant 18.5) = 0.989255\ldots\) imply M1

A1FT: A0 if their \(0.0107 \gt 0.025\) or their \(0.989 \lt 0.975\)

A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest

Notes

M1: must see some reasoning; allow explanation in words

A1: if unsupported, allow SC1 for correct answer

Notes

B1FT: their \(7p\)

Notes

M1: FT their \(3p\); allow M1 for \((\textit{their }0.3)^2 \times (\textit{their }0.7)^{28}\)

A1: not from wrong working; if unsupported, allow SC1 for correct answer

Notes

B1: do not allow
eg the sample is too small
eg the sample is too small to be representative

Notes

B1: allow equivalent in words;
do not allow percentages

B1: or \(p\) is the proportion of babies that are born without wisdom teeth
B1B1 if other symbol instead of \(p\) used if correctly defined

M1*: or critical region is \(X \geqslant k\) found from calculation of probability; allow \(k = 58\), 59 or 60

A1: or critical region is \(X \geqslant 59\) from 0.0475…. or 0.048

M1dep*: or 61 correctly compared with their 59;
allow their 0.978 correctly compared with 0.95

A1FT: A0 if their \(0.022 \gt 0.05\) or \(61 \lt\) their 59

A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest

Notes

B1: allow eg Nina’s, because with a larger sample size the probabilities get closer to the true probabilities oe

Notes

M1: multiply by \(k\) or by 120; may be embedded

M1: multiplying by both \(k\) and 120

Notes

Alternatively

M1: or \(\frac{kf}{120} = \frac{32}{120}\)
(from \(11f = 120 - 32 = 88\) so \(f = 8\) and so \(kp = \cdots\))

A1: \(k = 4\)

Notes

M1: \(Y\) is the number of 12s obtained in 30 rolls;

A1: allow B2 for 0.1628 – 0.163 unsupported

Notes

M1: use of \(\mathrm{B}(10, p)\) FT their \(p\) to find at least 2 probabilities; may be unsimplified
must see at least 2 probabilities for M1

M1: multiplying their binomial probabilities by 100
must see at least 2 expected values calculated for M1

A1: all correct to 1 dp or better; must add up to 100

Notes

B1: FT dependent on award of M1M1 in (c) and all frequencies calculated

Notes

M1: if M0, allow SC1 for awrt 0.0575… or 0.943
NB 0.975178058

Notes

M1: allow M1 if 2 omitted

A1: eg \(2 \times \frac{0.03}{q} + q = 0.5\) or \(2p + \frac{0.03}{p} = 0.5\)
NB if 2 omitted, A0 for \(2p^2 - 0.5p + 0.06 = 0\) or \(2q^2 - q + 0.24 = 0\) which have no real roots (corrected from the printed mark scheme: printed as \(2p^2 - 0.5p + 0.6 = 0\) or \(2q^2 + q + 0.24 = 0\))

A1: may be implied by eg \(q = 0.2\) or 0.3 and \(2p = 0.3\) or 0.2

Notes

A1: FT their \(q\) where \(0 < q < 1\)

Notes

B1: both hypotheses; allow equivalent in words or eg \(\mathrm{P}(1) = 0.2\)
allow any parameter as long as clearly defined as probability

M1: \(k = 27, 28\) or 29
M0 for \(\mathrm{P}(X = k)\)
NB \(\mathrm{P}(X = 28) = 0.014\ldots\) \(\mathrm{P}(X = 27) = 0.020168\ldots\)

A1: or critical region is \(X \geqslant 28\)

M1: or 28 is in critical region
FT their probability, dependent on award of first M1

A1: must have the correct probability or correct critical region for the last two A marks

A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; A0 if answer spoiled