Notes

(i) B1: for awrt 0.270 (accept just 0.27)

(ii) B1: for awrt 0.607

Notes

M1: for standardising 170 with \(\mu\) and \(\sigma\) and set equal to a \(z\) value \(0 < |z| < 0.3\)

M1: for standardising 175 with \(\mu\) and \(\sigma\) and set equal to a \(z\) value \(1 < |z| < 2\)

M1: for solving their 2 linear equations, reducing to a linear equation in one variable
This can be implied by two answers that are correct to 2sf.

To award the A marks they must have scored 1^st M1 and 2^nd M1

A1: for = \(\sigma\) awrt 3.88

A1: for = \(\mu\) awrt 171 (Must not be from incorrect working e.g. \(\sigma < 0\))

Ans only: We need to see the linear equations in \(\mu\) and \(\sigma\) to start awarding marks.
If 1^st M1 and 2^nd M1 are scored and correct answers are seen we can award 5 marks.

MR: (15% as 0.015) 2^nd M1 for \(z = 2.17\ldots\), 3^rd M1 implied by \(\sigma = 2.06\ldots\) and \(\mu = 170.5\ldots\)
i.e. Maximum M1M1M1A0A0

Notes

M1: for correct probability for women or correct probability for men
or clear intention to multiply \(\mathrm{P}(170 < F < 175)\) by \(\mathrm{P}(170 < M < 175)\)

A1ft: for using \(\mathrm{P}(170 < F < 175) \times \mathrm{P}(170 < M < 175)\) with one probability correct to 2sf

A1: for awrt 0.021. Correct answer only is 3 marks.

MR: (if used in (b)) \(\mathrm{P}(170 < F < 175) = 0.585\) (or awrt 0.58 or 0.59) answer awrt 0.027 or 0.028
All 3 marks can be scored here.

Notes

B1 for awrt 0.0912 (from calculator)

Notes

B1 for a suitable reason mentioning or implying \(H\) and \(T\) are independent
Allow: e.g. “they”/ “each event”/ “\(\mathrm{P}(H)\) and \(\mathrm{P}(T)\)”/ “the variables” and “independent”
B0 for “the results” / “the values” are independent.
Ignore other comments that are not incorrect or contradictory.

Notes

1^st M1 for using model for \(T\) to attempt to find \(\mathrm{P}(T \lt 300)\) e.g. sight of 0.124 or better or sight of \(\pm\left(\dfrac{300 - 330}{26}\right)\) or \(\pm\left(\dfrac{5 - 5.5}{0.433\ldots}\right)\) or \(Z = \pm\,1.15(3\ldots)\)

2^nd M1 for multiplying their two probabilities together ft part (a) and their \(\mathrm{P}(T \lt 300)\) provided both values are probabilities. NB M0M1 is possible here

A1 for awrt 0.0113   [Correct answer with no incorrect working 3/3]

Notes

1^st M1 for standardising 16.3 and setting equal to \(z\) value where \(0.5 \lt |z| \lt 0.6\)

2^nd M1 for standardising 29 and setting equal to \(z\) value where \(1 \lt |z| \lt 1.5\)

3^rd M1 dep on 1^st or 2^nd M1 for solving their two linear eq’ns – reach an eq’n in one variable
May be implied by sight of \(\sigma = 7\) (or better) or \(\mu = 20\) (or better)

For 1^st A mark we must also see one of \(-0.5244\) or 1.2816 (or better) used in their equ’ns OR both \(z\) values correct to 3dp i.e. \(-0.524\) and 1.282

1^st A1 for \(\sigma =\) awrt 7.03 (but see 3^rd case below)

2^nd A1 for \(\mu =\) in [19.95, 20.0] (i.e shouldn’t see something rounding down to 20.0)
Allow 20 from equations with suitable \(z\) values (see examples below)

NB Use of \(-0.524\) and 1.28 [would give 7.0399… and 19.988…] and scores M3A0A1
Use of \(-0.524\) and 1.2816 [would give 7.033… and 19.99…] and scores M3A1A1
Use of \(-0.5244\) and 1.28 [would give 7.038… and 19.99 …] and scores M3A1A1

Both \(z\) values correct to 3dp i.e. \(-0.524\) and 1.282 [should give 7.032 and 19.984] scores A1A1

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

0.3875

Typical solution

0.8427

Typical solution

(c)(i)

0.0033

(c)(ii)

\(\mathrm{P}(F = 0.6) = 0\) since \(F\) is continuous.

Typical solution

The oxides of nitrogen emissions by cars in 2000 are greater on average and less varied than those in 2010.

Typical solution

There are some blanks on the oxides of nitrogen emissions in the LDS.

Typical solution

There are no Nissan cars in the LDS.

Typical solution

Typical solution

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

There is insufficient evidence to suggest that the mean cholesterol level has reduced after taking the dietary supplement.

Typical solution

\(a = 6\) and \(b = 12\)

Typical solution

Typical solution

0.5

Typical solution

0.0668

Typical solution

0.9876

Typical solution

Minimum length is 43.4 cm

Typical solution

\(X\) = length of new-born baby

Reject \(\mathrm{H_0}\)

There is sufficient evidence to suggest that the mean length of a new-born baby at the clinic in 2020 has increased compared to 2019.

Typical solution

(i)

0.0375

(ii)

0.2467

(iii)

0.9601

Typical solution

Typical solution

1

Typical solution

Reject \(\mathrm{H}_0\)

There is sufficient evidence to suggest that the mean daily mass of aluminium cans recycled has changed.

Typical solution

Sample mean could be different.

The result could be different so the claim could be wrong.

Typical solution

Height is continuous data and 95% of heights lies within two standard deviations of the mean so normal may be a suitable model.

Typical solution

(i)

0

(ii)

0.335

(iii)

Typical solution

Mean = 1.73

Standard deviation \(= \sqrt{\dfrac{2.81}{40}}\)

\(=\) 0.265

Typical solution

Summer athletes are taller on average than Winter athletes.

Summer athletes’ heights are less varied than the heights of Winter athletes.

Typical solution

\(X\) = working hours per week

Reject \(H_0\)

There is sufficient evidence to suggest that the mean working hours have increased.

Typical solution

0.59

Notes

M1: May be implied by A1. Allow this mark for awrt 0.32

A1: awrt 0.104

Notes

M1: Must be seen. May see \(\sigma^2 = 0.00036\).
Condone \(\bar{X} \sim \mathrm{N}\left(0.36, \frac{0.12}{\sqrt{40}}\right)\) for this mark only (i.e. M1A0).
Accept \(\mu = 0.36\) and \(\sigma^2 = \frac{0.12^2}{40}\) or \(\sigma = \frac{0.12}{\sqrt{40}}\).

A1: awrt 0.0175 Dep on correct distribution seen.
Answer only without working (correct distribution) scores 0/2.

Notes

Condone use of \(X\) for first four marks (but not final A1). Condone misuse of \(\Phi^{-1}\) but only if correct distribution and correct values seen.

M1: Must see either of these steps (could be shown on a graph if accompanied by correct \(a\) to 1sf), or use of symmetry soi if \(b\) found first (i.e. \(a = 0.36 - |\text{‘}b\text{’} - 0.36|\)).

M1: Or \(\mathrm{P}(\bar{X} \lt b)\,[= 1 - 0.025] = 0.975\)
Must see either of these steps (could be shown on a graph if accompanied by correct \(b\) to 1sf), or use of symmetry soi if \(a\) found first (i.e. \(b = 0.36 + |\text{‘}a\text{’} - 0.36|\)).

A1: Must be given correctly using \(\bar{X}\) for this mark.
Accept \(\bar{X} \gt 0.323\) and \(\bar{X} \lt 0.397\)
Allow \(\leqslant\) for \(\lt\)

SCB2 for 0.323 and 0.397 without working (B1 for each max. 2/5)

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

B1: awrt 0.886 (Condone 88.6%)

Notes

M1*: Attempted, values not required and allow truncated or rounded values e.g. 0.84… etc. throughout.
May be implied by \(\pm 0.84162\) & \(\pm 0.52440\)

A1: Allow \(\Phi(0.8)\) and \(\Phi(0.3)\) – or use of these expressions throughout (need not see values for \(\Phi\) but if given they must be correct)

M1 dep*: One correct equation in one unknown (FT their \(\Phi\) values) (may be implied by correct answers).

A1: Correct answers to 2sf (500 and 2.2) or better

A1: Correct answers to exactly 2dp, cao

SCB1 for one correct answer to 2dp (max 4/5) or if one/both answers are given with no/insufficient working (max 1/5).

Notes

(c)(i) B1: Allow 0 to 1, 3 to 4

(c)(ii) B1FT: \(\dfrac{\text{Their ‘}3.5\text{’} - \text{‘}0.5\text{’}}{2}\) or \(|2 - \text{‘}3.5\text{’}|\) or \(|2 - \text{‘}0.5\text{’}|\)
FT their points of inflection (Note that \(\mu \approx 2\))
Need value and explanation (calculation alone is not sufficient)

Notes

B1: Allow \(\frac{5}{6}\) or 0.84 (2 sf) (from “rule-of-thumb”)
awrt 0.84

Notes

M1: One of these attempted. Or \(\mathrm{P}\left(Z \lt \frac{45 - \mu}{\sigma}\right) = 0.8\) seen or this standardised form clearly shown on a diagram

A1: Not a required answer so allow truncated e.g. 0.841, 0.84…

A1: Not a required answer so allow truncated e.g. −0.52…

M1dep: dep previous M1. Attempt to solve their equations simultaneously, any method (may be implied by correct answers)

A1: For obtaining both values in the given intervals.

A1: For both \(\mu\) and \(\sigma\) correct to 3sf BC cao
(\(\mu = 32.6778\), \(\sigma = 14.6411\))
SC B1 for either \(\mu = 32.7\) (3sf) or \(\sigma = 14.6\) (3sf) if neither A mark gained, but must be correct to 3sf. (max [5/6])

Notes

M1: oe but signs must be correct (e.g. accept \(\frac{b - 10}{2} = \frac{12 - c}{3}\))

A1: oe but must have \(b\) as the subject

Notes

B1: For a correct assumption, in context, that is necessary and specific to ‘the population’ or ‘the river’.
Acceptable answers include:

• sd for this river is the same as for the UK
• assume the population of this river is normally distributed
• mass of fish in this river is normally distributed

Do not accept answers that are:

• referencing the sample (e.g. bias) – question states random
• not about this river e.g. ‘in the UK’
• generic e.g. ‘masses of fish are normally distributed’

Candidates must show recognition that the assumption is needed about the (sub-)population of this river. The mean and sd for the whole of the UK are given in the question.
Condone the sample mean of fish from this river \(\bar{X} \sim N\) but references to \(X\) or \(\bar{X}\) alone are not enough unless defined in context.
Ignore all else (e.g. ignore any statements about the mean)

B1B1: Subtract B1 for each error eg:

• 2-tail: B1B0
• undefined \(\mu\): B1B0
• not in terms of parameter: B1B0
• \(\mu\) = sample mean implied: B1B0
• Not include value 4.2: B0B0
• eg \(\mathrm{H}_0 = 4.2\) etc: B0B0

Allow any letter for \(\mu\) (except \(X, \bar{X}\): B0B0)
Condone ‘average’ for ‘mean’ but do not accept definitions for \(\mu\) that are clearly not about this river e.g. ‘the UK’ B1B0

M1*: This mark may be implied by the correct value of 0.0548 or 0.945 (correct to 2sf) (even if within incorrect statement eg \(\mathrm{P}(X = 4.16) = 0.0548\)) Condone \(\gt, =, \geqslant, \leqslant\)

A1: BC (awrt 0.055 2sf)

A1FT: FT correct comparison for their value as long as consistent with their test (e.g. 2-tail 0.025 or 0.975) Must be seen, allow on diag
A0 if the comparison is not for their value (e.g. if miscopied)

Alternative (critical value)

A1*: cao

A1dep*: dep A1* Must be seen

Alternative (test statistic)

A1*: cao

A1dep*: dep A1* Must be seen

M1dep*: dep M1* Correct conclusion about \(\mathrm{H}_0\) for their comparison, provided they have compared with an appropriate value.
Accept Reject \(\mathrm{H}_1\) or Insufficient (or No) evidence to reject \(\mathrm{H}_0\)

A1: www (all preceding calculations must be correct, i.e. dependent on all previous M and A marks)
Must be in context and not definite.
Acceptable answers include:

• ‘no evidence that (mean) mass is less’
• ‘insufficient evidence to say that the environmentalist is correct’
• ‘insufficient evidence that the fish in this river are smaller’

Do not accept:

• ‘therefore the mean is 4.2’ (definite)
• ‘the mean has decreased’ (incorrect reference to change over time)
• ‘the mean is not less’ or ‘the mean is the same’ (insufficient evidence does not mean that the statement for \(\mathrm{H}_0\) is true)

Notes

NB. Use of a “continuity correction” loses 1st A1 only

B1 B1: Allow other letter (including \(X\)) only if clearly defined
Subtract B1 for each error eg:

\(\mu =\) sample mean implied B0 & (B1 or B0)

M1*: Correct distribution and \(\overline{X}\) (allow 3359.5, 3360.5, 3659) stated or implied eg by 0.0297 or 0.0307 or 0.0286
even if within incorrect statement eg \(\mathrm{P}(X = 3360) = 0.0297\)
Allow \(450^2 \div \sqrt{200}\) or \(450^2 \div 200^2\)
Not 0.0297 from \(\mu = 3360\), \(\mathrm{P}(\overline{X} \lt 3300)\)

A1: BC cao

A1: Explicit comparison
Allow compare (any value \(\leqslant\) 0.35) with 0.025

M1dep: Allow Reject \(\mathrm{H}_1\) or Accept \(\mathrm{H}_0\)
Dep M1A1A1 or M1A0A1 or M1A1A0
Dep also on comparing like with like, eg not \(0.970 \gt 0.025\)
May be implied by their conclusion, if M1 criterion is met
ft with opposite conclusion if, eg \(\mathrm{P}(\overline{X} \gt 3360) = 0.024\)

A1f: Dep M1A1A1 or M1A0A1 or M1A1A0
In context, not definite; eg not “Mean mass not > 3300 g”
Not “There is evidence that mean mass hasn’t increased”
Not “This suggests that mean mass hasn’t increased”
Not “Unlikely (or unsure) that mean has increased”
ft with opposite conclusion if, eg \(\mathrm{P}(\overline{X} \gt 3360) = 0.024\)

Alternative method 1 for M1A1A1

M1*: Correct distribution and \(\overline{X}\) (allow \(\overline{X} = 3359.5\) or 3360.5) stated or implied eg by 0.970 or 0.969 or 0.971
even if within incorrect statement eg \(\mathrm{P}(X = 3360) = 0.970\)
Allow \(450^2 \div \sqrt{200}\) or \(450^2 \div 200^2\)

A1: BC cao

A1: Explicit comparison
Allow compare (any value \(\geqslant\) 0.65) with 0.975

Alternative method 2 for M1A1A1

M1*: May be implied, eg by 3362
Allow \(450^2 \div \sqrt{200}\) or \(450^2 \div 200^2\)

A1: Explicit comparison of their \(a\) with 3360

Alternative method 3 for M1A1A1

M1*: Allow 3359.5 or 3360.5 May be implied, eg by 1.89
Allow \(450^2 \div \sqrt{200}\) or \(450^2 \div 200^2\)

A1: cao

A1: Explicit comparison

Alternative scheme for incorrect method using 2-tails

Exemplars for Q11

Notes

M1: attempted, using any units, eg small squares or cm²

A1: Correct calculation seen and answer 0.147 seen
Not any method starting with 0.147, eg \(0.147 \times 150 = 22.05\)

Notes

(b)(i)

B1: or Roughly symmetrical and peaks in middle or has one peak and tails off at each end, or drops off either side
All 3 of these must be seen (except “Bell-shaped” scores B1)
Not “Shape is like normal curve”   Ignore all else

(b)(ii)

B1: or calculate mean and obtain \(\frac{5915}{150}\) or 39.4
Allow 40 has the highest frequency or frequency density
Ignore all else

B1: or calculate sd and obtain 10.3
Most data is within \(6\sigma\)
Must see correct fraction and \(\approx \frac{2}{3}\), or 68% or 0.68

Notes

B1: BC

Notes

B1: Allow \(39.1 \leqslant m \leqslant 39.7\) Ignore method   BC

B1: Allow \(105.5 \leqslant s^2 \leqslant 108.5\) or \(10.27 \leqslant s \leqslant 10.42\) Ignore method
(Use of denominator \(n\) or \((n - 1)\) is OK for full marks)
OR if neither mark scored,
M1 for attempting find frequencies or areas (NOT heights) or at least five of these seen: 4, 22, 23, 28, 29, 22, 20, 2
or 20, 110, 115, 140, 145, 110, 100, 10

B2: cao Correct with unclear or no working; B1B1B1B1
or B1 for 0.145 to 0.158

NB No retrospective marks if 0.151 seen in table for (e)(i)

Notes

(e)(i)

No FT

B1 for middle row correct \(\pm 0.001\)   NB

B1 for bottom row correct \(\pm 0.003\)   NB

(e)(ii)

Allow “more accurate” or “less accurate” or similar

BUT SC: “Both less good fit for >60” alone: B1 only
NOT “Both are fairly good fit” B0B0
Ignore all else   NB No ft

Notes

or, eg Would take too long to contact all students.
NOT “Easier”

Notes

or: Different years may have different numbers of students

NOT “It’s more representative” or “Takes all students into account”
“You get a range of people” “It avoids bias”

Notes

B1: Allow “30 is more than 2 sds away from the mean”

B1: Allow “30 is less than 1.5×IQR from UQ”

B1: or eg "It depends which definition you use."
Any comment implying uncertainty
Ignore comments about mean ± 3 sds or mean ± 1 sd

Notes

NB Allow 2 sf throughout

B1 B1: Allow other letters, not \(X\) unless defined. Not \(\overline{X}\)
B1B0 for 1 error eg 2-tail or:

• 2-tail: B1B0
• undefined \(\mu\): B1B0
• not in terms of parameter: B1B0
• \(\mu\) = sample mean implied: B1B0
• Not include value 20: B0B0
• eg \(\mathrm{H}_0 = 20\) etc: B0B0

M1: Correct distribution and value of \(\overline{X}\).
stated or implied eg by 0.0326 or 0.967 or 20.9 or 1.84 or 0.000335 even if within incorrect statement eg \(\mathrm{P}(X = 21) = 0.0326\)
Condone \(\dfrac{4.2^2}{\sqrt{60}}\) or \(\dfrac{4.2^2}{60^2}\) or \(\dfrac{4.2}{60}\)

A1: BC Allow 2 sf, ie 0.033

A1: Dep 0.0326 or 1.84 or 0.9674 or \(\mathrm{P}(X \gt 21)\) or \(\mathrm{P}(X \geqslant 21)\) soi
Must compare like with like,
eg NOT prob cf \(z\)-value or large prob cf small prob or CV cf wrong end of acceptance region

M1: Dependent on clearly valid comparison of like with like.
Dep 0.0326 or 0.9674 or 20.9 or 1.84 or \(\mathrm{P}(X \gt 21)\) or \(\mathrm{P}(X \geqslant 21)\) soi (corrected from the printed mark scheme: it printed “0.0485 or 0.951”, values that do not arise in this question)
May be implied by conclusion,
eg “There is evidence that mean time is \(\gt 20\) hours” M1A1

A1f: In context, not definite;
eg “Mean time is \(\gt 20\) hours": A0
But “There is evidence to reject \(\mathrm{H}_0\) and that mean time is \(\gt 20\) h” M1A1
Allow opposite conclusion, ft their values, if above conditions met and loses 2^nd A1. But potentially can score all the other 6 marks

Alternative methods for M1A1A1

For each M1: Condone \(\dfrac{4.2^2}{\sqrt{60}}\) or \(\dfrac{4.2^2}{60^2}\) or \(\dfrac{4.2}{60}\)
\(\mathrm{P}(\overline{X} \lt 21) = 0.9674\): BC

Notes

DR: This question included the instruction: In this question you must show detailed reasoning.

B1: BC allow 347.3 or 347; allow eg mean or \(\bar{X}\), but no other symbol if seen

B1: NB 88.6371428571; may see 9.4147(3010007); accept awrt 88.6 or awrt 9.41; do not allow 86.86 or 9.320; allow eg \(\sigma^2\) or VAR[\(X\)]

B1: allow use of other parameters (but not \(\bar{x}\) or \(\bar{X}\)) if defined as [population] mean mass;
allow if hypotheses stated in words: may earn B1B0 if “population” omitted

B1: condone weight instead of mass

B1: FT their \(s^2\) correct to 3 sf or better;
may see N(350, 1.77(274285741)) or \(\mathrm{N}\left(350, \dfrac{9.415^2}{50}\right)\) or \(\mathrm{N}(350, 1.331^2)\)

M1*: allow M1 for correct probability found from \(\mathrm{N}\left(350, \dfrac{\text{their awrt } 88.6}{50}\right)\);
must see their distribution if value incorrect;
condone slip in notation eg use of \(\mathrm{P}(X \lt 347.34)\); allow for use of 347 or 347.3 instead of 347.34

M1dep*: allow eg 2.28% \(\gt\) 1%

A1FT: A0 if their 0.0229 \(\lt\) 0.01

A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; this mark is dependent on the award of all preceding marks apart from the fourth and fifth B marks ; condone weight instead of mass; must come from correct probability: \((\mathrm{P}(\bar{X} \lt 347.34) =)\) awrt 0.023

Alternatively, for the last 4 marks

M1*: allow M1 for critical region found from \(\mathrm{N}\left(350, \dfrac{\text{their awrt } 88.6}{50}\right)\); must see their distribution if value incorrect; condone slip in notation eg use of \(X\)

M1dep*: allow 347 or 347.3 compared with 346.9(0)

A1FT: A0 if their 347.34 \(\lt\) their 346.9(0)

A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; this mark is dependent on the award of all preceding marks apart from the fourth and fifth B marks; must have correct critical region

Alternatively, for the last 4 marks

M1*: allow M1 for \(z\) found from \(\mathrm{N}\left(350, \dfrac{\text{their awrt } 88.6}{50}\right)\); must see their distribution if value incorrect; allow for use of 347 or 347.3 instead of 347.34
NB may see awrt \(-2.03\) or awrt \(-2.25\) from 347.3 or 347

M1dep*: their \(z\) correctly compared with \(-2.326347877\ldots\) to 2 or more dp

A1FT: A0 if their \(-1.998 \lt -2.33\)

A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; this mark is dependent on the award of all preceding marks apart from the fourth and fifth B marks; \(z\) must be correct

Notes

(ii) B1: NB \(\sqrt{14.289} = 3.7800\ldots\)
may see \(\sqrt{\frac{6247066.6}{55} - 337^2} \times \sqrt{\frac{55}{54}}\) or \(\sqrt{\frac{6247066.6}{54} - \frac{55 \times 337^2}{54}}\) oe;
AG
must see substitution of at least three of 6247066.6, 337, 55 and 54

Notes

M1: may be implied by 0.032; allow a more precise value for 3.78 if found in part (a)(ii)
NB eg N(−9E999,330,337,3.78);
NB may see \(\sigma^2 = \frac{643}{45}\) or \(z = \frac{330-337}{3.78}\ (= -1.85\ldots)\)

A1: mark the final answer

Notes

B1: to 2 or more dp

A1: must be correct to 3 sf
A0 for \(\mu \geqslant 339\) or \(\mu \gt 339\)

Alternatively using calculator

M1: allow slip in calculation if intent is clear; allow for any value between 338 and 338.78 inclusive
must see correct distributions if probabilities are wrong

M1: allow for any value between 338.8 and 339.5
NB critical value is 338.79

A1: must be correct to 3 sf; A0 for \(\mu \geqslant 339\) or \(\mu \gt 339\)

Notes

B1: do not allow \(\bar{X}\) or \(X\), but allow other symbol if defined as [population] mean volume;
allow equivalent in words

M1*: may be implied by \(0.0477 - 0.048\)
may see N\((340, 0.378^2)\)
may see \(\sigma^2 = \frac{643}{4500}\)

A1: allow slip such as \(X\) for \(\bar{X}\) but do not allow \(\mu\)

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

B1: do not allow \(\bar{X}\) or \(X\), but allow other symbol if defined as [population] mean volume;
allow equivalent in words

M1*: may be implied by \(339.378 - 339.38\)
may see N\((340, 0.378^2)\)
may see \(\sigma^2 = \frac{643}{4500}\)

A1: or [critical value is \(\bar{X} =]\ 339.378 - 339.38\);
allow slip such as \(X\) for \(\bar{X}\) but do not allow \(\mu\)

M1dep*: allow eg so 339.37 is in the critical region if critical region explicitly identified

A1FT: A0 if \(339.37 \gt\) their 339.378

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 standard Normal distribution

B1: do not allow \(\bar{X}\) or \(X\), but allow other symbol if defined as [population] mean volume;
allow equivalent in words

M1*: may be implied by \(z = -1.667\)
may see N\((340, 0.378^2)\)
may see \(\sigma^2 = \frac{643}{4500}\)

A1FT: A0 if their \(z \gt -1.64485\)

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: mark the final answer

Notes

B1: ignore extra comments unless they contradict an otherwise correct answer

Notes

M1: NB 62.15567…may be implied by sd = 7.88 or 7.89
NB 0.263047…from \(\sqrt{62.2}\) or 0.263133… from 7.89
NB 0.262871… from 7.88, 0.262973…from unrounded sd

A1: allow B2 for correct answer unsupported

Notes

A1: may be given to 2 sf; allow B2 for correct answer unsupported

Notes

M1: or \(0.18 \times \sqrt{62.2}\)

A1: allow eg \(1.42^2\) for variance

Notes

B1: allow any other symbol except \(\bar{x}\) or \(\bar{X}\), as long as it is correctly defined;
allow hypotheses stated in words

B1: allow weight;
correct definition of \(\mu\) may be embedded in hypotheses written out as a sentence;
do not allow \(\bar{x}\) or \(\bar{X}\)

Notes

B1: Normal distribution with correct mean or variance
allow variance = awrt \(4.95 \times 10^{-6}\) or awrt \(0.00222^2\)

B1: all correct, but allow full credit if no symbol used;
allow symbol other than \(\bar{X}\) if correctly defined as sample mean, but do not allow \(\mu\)

Notes

B1: FT other correctly defined symbol

Notes

M1: condone \(p =\) awrt \(0.00007 < 0.05\) oe NB 0.0000740
or \(z =\) awrt \(-3.79 < -1.645\) oe from use of \(\bar{X} \sim \mathrm{N}\left(0.14,\ \frac{0.0198^2}{80}\right)\)

A1: allow accept \(\mathrm{H_1}\) or result is significant

A1: allow weight; do not allow eg conclude / prove / indicate or other assertive statement instead of suggest

Notes

B1: both hypotheses in terms of \(\mu\)

B1: allow \(\mu\) is the population mean weight of a bag of flour

Notes

M1: \(a\) and \(b\) are numerical values

A1: allow \(0.0424^2\) for variance
A0 if answer spoiled by wrong variable quoted eg \(\mu \sim \mathrm{N}\left(1.5, \frac{0.24^2}{32}\right)\) or \(X \sim \mathrm{N}\left(1.5, \frac{0.24^2}{32}\right)\);
allow only \(\bar{X}\) oe if variable included

Notes

M1: or \(1.44 \gt 1.43(02586\ldots)\)
NB 1.43(02586…) is from InvNorm(0.05, 1.5, 0.0424)

A1: allow accept \(\mathrm{H_0}\) or not significant or reject \(\mathrm{H_1}\)

A1: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest
if calculated values are used full marks may be awarded for awrt 0.07865 or 0.0786, or \(-1.415 \leqslant z \leqslant -1.414\);
otherwise award a maximum of M1A1 for 0.07…or −1.4…
other calculated values score M0

Notes

B1: B0 for 8.757 explicitly rounded to 8.8

Notes

M1: allow M1 for \(8.8^2\) or \(\sqrt{8.8}\)

A1: \(a = 112.4\) or 112 and \(b = 8.8\) or \(2.97^2\)

Notes

M1: may see \(\mathrm{N}(-\infty, 104.5, 112.4, \sqrt{8.8})\)
NB 0.00387 or 0.0063(06) implies M1
NB 0.00573 or 0.00914 implies M1
NB 0.00379(69..) or 0.00619(81…) may imply M1 FT use of variance = 8.757
NB 0.200(199…) and 0.184(665…) may imply M1 FT use of sd = 8.8
if probability is correctly found to be 0 eg from use of \(\mathrm{N}\left(112.4, \frac{8.8}{205}\right)\) allow M1 only – no further marks available

M1: or compare \(\frac{1}{205}\ (\approx 0.00488)\) with their non-zero 0.00387

A1: or probabilities similar so consistent oe

Alternatively

M1: FT their distribution

Notes

M1: NB awrt 0.0915 or awrt 0.0807 implies M1

A1: unsupported answers score M0

Notes

M1: area identified which is symmetrical about the mean

A1: \(x = \mu \pm \sigma\) at points of inflection

Notes

B1: both hypotheses; allow other parameters (but not \(X\) or \(\bar{X}\)) if defined as mean

Notes

M1: may see \(1.645 = \dfrac{\bar{X} - 161.6}{\sqrt{\frac{1.96}{200}}}\)
\(\bar{X} = 161.76\ldots\) implies M1 only

A1: allow if inequality not strict
NB 161.76283…
allow eg sample mean > 161.8

Notes

M1: FT their 161.8 (must be greater than 161.6) or allow if their \(P(\bar{X} > 161.9) [= 0.00122] < 0.05\) or their \(z > 1.645\) is considered
critical region must be an upper tail; their probability must be less than 0.5

A1: FT their calculated critical region or their calculated probability,

B1: ignore comments about rejecting the null hypothesis oe

Notes

M1: \(b\) may be incorrect numerical value

Notes

B1: \(N \times P = 120\) where \(N\) and \(P\) are integers greater than 1.
eg 20 and 6 or 15 and 8

Notes

B1: allow any parameter apart from \(\bar{x}\) for population mean as long as clearly defined as (population) mean

after B marks M1A1M1A1A1 may be earned if working with 2.7054 rounded to 2 or more sf

M1: condone use of 2.6941 (may be rounded) instead of 2.7054 for M1, may be implied by 0.32076 or CR is \(\bar{X} < 4.51797\ldots\), but A1 not available
or \(z = \dfrac{4.8855 - 5}{\frac{2.7054}{\sqrt{120}}}\) for M1

A1: or CR is \((\bar{X}) < 4.5 - 4.6\)
awrt \(-0.46\) A1 (may be implied by \(-0.466\) or \(-0.465567\ldots\) if 2.6941 used)

M1: comparison of their probability with 0.025 or comparison of 4.8855 with their critical value from use of 0.025, as long as previous M1 awarded
their \(-0.4636 > -1.96\) oe (corrected from the printed mark scheme: printed as \(-0.4436\))
M1 dep on award of previous M1

A1FT: may be embedded in conclusion in context

A1FT: do not allow eg conclude / prove / indicate or other assertive statement instead of suggest; A0 if answer spoiled

Notes

B1: from \(5 \times 1.8\)

Notes

M1: alternatively select every 25^th number on the list
alternatively select every 24.8^th value on list, rounding as appropriate

A1: alternatively start randomly between \(n = 1\) and \(n \geqslant 25\), and cycle through the list again, stopping when 200 have been selected
alternatively start randomly with any value on list. Cycle through the list repeatedly until 200 items have been selected

E1: alternatively allow any process where each member of the population has an equal chance of being selected

E1: allow any process where each possible sample has an equal chance of being selected

Notes

B1: must refer to shape and closer to Normal shape for larger sample

Notes

M1: condone use of \(6.5717^2\) or parameters rounded to 3 sf
M0 if continuity correction used or eg \(\mathrm{P}(X > 64)\) found

Notes

B1: allow any sensible reason in context
do not allow eg mean and sd may be different