Questions - A-Level Maths

OCR Further Pure Core 2 2021 June Q2

2 In this question you must show detailed reasoning.
Show that $\int _ { 5 } ^ { \infty } ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = 1$.
$3 A$ is a fixed point on a smooth horizontal surface. A particle $P$ is initially held at $A$ and released from rest. It subsequently performs simple harmonic motion in a straight line on the surface. After its release it is next at rest after 0.2 seconds at point $B$ whose displacement is 0.2 m from $A$. The point $M$ is halfway between $A$ and $B$. The displacement of $P$ from $M$ at time $t$ seconds after release is denoted by $x \mathrm {~m}$.

Sketch a graph of $x$ against $t$ for $0 \leqslant t \leqslant 0.4$.
Find the displacement of $P$ from $M$ at 0.75 seconds after release.

OCR Further Pure Core 2 2021 June Q4

4 In an Argand diagram the points representing the numbers $2 + 3 \mathrm { i }$ and $1 - \mathrm { i }$ are two adjacent vertices of a square, $S$.

Find the area of $S$.
Find all the possible pairs of numbers represented by the other two vertices of $S$.

OCR Further Pure Core 2 2021 June Q5

23 marks

5 In this question you must show detailed reasoning.
The diagram below shows the curve $r = \sqrt { \sin \theta } \mathrm { e } ^ { \frac { 1 } { 3 } \cos \theta }$ for $0 \leqslant \theta \leqslant \pi$.
\includegraphics[max width=\textwidth, alt={}, center]{58789f16-bfc3-4f21-b7c4-abca4f549fc7-03_574_878_276_477}

Find the exact area enclosed by the curve.

Show that the greatest value of $r$ on the curve is $\sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 6 } }$. Total Marks for Question Set 4: 37 \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available.
M
A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B
Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
g For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero.

Abbreviations

Question

Answer

Marks

AO

Guidance

\multirow[t]{7}{*}{2}

\multirow{7}{*}{}

DR

\multirow{6}{*}{

B1

М1

A1

B1

A1

}

\multirow{4}{*}{

1.1a

2.1

1.1

}

\multirow[b]{3}{*}{Consideration of a finite upper limit}

\multirow{7}{*}{Can be seen as part of limit of both terms, but must be explicitly shown as zero}

$\int ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = - 2 ( x - 1 ) ^ { - \frac { 1 } { 2 } } ( + c )$

$\int _ { 5 } ^ { N } ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = \left[ - 2 ( x - 1 ) ^ { - \frac { 1 } { 2 } } \right] _ { 5 } ^ { N }$ $- \frac { 2 } { \sqrt { N - 1 } } + \frac { 2 } { \sqrt { 5 - 1 } }$

$\lim _ { N \rightarrow \infty } \frac { 1 } { \sqrt { N - 1 } } = 0$ oe

2.1

Not just eg $\frac { 1 } { \infty } = 0$

$\int _ { 5 } ^ { \infty } ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = \lim _ { N \rightarrow \infty } \left\{ - \frac { 2 } { \sqrt { N - 1 } } + \frac { 2 } { \sqrt { 5 - 1 } } \right\} = 1$

2.2a

AG. Convincing argument equating improper integral to solution

[5]

Question

Answer

Marks

AO

Guidance

3

(a)

\includegraphics[max width=\textwidth, alt={}]{58789f16-bfc3-4f21-b7c4-abca4f549fc7-10_450_671_93_559}

B1

1.2

3.4

3.1b

At least one cycle of $A \cos \omega t$ graph

Amplitude 0.1

Period 0.4

Intersect with the $t$-axis at 0.1 and 0.3 (values must be indicated or implied unambiguously (eg by tickmarks and a single value))

graph must instantaneously horizontal at top/bottom, continuous and not vertical at any point.

Ignore any graph outside [0, 0.4] Non-inverted cos graph can still get 4/4

(b)

So $x = \pm 0.1 \cos ( 5 \pi \mathrm { t } )$ or $x = 0.1 \sin \left( 5 \pi t \pm \frac { 1 } { 2 } \pi \right)$ when $t = 0.75$ or 0.35 $x = - \frac { \sqrt { 2 } } { 20 } ( = - 0.0707 \text { to } 3 \mathrm { sf } )$

M1

A1

[2]

3.1b

3.4

or by argument from sketch

Condone amplitude of 0.2 for M1

4

(a)

\(\begin{aligned}

( 2 + 3 i ) - ( 1 - i ) ( = \pm ( 1 + 4 i ) ) \text { soi }

| 1 + 4 i | = \sqrt { 1 ^ { 2 } + 4 ^ { 2 } }

17 \end{aligned}\)

B1

M1

A1

[3]

1.1

2.2a

1.1

Either way round

Can be implied by vector

Or finding the square of their side

(b)

$( \pm \mathrm { i } ) \times ( \pm ( 1 + 4 \mathrm { i } ) )$

$( 2 + 3 i ) \pm i ( 1 + 4 i )$ and $( 1 - i ) \pm i ( 1 + 4 i )$

So vertices at - 3 and $- 2 + 4 \mathrm { i }$

Or at $5 - 2 \mathrm { i }$ and $6 + 2 \mathrm { i }$

*M1

dep*M1

A1

[4]

3.1a

2.2a

3.2a

Method to find a complex number representing perpendicular side. Can be implied by $\pm ( 4 - \mathrm { i } )$ Method to find both pairs of numbers

Both clearly paired and in complex number form for final A1

Or vector form if take geometric approach

If M1M0A0A0 then add SC1 for any two correct vertices

Question

Answer

Marks

AO

Guidance

5

(a)

DR \(\begin{aligned}

\frac { 1 } { 2 } \int \left( \sqrt { \sin \theta } e ^ { \frac { 1 } { 3 } \cos \theta } \right) ^ { 2 } \mathrm {~d} \theta

\mathrm {~A} = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } \sin \theta \mathrm { e } ^ { \frac { 2 } { 3 } \cos \theta } \mathrm {~d} \theta

= \frac { 1 } { 2 } \times - \frac { 3 } { 2 } \left[ \mathrm { e } ^ { \frac { 2 } { 3 } \cos \theta } \right] _ { 0 } ^ { \pi }

\frac { 3 } { 4 } \left( \mathrm { e } ^ { \frac { 2 } { 3 } } - \mathrm { e } ^ { - \frac { 2 } { 3 } } \right) \end{aligned}\)

M1

*A1

dep*M1

A1

[4]

3.1a

2.1

1.1a

1.1

Correct form, in terms of $\theta$,

Integrand has been squared out. Must include limits (can be seen later)

Might be as result of substitution Allow coefficient error for M1 isw

M1 can be implied by 1.0757 ... BC

eg $\frac { 3 } { 4 } \left[ \mathrm { e } ^ { u } \right] _ { - \frac { 2 } { 3 } } ^ { \frac { 2 } { 3 } }$ or $\frac { 3 } { 4 } \left[ \mathrm { e } ^ { \frac { 2 } { 3 } u } \right] _ { - 1 } ^ { 1 }$ oe

(b)

DR $\frac { \mathrm { d } r } { \mathrm {~d} \theta } = \frac { 1 } { 2 } \cos \theta ( \sin \theta ) ^ { - \frac { 1 } { 2 } } e ^ { \frac { 1 } { 3 } \cos \theta } +$ $( \sin \theta ) ^ { \frac { 1 } { 2 } } \left( - \frac { 1 } { 3 } \sin \theta \right) e ^ { \frac { 1 } { 3 } \cos \theta }$

$\frac { \mathrm { d } r } { \mathrm {~d} \theta } = \frac { 1 } { 6 } ( \sin \theta ) ^ { - \frac { 1 } { 2 } } e ^ { \frac { 1 } { 3 } \cos \theta } \left( 3 \cos \theta - 2 \sin ^ { 2 } \theta \right)$

$\frac { \mathrm { d } r } { \mathrm {~d} \theta } = 0 \Rightarrow 3 \cos \theta - 2 \sin ^ { 2 } \theta = 0$

$2 \cos ^ { 2 } \theta + 3 \cos \theta - 2 = 0$

$\cos \theta = \frac { 1 } { 2 } , - 2$

$\cos \theta \neq - 2$

$\Rightarrow \sin \theta = \frac { \sqrt { 3 } } { 2 } \Rightarrow r = \sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 3 } \times \frac { 1 } { 2 } } = \sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 6 } }$

*M1

A1

dep*M1

M1

*A1

dep*A1

3.1a

1.1

2.2a

2.1

1.1

2.3

2.2a

Attempt to differentiate using product and chain rules.

Setting $r ^ { \prime }$ to zero and factorising/cancelling to produce a quadratic equation in $\cos$ and/or sin Use of $\cos ^ { 2 } + \sin ^ { 2 } = 1$ to find 3 term quadratic equation in $\cos \theta$.

Solving quadratic correctly Explicitly rejecting root

AG. At least one intermediate step must be seen.

Must be in the form $u v ^ { \prime } + u ^ { \prime } v$ with at most one of $u , v , u ^ { \prime }$ or $v ^ { \prime }$ incorrect or omitted

Or could be in $\sin ^ { 2 } \theta$; $4 \sin ^ { 4 } \theta + 9 \sin ^ { 2 } \theta - 9 = 0$

$\sin ^ { 2 } \theta = \frac { 3 } { 4 } , - 3$

Rejects $\sin ^ { 2 } \theta = - 3$ and $\sin \theta =$ $- \frac { \sqrt { 3 } } { 2 }$

Can be awarded even if rejection of root(s) was implicit.

OCR Further Pure Core 2 2021 June Q2

2 The equations of two intersecting lines $l _ { 1 }$ and $l _ { 2 }$ are
$l _ { 1 } : \mathbf { r } = \left( \begin{array} { l } 1
0
a \end{array} \right) + \lambda \left( \begin{array} { r } 2
1
- 3 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } 7
9
- 2 \end{array} \right) + \mu \left( \begin{array} { r } - 1
1
2 \end{array} \right)$
where $a$ is a constant.
The equation of the plane $\Pi$ is
r. $\left( \begin{array} { l } 1
5
3 \end{array} \right) = - 14$.
$l _ { 1 }$ and $\Pi$ intersect at $Q$.
$\zeta _ { 2 }$ and $\Pi$ intersect at $R$.

Verify that the coordinates of $R$ are $( 13,3 , - 14 )$.
Determine the exact value of the length of $Q R$.

OCR Further Pure Core 2 2021 June Q3

3 A capacitor is an electrical component which stores charge. The value of the charge stored by the capacitor, in suitable units, is denoted by $Q$. The capacitor is placed in an electrical circuit. At any time $t$ seconds, where $t \geqslant 0 , Q$ can be modelled by the differential equation $\frac { \mathrm { d } ^ { 2 } Q } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} Q } { \mathrm {~d} t } - 15 Q = 0$. Initially the charge is 100 units and it is given that $Q$ tends to a finite limit as $t$ tends to infinity.

Determine the charge on the capacitor when $t = 0.5$.
Determine the finite limit of $Q$ as $t$ tends to infinity. The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { r r } 0.6 & 2.4
- 0.8 & 1.8 \end{array} \right)$.
Find $\operatorname { det } \mathbf { A }$. The matrix $\mathbf { A }$ represents a stretch parallel to one of the coordinate axes followed by a rotation about the origin.
By considering the determinants of these transformations, determine the scale factor of the stretch.
Explain whether the stretch is parallel to the $x$-axis or the $y$-axis, justifying your answer.
Find the angle of rotation.

OCR Further Pure Core 2 2021 June Q5

19 marks

5 Two thin poles, $O A$ and $B C$, are fixed vertically on horizontal ground. A chain is fixed at $A$ and $C$ such that it touches the ground at point $D$ as shown in the diagram. On a coordinate system the coordinates of $A , B$ and $D$ are $( 0,3 ) , ( 5,0 )$ and $( 2,0 )$.
\includegraphics[max width=\textwidth, alt={}, center]{420598e3-4531-44da-8ae3-088e433f4c05-03_696_1338_1011_262} It is required to find the height of pole $B C$ by modelling the shape of the curve that the chain forms.
Jofra models the curve using the equation $y = k \cosh ( a x - b ) - 1$ where $k , a$ and $b$ are positive constants.

Determine the value of $k$.
Find the exact value of $a$ and the exact value of $b$, giving your answers in logarithmic form. Holly models the curve using the equation $y = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 3$.
Write down the coordinates of the point, $( u , v )$ where $u$ and $v$ are both non-zero, at which the two models will agree.

Show that Jofra's model and Holly's model disagree in their predictions of the height of pole $B C$ by 3.32 m to 3 significant figures. \section*{Total Marks for Question Set 5: 38} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Marks

AO

Guidance

1

DR $z = \frac { - - 20 \pm \sqrt { ( - 20 ) ^ { 2 } - 4 \times 4 \times 169 } } { 2 \times 4 }$

$z = \frac { 5 \pm 12 \mathrm { i } } { 2 }$ $r = \sqrt { \left( \frac { 5 } { 2 } \right) ^ { 2 } + \left( \frac { 12 } { 2 } \right) ^ { 2 } } = \frac { 13 } { 2 } \mathrm { oe }$

$\theta = \tan ^ { - 1 } \frac { 6 } { 2.5 }$ oe $\frac { 13 } { 2 } ( \cos ( - 1.18 ) + \mathrm { i } \sin ( - 1.18 ) )$

M1

1.1

Term by term substituting into formula.

If formula quoted, allow one slip ... Or correctly completes the square

Condone anything correct of the form $\frac { p \pm \sqrt { q } } { r }$

eg $4 \left( \left( z - \frac { 5 } { 2 } \right) ^ { 2 } - \frac { 25 } { 4 } \right) + 169 = 0$

B1ft

1.1

Ft workings from complex conjugate distinct pair (with real component)

M1

1.1

Attempting to find argument using trigonometry

A1

2.5

Angle must be in radians.

Argument could be 5.11 but both angles must be the same.

Not 5.10 (rounding error)

Not e.g. $\cos ( - 1.18 ) + \mathrm { i } \sin ( 5.11 )$

Question

Answer

Marks

AO

Guidance

\multirow[t]{3}{*}{2}

\multirow[t]{3}{*}{(a)}

\(\begin{aligned}

\left( \begin{array} { c } 13

3

- 14 \end{array} \right) \cdot \left( \begin{array} { l } 1

5

3 \end{array} \right) = 13 + 15 - 42 = - 14 ( \text { so } R \text { is on } \Pi )

\operatorname { eg } 7 - \mu = 13 \Rightarrow \mu = - 6 \Rightarrow

\mathbf { r } = \left( \begin{array} { c } 7

9

- 2 \end{array} \right) - 6 \left( \begin{array} { c } - 1

1

2 \end{array} \right) = \left( \begin{array} { c } 13

3

- 14 \end{array} \right) \quad \left( \text { so } R \text { is also on } l _ { 2 } \right) \end{aligned}\)

B1

1.1

AG. Intermediate working must be seen

AG. Or $9 + \mu = 3$ or $- 2 + 2 \mu = - 14$ but must be checked in other two equations.

Alternate method \(\begin{aligned}

\left( \begin{array} { l } 1

5

3 \end{array} \right) \cdot \left( \left( \begin{array} { c } 7

9

- 2 \end{array} \right) + \mu \left( \begin{array} { c } - 1

1

2 \end{array} \right) \right) = 46 + 10 \mu = - 14 \Rightarrow \mu = - 6

\mu = - 6 \Rightarrow

\mathbf { r } = \left( \begin{array} { c } 7

9

- 2 \end{array} \right) - 6 \left( \begin{array} { c } - 1

1

2 \end{array} \right) = \left( \begin{array} { c } 13

3

- 14 \end{array} \right) \text { so } R \text { is } ( 13,3 , - 14 ) \end{aligned}\)

М1

A1

AG. Substituting in expression of the point into the equation of the plane to find a value for $\mu$ AG.

Answer in vector form is acceptable.

[2]

(b)

\(\text { Since lines intersect } \left( \begin{array} { l } 1
0
a \end{array} \right) + \lambda \left( \begin{array} { c } 2
1
- 3 \end{array} \right) = \left( \begin{array} { c } 7
9
- 2 \end{array} \right) + \mu \left( \begin{array} { c } - 1
1
2 \end{array} \right)\)
for some $\lambda$ and $\mu$ \(\begin{aligned}	\text { so } 1 + 2 \lambda = 7 - \mu
	\lambda = 9 + \mu
	( a - 3 \lambda = - 2 + 2 \mu ) \end{aligned}\) $\Rightarrow \lambda = 5 , \mu = - 4$
so $a + 5 \times ( - 3 ) = - 2 + ( - 4 ) \times 2 \Rightarrow a = 5$

M1

3.1a

Equating the lines and deriving 2 useful equations. Ignore attempts at $z$ coefficient equation

Can be BC

\includegraphics[max width=\textwidth, alt={}]{420598e3-4531-44da-8ae3-088e433f4c05-09_1607_2571_86_239}

Question

Answer

Marks

AO

Guidance

4

(a)

$\operatorname { det } \mathbf { A } ( = 0.6 \times 1.8 - - 0.8 \times 2.4 ) = 3$

B1 [1]

1.1

(b)

Determinant of rotation $= 1$ Determinant of rotation × determinant of stretch $= 1 \times \mathrm { sf } = 3 \Rightarrow \mathrm { sf } = 3$

B1 B1

[2]

1.1 2.2a

(c)

Since the second column of A contains entries bigger than 1 (in magnitude) the stretch must be parallel to the $y$-axis.

B1

[1]

2.4

Or any correct, complete explanation.

\(\begin{gathered} \left( \begin{array} { c c } \cos \theta

- \sin \theta

\sin \theta

\cos \theta \end{array} \right) \left( \begin{array} { l l } 1

0

3 \end{array} \right)

= \left( \begin{array} { c c } \cos \theta

- 3 \sin \theta

\sin \theta

3 \cos \theta \end{array} \right) \end{gathered}\)

(d)

$\sin \theta = - 0.8$ and $\cos \theta = 0.6$ oe awrt $- 53 ^ { \circ }$ (or - 0.93 rads)

M1

A1

[2]

2.2a

1.1

Condone if only one equation or $53 ^ { \circ }$ ( 0.93 rads) clockwise or $307 ^ { \circ }$ (5.36 rads) (anticlockwise).

5

(a)

Min value of cosh is 1 (and point on ground is at the minimum) $( \text { so } 0 = k \times 1 - 1 \Rightarrow ) k = 1$

M1

A1

[2]

2.2a

Using minimum point of curve and knowledge of cosh graph

Could be derived by differentiation

If zero scored then sc1 for $\mathrm { k } = 1$ www

(b)

\(\begin{aligned}

\text { Passes through } ( 0,3 ) = > 3 = \cosh ( - b ) - 1

= > b = - \cosh ^ { - 1 } ( 3 + 1 )

b = ( \pm ) \ln \left( 4 + \sqrt { } \left( 4 ^ { 2 } - 1 \right) \right)

= > b = \ln ( 4 + \sqrt { } 15 )

\text { Passes through } ( 2,0 ) = > 0 = \cosh ( 2 a - b ) - 1

\Rightarrow b = 2 a

= > a = 1 / 2 \ln ( 4 + \sqrt { } 15 ) \end{aligned}\)

*M1

dep*M1

A1

M1

A1

[5]

3.3

3.1a

1.1

3.3

1.1

Use of ( 0,3 ) to derive an expression for $b$ Correct numerical use of formula

Use of $( 2,0 )$ to derive $b = 2 a$

$\operatorname { accept } \cosh ( - b ) = \frac { 4 } { k }$

Or rearranges.

Could be from (a). Allow ft

(c)

(By symmetry of both;) (4,3)

B1

[1]

2.2a

(d)

Holly's model; $d _ { \mathrm { H } } = 6.75$

Jofra's model: $d _ { \mathrm { J } } = \cosh ( 5 a - b ) - 1$

AG

$d _ { \mathrm { J } } - d _ { \mathrm { H } } = 10.067 \ldots - 6.75 = 3.32 ( 3 \mathrm { sf } )$

B1

M1

A1

[3]

3.4

1.1

Use of $x = 5$ with their values of $a$ and $b$ to predict $d$. Must have - 1

From correct values

Condone 27/4 $a = 1.0317 \ldots , b = 2.0634 \ldots , d _ { J } = 10.067 \ldots$ Condone 10.07 only if clear evidence of production

OCR Further Pure Core 2 2021 June Q1

1 In this question you must show detailed reasoning.
The roots of the equation $3 x ^ { 3 } - 2 x ^ { 2 } - 5 x - 4 = 0$ are $\alpha , \beta$ and $\gamma$.

Find a cubic equation with integer coefficients whose roots are $\alpha ^ { 2 } , \beta ^ { 2 }$ and $\gamma ^ { 2 }$.
Find the exact value of $\frac { \alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } } { \alpha \beta \gamma }$.

OCR Further Pure Core 2 2021 June Q2

2 In this question you must show detailed reasoning.

Use partial fractions to show that $\sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }$.
Write down the value of $\lim _ { n \rightarrow \infty } \left( \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } \right)$.

OCR Further Pure Core 2 2021 June Q3

3 The equation of a curve in polar coordinates is $r = \ln ( 1 + \sin \theta )$ for $\alpha \leqslant \theta \leqslant \beta$ where $\alpha$ and $\beta$ are non-negative angles. The curve consists of a single closed loop through the pole.

By solving the equation $r = 0$, determine the smallest possible values of $\alpha$ and $\beta$.
Find the area enclosed by the curve, giving your answer to 4 significant figures.
Hence, by considering the value of $r$ at $\theta = \frac { \alpha + \beta } { 2 }$, show that the loop is not circular.

OCR Further Pure Core 2 2021 June Q4

4 In this question you must show detailed reasoning.
The complex number $- 4 + i \sqrt { 48 }$ is denoted by $z$.

Determine the cube roots of $z$, giving the roots in exponential form. The points which represent the cube roots of $z$ are denoted by $A , B$ and $C$ and these form a triangle in an Argand diagram.
Write down the angles that any lines of symmetry of triangle $A B C$ make with the positive real axis, justifying your answer.

OCR Further Pure Core 2 2021 June Q5

26 marks

5 Let $\mathrm { f } ( x ) = \sin ^ { - 1 } ( x )$.

1. Determine $f ^ { \prime \prime } ( x )$.
2. Determine the first two non-zero terms of the Maclaurin expansion for $\mathrm { f } ( x )$.
3. By considering the first two non-zero terms of the Maclaurin expansion for $\mathrm { f } ( \mathrm { x } )$, find an approximation to $\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x$. Give your answer correct to 6 decimal places.

By writing $\mathrm { f } ( x )$ as $\sin ^ { - 1 } ( x ) \times 1$, determine the value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x$. Give your answer in exact form. \section*{Total Marks for Question Set 6: 37} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Marks

AO

Guidance

1

(a)

DR \(\begin{aligned}

u = x ^ { 2 }

3 ( \sqrt { u } ) ^ { 3 } - 2 ( \sqrt { u } ) ^ { 2 } - 5 \sqrt { u } - 4 ( = 0 ) \end{aligned}\) $3 u \sqrt { u } - 5 \sqrt { u } = 2 u + 4 \Rightarrow u ( 3 u - 5 ) ^ { 2 } = ( 2 u + 4 ) ^ { 2 }$ \(\begin{aligned}

u \left( 9 u ^ { 2 } - 30 u + 25 \right) = 4 u ^ { 2 } + 16 u + 16 = >

9 u ^ { 3 } - 34 u ^ { 2 } + 9 u - 16 = 0 \end{aligned}\)

B1

М1

A1

3.1a

1.1

3.2a

Correct substitution chosen

Oe Attempting to make substitution

Rearranging and squaring bs to remove the square root(s) Rearranging to answer

or preparation for substitution by removing odd powers. $\operatorname { eg } x ^ { 2 } \left( 3 x ^ { 2 } - 5 \right) ^ { 2 } = \left( 2 x ^ { 2 } + 4 \right) ^ { 2 } \ldots$

...and then substituting $u ( 3 u - 5 ) ^ { 2 } = ( 2 u + 4 ) ^ { 2 }$

Equation can be in $x$

Alternative method

DR $\alpha ^ { 2 } \beta ^ { 2 } \gamma ^ { 2 } = ( \alpha \beta \gamma ) ^ { 2 } = \left( - \frac { - 4 } { 3 } \right) ^ { 2 } = \frac { 16 } { 9 }$

$\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }$ $= ( \alpha \beta + \beta \gamma + \gamma \alpha ) ^ { 2 } - 2 \alpha \beta \gamma ( \alpha + \beta + \gamma )$ $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = ( \alpha + \beta + \gamma ) ^ { 2 } - 2 ( \alpha \beta + \beta \gamma + \gamma \alpha )$

$u ^ { 3 } - \left( \left( \frac { 2 } { 3 } \right) ^ { 2 } - 2 \times \frac { - 5 } { 3 } \right) u ^ { 2 } + \left( \left( \frac { - 5 } { 3 } \right) ^ { 2 } - 2 \times \frac { 4 } { 3 } \times \frac { 2 } { 3 } \right) u - \frac { 16 } { 9 }$ $= u ^ { 3 } - \frac { 34 } { 9 } u ^ { 2 } + u - \frac { 16 } { 9 } = 0 \Rightarrow 9 u ^ { 3 } - 34 u ^ { 2 } + 9 u - 16 = 0$

B1

М1

A1

Writing the expression in terms of standard symmetrical forms

Writing the expression in terms of standard symmetrical forms Substituting in and rearranging to answer

Must include one intermediate step $\mathrm { NB } \sum \alpha = \frac { 2 } { 3 } , \sum \alpha \beta = - \frac { 5 } { 3 } , \alpha \beta \gamma = \frac { 4 } { 3 }$

Condone without factorisation of "2"

$\mathrm { NB } \sum \alpha ^ { 2 } = \frac { 34 } { 9 } , \sum \alpha ^ { 2 } \beta ^ { 2 } = 1$

[4]

(b)

DR $\frac { \sum \alpha ^ { 2 } \beta ^ { 2 } } { \alpha \beta \gamma } = \frac { \left( \frac { 9 } { 9 } \right) } { \left( \frac { 4 } { 3 } \right) } \text { or } \frac { 1 } { \left( \frac { 4 } { 3 } \right) }$

$= \frac { 3 } { 4 }$

M1

A1

[2]

3.1a

1.1

Their $\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }$ from part (a) over $\pm \frac { 4 } { 3 }$

Strict ft

Question

Answer

Marks

AO

Guidance

2

(a)

\(\begin{aligned}

\text { DR }

( r + 2 ) ( r - 1 ) \end{aligned}\) \(\begin{aligned}

\frac { A } { r - 1 } + \frac { B } { r + 2 }

A = 1 , B = - 1

= \end{aligned}\) \(\begin{array} { c c c c c c } \frac { 1 } { 4 }

-

\frac { 1 } { 7 }

\cdots

-

\frac { 1 } { n - 1 }

\frac { 1 } { 5 }

-

\frac { 1 } { 8 }

\frac { 1 } { n - 3 }

-

\frac { 1 } { n }

\frac { 1 } { 6 }

-

\frac { 1 } { 9 }

\frac { 1 } { n - 2 }

-

\frac { 1 } { n + 1 }

\frac { 1 } { 7 }

-

\cdots

\frac { 1 } { n - 1 }

-

\frac { 1 } { n + 2 } \end{array}\) \(\begin{aligned}

= \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }

= \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } \end{aligned}\)

A1

1.1

Correct factorisation of denominator soi Correct form for partial fractions

M1 can be ft from any $A , B$ having opposite signs

For M1, condone omission of $\frac { 1 } { 7 }$ or $- \frac { 1 } { n - 1 }$

Alternative Method \(\begin{aligned}

\therefore \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 5 } ^ { n } \frac { 1 } { r + 2 }

= \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 8 } ^ { n + 3 } \frac { 1 } { r - 1 }

= \sum _ { r = 5 } ^ { 7 } \frac { 1 } { r - 1 } - \sum _ { r = n + 1 } ^ { n + 3 } \frac { 1 } { r - 1 } \end{aligned}\)

М1

Using partial fractions, separating into two sums, re-indexing so that the summands have identical form and cancelling central terms.

Might see start and end terms explicitly. eg $\sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 8 } ^ { n + 3 } \frac { 1 } { r - 1 }$

Question

Answer

Marks

AO

Guidance

\multirow{2}{*}{}

\(\begin{aligned}

\therefore \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } =

= \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }

= \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } \end{aligned}\)

A1

AG.

Might see formal substitution of index. eg
Let $R = r + 3 \Rightarrow r + 2 = R - 1$ \(\begin{aligned}	\therefore \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 5 } ^ { n } \frac { 1 } { r + 2 }
	= \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { R = 8 } ^ { n + 3 } \frac { 1 } { R - 1 } \end{aligned}\)

[5]

(b)

$= \frac { 37 } { 60 }$ or awrt 0.617

B1

[1]

2.2a

3

(a)

\(\begin{aligned}

\ln ( 1 + \sin \theta ) = 0 \Rightarrow 1 + \sin \theta = 1 \Rightarrow \sin \theta = 0

\text { so } \alpha = 0 \text { and } \beta = \pi \end{aligned}\)

M1

A1 [2]

1.1a

2.2a

(b)

\(\begin{aligned}

A = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } ( \ln ( 1 + \sin \theta ) ) ^ { 2 } \mathrm {~d} \theta

= 0.4162 ( 4 \mathrm { sf } ) \text { cao } \end{aligned}\)

M1

A1 [2]

1.2

1.1

Correct formula for area with $r$ correctly substituted and their limits. Must be unambiguous but can be implied by correct answer/later work BC

Incorrect formula = M0A0 Condone missing $\mathrm { d } \theta$

(c)

$\theta = \frac { \pi } { 2 } \Rightarrow r = \ln 2 = 0.6931 ( 4 \mathrm { sf } )$ which would be the diameter, $D$, of the circle

But $A = 0.4162 ( 4 \mathrm { sf } ) = > D = 0.7280 ( 4 \mathrm { sf } )$ or $R = 0.3640 ( 4 \mathrm { sf } )$ so the curve is not circular

M1

A1

[2]

3.1a

3.2a

or radius $R = 0.3466 ( 4 \mathrm { sf } )$ condone correct $R$ or $D$ without reasoning

or $R = 0.3466 ( 4 \mathrm { sf } ) ($ or $D = 0.6931 ) \Rightarrow A = 0.3773 ( 4 \mathrm { sf } )$ which is not $0.4162 ( 4 \mathrm { sf } )$

It must be clear that the $r$ value would be the diameter of the circle; the calculation alone is insufficient for M1.

M1 can be implied by area given as $\pi \left( \frac { \ln 2 } { 2 } \right) ^ { 2 }$

Explanation must include comparison of $R$ 's, $D$ 's or $A$ 's and conclusion . Allow correct working to 3 sf .

Question

Answer

Marks

AO

Guidance

4

(a)

DR

\(\begin{aligned}

r ^ { 2 } = ( - 4 ) ^ { 2 } + ( \sqrt { 48 } ) ^ { 2 } \quad \text { or } ( r \cos \theta = - 4 \text { and }

r \sin \theta = \sqrt { 48 } ) \text { or } \tan \theta = - \sqrt { 3 } \text { oe }

r = 8 \left( \mathrm { ie } z = 8 \mathrm { e } ^ { \mathrm { i } \theta } \right) \quad \theta = 2 \pi / 3 \left( \mathrm { ie } z = r \mathrm { e } ^ { \mathrm { i } 2 \pi / 3 } \right)

\sqrt [ 3 ] { 8 } \text { or } 2

\frac { 2 \pi } { 9 } \text { soi }

\frac { 2 \pi } { 3 } + 2 \pi k \text { for } k = 1 \text { and } 2 \text { oe seen }

2 \mathrm { e } ^ { \frac { 2 } { 9 } \pi i } , 2 \mathrm { e } ^ { \frac { 8 } { 9 } \pi i } \text { and } 2 \mathrm { e } ^ { - \frac { 4 } { 9 } \pi i } \end{aligned}\)

A1

2.1

Correct use of relevant formula(e). Some working must be seen.

Correct answer with no working: M0A0 or eg $\theta = 8 \pi / 3$

B1ft

2.1

Argument of (principal) cube root is one third of their argument

M1

2.2a

Considering further arguments at angular distance $2 \pi$

A1

1.1

or eg $2 \mathrm { e } ^ { \frac { 2 } { 9 } \pi i } , 2 \mathrm { e } ^ { \frac { 8 } { 9 } \pi i }$ and $2 \mathrm { e } ^ { \frac { 14 } { 9 } \pi i }$

Must be in exponential form, not just $r =$ and $\theta =$. Do not condone any missing i's.

(b)

DR

The cube roots form an equilateral triangle which has (3) lines of symmetry, (one) through each vertex $\theta = \frac { 2 \pi } { 9 } , \theta = \frac { 8 \pi } { 9 } \text { and } \theta = - \frac { 4 \pi } { 9 } \text { soi }$

B1

2.2a

for one

for all three without extras

ft their angles if $2 \pi / 3$ apart.

If valid alternatives, must come from clear explanation/diagram

Question

Answer

Marks

AO

Guidance

5

(a)

(i)

\(\begin{aligned}

\mathrm { f } ^ { \prime } ( x ) = \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } } \text { from the formula book }

\text { so } \mathrm { f } ^ { \prime \prime } ( x ) = - \frac { 1 } { 2 } \cdot \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \cdot ( - 2 x )

= \frac { x } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \end{aligned}\)

М1

1.1

Formula from the Formula Booklet and attempt differentiation

To within sign error

(a)

(ii)

\(\begin{aligned}

f ( 0 ) = 0 , f ^ { \prime } ( 0 ) = 1 \text { and } f ^ { \prime \prime } ( 0 ) = 0

f ^ { \prime \prime \prime } ( x ) = \frac { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } - x \cdot \frac { 3 } { 2 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \cdot ( - 2 x ) } { \left( 1 - x ^ { 2 } \right) ^ { 3 } }

\text { so } f ^ { \prime \prime \prime } ( 0 ) = 1 \text { and } f ( x ) = x + \frac { 1 } { 6 } x ^ { 3 } + \ldots \end{aligned}\)

B1

M1

A1

[3]

1.1

3.1a

2.1

or $a _ { 0 } = 0 , a _ { 1 } = 1$ and $a _ { 2 } = 0$

Differentiate and simplify far enough to be able to justify value 1

Condone 3! In place of 6

Ignore sign error in $\mathrm { f } ^ { \prime \prime } ( x )$

Either full derivative or "zero term" denoted as such

Not BC. If M0 then SC1 for correct expansion

(a)

(iii)

\(\begin{aligned} \int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x \approx \int _ { 0 } ^ { \frac { 1 } { 2 } } x

+ \frac { 1 } { 6 } x ^ { 3 } \mathrm {~d} x

=

0.127604167 \ldots

=

0.127604 \text { to } 6 \mathrm { dp } \end{aligned}\)

M1

A1

[2]

1.1

Integral of their 2 term cubic with limits

Could be BC

(b)

\(\begin{aligned}

\int 1 \times \sin ^ { - 1 } x \mathrm {~d} x = x \sin ^ { - 1 } x - \int \frac { x } { \sqrt { 1 - x ^ { 2 } } } \mathrm {~d} x

= x \sin ^ { - 1 } x + \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } ( + \mathrm { c } )

\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) = \frac { \pi } { 12 } + \frac { \sqrt { 3 } } { 2 } - 1 \end{aligned}\)

M1

A1

[3]

3.1a

1.1

Attempt integration by parts

ignore limits. Formula for parts must be correct

OCR Further Statistics 2021 June Q1

28 marks

1 The performance of a piece of music is being recorded. The piece consists of three sections, $A , B$ and $C$. The times, in seconds, taken to perform the three sections are normally distributed random variables with the following means and standard deviations. \end{table}

Question

Answer

Mark

AO

Guidance

\multirow[t]{3}{*}{1}

\multirow[t]{3}{*}{(a)}

$A + B + C \sim \mathrm {~N} ( 701 , \ldots$

.. 419)

M1

1.1a

Normal, mean $\mu _ { A } + \mu _ { B } + \mu _ { C }$

\multirow{3}{*}{}

A1

1.1

Variance 419

$\mathrm { P } ( > 720 ) = 0.176649$

A1

1.1

Answer, 0.177 or better, www

\multirow[t]{2}{*}{1}

\multirow[t]{2}{*}{(b)}

$2 A + B \sim \mathrm {~N} ( 701,757 )$

M1

1.1a

Normal, same mean, $4 \sigma _ { A } { } ^ { 2 } + \sigma _ { B } { } ^ { 2 }$

\multirow{2}{*}{}

$\mathrm { P } ( > 720 ) = 0.244919$

A1 [2]

1.1

Answer, art 0.245

\multirow{2}{*}{2}

\multirow{2}{*}{(a)}

$\frac { { } ^ { 8 } C _ { 3 } \times { } ^ { 20 } C _ { 5 } } { { } ^ { 28 } C _ { 8 } }$

M1 A1

3.1b 1.1

(Product of two ${ } ^ { n } C _ { r }$ ) ÷ ${ } ^ { n } C _ { r }$ At least two ${ } ^ { n } C _ { r }$ correct

\multirow[t]{2}{*}{Or $\frac { 8 } { 28 } \times \frac { 7 } { 27 } \times \frac { 6 } { 26 } \times \frac { 20 } { 25 } \times \ldots \times \frac { 16 } { 21 } \times { } ^ { 8 } C _ { 3 } = 0.27934 \ldots$}

$\frac { 56 \times 15504 } { 3108105 } = 0.27934 \ldots$

A1 [3]

1.1

Any exact form or awrt 0.279

2

(b)

× B × B × B × B × B × B × B × B x

GGG in one $\mathrm { x } , \mathrm { G }$ in another: $9 \times 8$ $\div \frac { 12 ! } { 8 ! \times 4 ! }$ $= \frac { 72 } { 495 } = \frac { 8 } { 55 } \text { or } 0.145 \ldots$

M1 A1

3.1b 2.1

Or e.g. $\frac { 10 ! } { 8 ! } - 2 \times 9$

Divide by ${ } _ { 12 } \mathrm { C } _ { 4 }$ oe

Or, e.g. find ${ } _ { 12 } \mathrm { C } _ { 4 }$ - (\# (all separate) +\#(all together) $+ \# ( 2,1,1 ) \times 3 +$ \#(2,2))

М1

1.1

A1

1.1

[4]

Question

Answer

Mark

AO

Guidance

\multirow{7}{*}{3}

\multirow{7}{*}{(a)}

$\mathrm { H } _ { 0 } : \mu = 700$

B2

1.1

One error, e.g. no or wrong

Ignore failure to define $\mu$

$\mathrm { H } _ { 1 } : \mu < 700$ where $\mu$ is the mean reaction

1.1

letter, $\neq$, etc : B1

here

$\bar { x } = 607$

М1

3.3

Find sample mean

$z = - 1.822$ or $p = 0.0342$ or $\mathrm { CV } = 616.05 \ldots$

A1

3.4

Correct $z , p$ or CV

$z < - 1.645$ or $p < 0.05$ or $607 < \mathrm { CV }$

A1

1.1

Correct comparison

Reject $\mathrm { H } _ { 0 }$

M1ft

1.1

Correct first conclusion

Needs correct method, like-

Significant evidence that mean reaction times

A1ft

2.2b

Context, not too definite (e.g. not "international athletes' reaction times are shorter"

ft on their $z , p$ or CV

3

(b)

(i)

Uses more information (e.g. magnitudes of differences)

B1 [1]

2.4

\multirow{5}{*}{3}

\multirow{5}{*}{(b)}

\multirow{5}{*}{(ii)}

$\mathrm { H } _ { 0 } : m = 700 , \mathrm { H } _ { 1 } : m < 700$ where $m$ is the median reaction time for all international athletes

B1

2.5

Same as in (i) but different letter or "median" stated

$W _ { - } = 18$

$W _ { + } = 3$ so $T = 3$

For both, and $T$ correct

$n = 6 , \mathrm { CV } = 2$

A1

1.1

Correct CV

Do not reject $\mathrm { H } _ { 0 }$. Insufficient evidence that median reaction times of international athletes are shorter

A1ft [6]

2.2b

In context, not too definite

FT on their $T$

3

(c)

They use different assumptions

B1 [1]

2.3

Not "one is more accurate"

Question

Answer

Mark

AO

Guidance

4

(a)

\(\begin{aligned}

\int _ { 0 } ^ { a } x \frac { 2 x } { a ^ { 2 } } d x = 4

{ \left[ \frac { 2 x ^ { 3 } } { 3 a ^ { 2 } } \right] = 4 }

\frac { 2 } { 3 } a = 4 \Rightarrow a = 6 \end{aligned}\)

M1

B1

A1 [3]

3.1a

1.1

2.2a

4

(b)

$\mathrm { F } ( x ) = \frac { x ^ { 2 } } { 36 }$
Let the CDF of $M$ be $\mathrm { H } ( m )$. Then $\mathrm { H } ( m ) = \mathrm { P } ($ all observations less than $m )$ $= [ \mathrm { P } ( X \leqslant m ) ] ^ { 5 }$ $= \left[ \frac { m ^ { 2 } } { 36 } \right] ^ { 5 }$
\(\mathrm { H } ( m ) = \begin{cases} 0	m < 0 ,
\frac { m ^ { 10 } } { 60466176 }	0 \leq m \leq 6 ,
1	m > 6 . \end{cases}\)

M1 A1ft

M1

A1

[8]

1.1
1.1
2.1
3.1a
2.2a
2.1
2.1
1.2

Find $\mathrm { F } ( x ) ; = \frac { x ^ { 2 } } { a ^ { 2 } }$

Correct basis for CDF of $m$

Correct function, any letter Range $0 \leq m \leq 6$

Letter not $x$, and 0, 1 present

ft on their $a$

Allow

OCR Further Statistics 2021 June Q1

1 A set of bivariate data ( $X , Y$ ) is summarised as follows.
$n = 25 , \Sigma x = 9.975 , \Sigma y = 11.175 , \Sigma x ^ { 2 } = 5.725 , \Sigma y ^ { 2 } = 46.200 , \Sigma x y = 11.575$

Calculate the value of Pearson's product-moment correlation coefficient.
Calculate the equation of the regression line of $y$ on $x$. It is desired to know whether the regression line of $y$ on $x$ will provide a reliable estimate of $y$ when $x = 0.75$.
State one reason for believing that the estimate will be reliable.
State what further information is needed in order to determine whether the estimate is reliable.

OCR Further Statistics 2021 June Q2

2 The average numbers of cars, lorries and buses passing a point on a busy road in a period of 30 minutes are 400,80 and 17 respectively.

Assuming that the numbers of each type of vehicle passing the point in a period of 30 minutes have independent Poisson distributions, calculate the probability that the total number of vehicles passing the point in a randomly chosen period of 30 minutes is at least 520 .
Buses are known to run in approximate accordance with a fixed timetable. Explain why this casts doubt on the use of a Poisson distribution to model the number of buses passing the point in a fixed time interval. The greatest weight $W \mathrm {~N}$ that can be supported by a shelving bracket of traditional design is a normally distributed random variable with mean 500 and standard deviation 80 . A sample of 40 shelving brackets of a new design are tested and it is found that the mean of the greatest weights that the brackets in the sample can support is 473.0 N .
Test at the $1 \%$ significance level whether the mean of the greatest weight that a bracket of the new design can support is less than the mean of the greatest weight that a bracket of the traditional design can support.
State an assumption needed in carrying out the test in part (a).
Explain whether it is necessary to use the central limit theorem in carrying out the test.

OCR Further Statistics 2021 June Q4

4
The random variable $D$ has the distribution $\operatorname { Geo } ( p )$. It is given that $\operatorname { Var } ( D ) = \frac { 40 } { 9 }$.
Determine

$\operatorname { Var } ( 3 D + 5 )$,
$\mathrm { E } ( 3 D + 5 )$,
$\mathrm { P } ( D > \mathrm { E } ( D ) )$.

OCR Further Statistics 2021 June Q5

48 marks

5 A university course was taught by two different professors. Students could choose whether to attend the lectures given by Professor $Q$ or the lectures given by Professor $R$. At the end of the course all the students took the same examination. The examination marks of a random sample of 30 students taught by Professor $Q$ and a random sample of 24 students taught by Professor $R$ were ranked. The sum of the ranks of the students taught by Professor $Q$ was 726 . Test at the $5 \%$ significance level whether there is a difference in the ranks of the students taught by the two professors.
[0pt] [10] Total Marks for Question Set 3: 38 \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Question

Answer

Mark

AO

Guidance

1

(a)

0.8392...

B1 [1]

1.1

Awrt 0.839

\(\begin{aligned} S _ { x x }

= 1.7449 \ldots , S _ { y y } = 41.2 \ldots ,

S _ { x y }

= 7.116 \ldots \end{aligned}\)

1

(b)

$y = - 1.180 + 4.0781 x$

B1

[2]

1.1

Both coeffs, awrt -1.18 and 4.08

Letters correct, needs 1 correct coefficient

1

(c)

Value of PMCC suggests that there is strong correlation, or 0.75 shown close to mean 0.399

B1

[1]

3.5a

E.g. " $r$ high so points lie close to line". " $r$ is high" alone is enough.

No wrong extras

Not "0.75 is close to mean", unless properly justified, e.g. SD (= 0.264) calculated

1

(d)

Whether $x = 0.75$ is within the data range

B1

[1]

3.5b

E.g. "maximum and minimum values of $x$ "; not "all data points".

No wrong extras

Or clear reference to interpolation. NB: 95\% CI for $x$ is ( $- 0.156,0.954$ )

2

(a)

Po(497)

$\mathrm { P } ( \geq 520 ) = 1 - \mathrm { P } ( \leq 519 )$ used correctly

$= 0.1564 \ldots$

B1

M1

A1 [3]

1.1

1.1a

1.1

Stated or implied

Allow 0.146(08) from 1 $\mathrm { P } ( \leq 520 )$

In range [0.156,0.157]

SC: Normal approx.:

N(497, 497) B1

In range [0.156, 0.157]: B2

2

(b)

Occurrence of a bus is not a random event if it runs on or close to a schedule.

B1

[1]

2.4

Needs context (not just "events").

Allow just "buses not random", or "buses not independent because time between buses is regulated"

Not "not independent" without such justification. Not "not constant rate". No extras.

Question

Answer

Mark

AO

Guidance

\multirow[t]{4}{*}{3}

(a)

$\mathrm { H } _ { 0 } : \mu = 500 , \mathrm { H } _ { 1 } : \mu < 500$

B1

1.1

One error, e.g. $\mathrm { H } _ { 1 } : \mu \neq 500$, or $\mu$ not defined, or all in words: B1

$x$ or $\bar { x } : 0$ unless defined as population mean (then B1)

\(\begin{aligned}

\bar { X } \sim \mathrm {~N} \left( 500 , \frac { 80 ^ { 2 } } { 40 } \right) = \mathrm { N } ( 500,160 ) \text { and } \bar { X }

\mathrm { P } ( \bar { X } < 473 ) = 0.01640 \text { or } z = - 2.13 ( 45 )

\text { or } \mathrm { CV } = 470.6 \end{aligned}\)

М1

3.3

$p$ or $z$ correct to 3 sf .

Can be implied by 0.0164, 0.9836, 0.433, 0.198, 0.000 but not 0.3679 or 0.00127

$p > 0.01$ or $z > - 2.326 \quad$ or $473 > 470.6$

A1

1.1

Compare $p$ with 0.01 or $z$ with -2.326, or 2.326 used in CV

Must be like-with-like, Not e.g. 0.9836 > 0.01 or $p < 2.326$

Do not reject $\mathrm { H } _ { 0 }$. Insufficient evidence that greatest weight that new design can support is less than the greatest weight that the traditional design can support.

M1ft

A1ft [7]

1.1

Correct first conclusion, needs correct method and like-with-like, ft on test statistic if method correct Contextualised, not too definite

But BOD if no explicit comparison of $p$ with 0.01 Not "the new design does not have a smaller greatest weight . . ."

3

(b)

Standard deviation/variance remains unchanged, or sample must be random

B1 [1]

1.2

No extras. Not "same distribution".

Not "assume normal"; this is not needed

3

(c)

Either: Yes as we do not know that the distribution of weights for the new design is normal

Or: $\quad$ No as the population distribution known to be normal

B1 [1]

2.1

Allow "population distribution assumed to be normal". No extras, e.g. "and sample size is large".

Allow "yes as we do not know that the distribution for the new design is normal" only if clearly refers to the new design only

Question

Answer

Mark

AO

Guidance

4

(a)

$9 \times \frac { 40 } { 9 } = 40$

B1 [1]

1.1

40 or awrt 40.0 only

4

(b)

$\frac { 1 - p } { p ^ { 2 } } = \frac { 40 } { 9 }$

М1

3.1b

Use correct formula for variance

SC: insufficient working, $\frac { 3 } { 8 }$ only: M0B1 for $\frac { 3 } { 8 }$, then B0

\(\begin{aligned}

\mathrm { E } ( D ) = 1 / p \quad \left[ = \frac { 8 } { 3 } \right]

\mathrm { E } ( 3 D + 5 ) = 3 \times \frac { 8 } { 3 } + 5 \quad [ = 13 ] \end{aligned}\)

B1ft

2.3

- formula for $\mathrm { E } ( D )$

Allow for explicit rejection of a solution even if both are wrong

$p$ doesn't need to be between 0 and 1 for either of these marks

A1ft [6]

1.1

$3 \times ($ their $\mathrm { E } ( D ) ) + 5$

SC: $\frac { 1 - p } { p ^ { 2 } } = 40$ (their 40), $p = \frac { - 1 \pm \sqrt { 161 } } { 80 }$, reject negative solution, $\mathrm { E } ( D ) = \frac { 1 + \sqrt { 161 } } { 2 } = 6.844 , \mathrm { E } ( 3 D + 5 ) = 25.53 : \quad \mathrm { M } 1 , \mathrm { M } 1 \mathrm {~A} 0 , \mathrm {~B} 1 , \mathrm {~B} 2$ total $5 / 6$

4

(c)

\(\begin{aligned}

\mathrm { P } ( D > \mathrm { E } ( D ) ) = \mathrm { P } ( D \geq 3 )

= ( 1 - p ) ^ { 2 }

= \frac { 25 } { 64 } \text { or } 0.390625 \end{aligned}\)

M1ft М1

A1 [3]

3.1a 1.1a

1.1

Convert inequality to integer, their $[ 1 / p ] + 1$, allow >

$( 1 - p ) ^ { r } , \mathrm { ft }$ on their $p , r$, e.g. 8/3 or 13

Allow $( 1 - p ) ^ { 3 } = 125 / 512$ or 0.244

Answer, exact or art 0.391, www

Not their 13

$( 1 - p ) ^ { 8 / 3 } [ 0.286 ]$ : M0M1A0

Need $0 < p < 1$ here

Allow $( 1 - p ) ^ { 6 } = 0.3876$ from SC above

Question

Answer

Mark

AO

Guidance

\multirow[t]{10}{*}{5}

\multirow{10}{*}{}

$\mathrm { H } _ { 0 } : m _ { Q } = m _ { R } , \mathrm { H } _ { 1 } : m _ { Q } \neq m _ { R }$, where $m _ { Q }$ and $m _ { R }$ are the medians of the rankings given to $Q$ and

B1

1.1

Allow $m$ undefined. If verbal, must mention medians, $m$ or distribution. Allow $m _ { d } = 0$ as opposed to $m Q = m _ { R }$

Not anything that might be $\mu$ unless symbol clearly defined as median. Not "there is no difference in the ranks ..."

Sum of ranks $= 1 / 2 \times 54 \times 55 = 1485$

М1

1.1

Find sum of ranks

$R _ { m } = 1485 - 726 = 759 \quad$ [or 561]

A1

1.1

Correct value of $R _ { m }$ seen

Allow even if 726 used later

\(\begin{array} { r } R _ { m } \sim \mathrm {~N} ( 660 ,

\quad \ldots 3300 ) \end{array}\)

М1

A1

3.1b

3.3

normal, mean their $\frac { 1 } { 2 } \times 24 \times$ 55

Allow SD/Var muddle

\(\begin{aligned} \mathrm { P } \left( R _ { m } \geq 759 \right)

= 0.0432 \text { (3 s.f.) }

{ [ \text { or } z }

= 1.715 ] \end{aligned}\)

М1

A1

3.4

1.1

Both parameters correct Standardise, their $R _ { m }$

Correct test statistic (0.0432) 0.0424 or 0.0416 (no/wrong cc): M1A0

(Same for $\mathrm { P } \left( R _ { m } \leq 561 \right)$ Allow $z \square \in [ 1.71,1.715 ]$, allow $z = 1.72$ only if cc demonstrated correct

Alternatively: $\operatorname { CV } 660 + 1.96 \sqrt { } 3300 [ = 772.6 ]$

758.5 < 772.6

M1 A1

Not 759 - or 726 - ...; not wrong tail for comparison, but allow ± Needs correct cc

Or 561.5 > 547.4 Wrong $z$-value: M1A1ft B0

$p > 0.025,2 p > 0.05 , z < 1.96$, or 1.96 used in CV

B1

1.1

Explicit correct comparison

Needs like-with-like (e.g. $p$ must be < 0.5)

Do not reject $\mathrm { H } _ { 0 }$. Insufficient evidence of a difference between the ranks.

M1ft

A1ft [10]

1.1

2.2b

Correct first conclusion, needs correct method and like-with-like Contextualised, not too definite

ft on wrong ts, or 1-tail/2-tail confusions, e.g. $p$ compared with 0.05 or not explicit, or $z \geq 1.645$

\includegraphics[max width=\textwidth, alt={}]{6cdb3135-90ca-42f1-bab1-a4b35451cea2-10_54_1750_1703_611}

OCR Further Statistics 2021 June Q1

1
The continuous random variable $X$ has the distribution $\mathrm { N } ( \mu , 30 )$. The mean of a random sample of 8 observations of $X$ is 53.1 . Determine a $95 \%$ confidence interval for $\mu$. You should give the end points of the interval correct to 4 significant figures.

OCR Further Statistics 2021 June Q2

2 A book collector compared the prices of some books, $\pounds x$, when new in 1972 and the prices of copies of the same books, $\pounds y$, on a second-hand website in 2018.
The results are shown in Table 1 and are summarised below the table. \begin{table}[h]

Book	A	B	C	D	E	F	G	H	I	J	K	L
$x$	0.95	0.65	0.70	0.90	0.55	1.40	1.50	0.50	1.15	0.35	0.20	0.35
$y$	6.06	7.00	2.00	5.87	4.00	5.36	7.19	2.50	3.00	8.29	1.37	2.00

\captionsetup{labelformat=empty} \caption{Table 1}

\end{table} $$n = 12 , \Sigma x = 9.20 , \Sigma y = 54.64 , \Sigma x ^ { 2 } = 8.9950 , \Sigma y ^ { 2 } = 310.4572 , \Sigma x y = 46.0545$$

It is given that the value of Pearson's product-moment correlation coefficient for the data is 0.381 , correct to 3 significant figures.
1. State what this information tells you about a scatter diagram illustrating the data.
2. Test at the $5 \%$ significance level whether there is evidence of positive correlation between prices in 1972 and prices in 2018.
The collector noticed that the second-hand copy of book J was unusually expensive and he decided to ignore the data for book J. Calculate the value of Pearson's product-moment correlation coefficient for the other 11 books.

OCR Further Statistics 2021 June Q3

3 The numbers of CD players sold in a shop on three consecutive weekends were 7,6 and 2 . It may be assumed that sales of CD players occur randomly and that nobody buys more than one CD player at a time. The number of CD players sold on a randomly chosen weekend is denoted by $X$.

How appropriate is the Poisson distribution as a model for $X$ ? Now assume that a Poisson distribution with mean 5 is an appropriate model for $X$.
Find
1. $\mathrm { P } ( X = 6 )$,
2. $\mathrm { P } ( X \geqslant 8 )$. The number of integrated sound systems sold in a weekend at the same shop can be assumed to have the distribution $\operatorname { Po } ( 7.2 )$.
Find the probability that on a randomly chosen weekend the total number of CD players and integrated sound systems sold is between 10 and 15 inclusive.
State an assumption needed for your answer to part (c) to be valid.
Give a reason why the assumption in part (d) may not be valid in practice.

OCR Further Statistics 2021 June Q4

38 marks

4 The continuous random variable $X$ has probability density function $$\mathrm { f } ( x ) = \begin{cases} \frac { k } { x ^ { n } } & x \geqslant 1
0 & \text { otherwise } \end{cases}$$ where $n$ and $k$ are constants and $n$ is an integer greater than 1 .

Find $k$ in terms of $n$.
1. When $n = 4$, find the cumulative distribution function of $X$.
2. Hence determine $\mathrm { P } ( X > 7 \mid X > 5 )$ when $n = 4$.

Determine the values of $n$ for which $\operatorname { Var } ( X )$ is not defined. \section*{Total Marks for Question Set 5: 38} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
When a value is not given in the paper accept any answer that agrees with the correct value to $\mathbf { 3 ~ s } . \mathbf { f }$. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.

Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of $9.80,9.81$ or 10 for $g$ should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:

If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.

For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Abbreviations}

Abbreviations used in the mark scheme	Meaning
dep*	Mark dependent on a previous mark, indicated by . The may be omitted if only one previous M mark
cao	Correct answer only
ое	Or equivalent
rot	Rounded or truncated
soi	Seen or implied
www	Without wrong working
AG	Answer given
awrt	Anything which rounds to
BC	By Calculator
DR	This question included the instruction: In this question you must show detailed reasoning.

\end{table}

Answer

Mark

AO

Guidance

\multirow{2}{*}{}

\multirow[t]{2}{*}{

$53.1 \pm 1.96 \sqrt { \frac { 30 } { 8 } }$

(49.30, 56.90)

}

M1

3.3

Square root correct Awrt 1.96 used, can be implied

\multirow{2}{*}{Allow e.g. (49.30, 56.9)}

A1 [4]

3.4

Both, only these numbers (4 sf needed at least once)

2

(a)

(i)

The points do not lie very close to a straight line

B1 [1]

1.1

Or equivalent. Must refer to diagram, not just to "correlation"

Ignore extras unless wrong

\multirow{3}{*}{}

\multirow[t]{3}{*}{(ii)}

$\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0$, where $\rho$ is the population pmcc between prices in 1972 and prices in 2018

B2

1.1 2.5

One error, e.g. $\rho$ not defined, B1 (but allow "population" not stated)

$\mathrm { H } _ { 0 } : r = 0 , \mathrm { H } _ { 1 } : r > 0$ : same scheme, but B 2 needs "population" pmcc Compare with 0.497(3)

$\mathrm { H } _ { 0 }$ : no correlation, $\mathrm { H } _ { 1 }$ : positive correlation: B 1

1.1

2.2b

FT on CV 0.5760 only

$0.381 < 0.4973$

Do not reject $\mathrm { H } _ { 0 }$.

There is insufficient evidence of (positive) correlation between prices in the two years.

Exx:

$\alpha$ : Insufficient evidence to reject $\mathrm { H } _ { 0 }$. No correlation between ...

$\beta$ : Wrong first conclusion, correct interpretation:

$\gamma$ : Hypotheses wrong way round:

M1

M1ft

A1ft [5]

Correct first conclusion, needs like-with-like In context, not too definite

M1A1 (bod)

M0A0

Maximum M1M1

2

(b)

0.650

B2

[2]

3.1a

1.1

Full marks for correct answer by any method

SC: if B0 allow B1 for any 3 of 8.85, 46.35, 8.8725, 241.7331, 43.153

Question

Answer

Mark

AO

Guidance

\multirow{2}{*}{3}

\multirow{2}{*}{(a)}

\multirow{2}{*}{}

\multirow[t]{2}{*}{

B1

B1 [2]

}

\multirow[t]{2}{*}{

3.5b

}

"Events occur independently and at constant average rate": B0

Any reason for independence (or not)

... and for constant average rate (or not), in each case without misunderstanding of what they mean

SC: Mere assertion of both, properly contextualised: B1

SC: Variance $= 4.67$ which is closer to 5: B 1

SC: Considers only assumptions given in the question: B0

3

(b)

(i)

0.146(223) BC

M1

A1

[2]

3.4

1.1

Correct method stated or implied

Correct answer only, awrt 0.146

3

(ii)

0.133(372) BC

M1

A1

[2]

1.1

0.068: M1A0

(0.1337 give M1A1 BOD)

3

(c)

Po(12.2) $\mathrm { P } ( \leq 15 ) - \mathrm { P } ( \leq 9 ) \quad [ = 0.8296 - 0.2253 ]$

$= 0.604 ( 224 ) \quad \mathbf { B C }$

M1

A1 [3]

3.3

1.1

3.4

Stated or implied

Allow $\mathrm { P } ( \leq 16 )$ or $\mathrm { P } ( \leq 10 )$, e.g. 0.503 or 0.662

(M1M1A0)

Correct answer only, awrt 0.604

Allow this M1 also from $\lambda = 7.2 ( 0.187,0.110,0.189 )$

3

(d)

Sales of CD players and integrated systems need to be independent

B1

[1]

1.1

Need "independent" or "not related" clearly referred to the two types of machine.

Not just "purchases independent" or "distributions independent"

\multirow{2}{*}{3}

\multirow{2}{*}{(e)}

\multirow{2}{*}{}

\multirow[t]{2}{*}{B1 [1]}

\multirow[t]{2}{*}{3.5b}

Any reason for nonindependence of sales of CD players and integrated sound systems

Can get B0B1 provided they are focussing on independence

If a customer buys a CD player they probably won't (or will) buy an integrated system as well

Exx:

α: May buy both so not independent: B0

$\beta$ : Often bought together: B1

$\gamma$ : $\quad$ Context misunderstood: can get B1

e.g. CDs/CD players, or assuming that integrated systems don't include CD players

Question

Answer

Mark

AO

Guidance

4

(a)

\(\begin{aligned}

\int _ { 1 } ^ { \infty } k x ^ { - n } \mathrm {~d} x = \left[ \frac { k } { ( 1 - n ) x ^ { n - 1 } } \right] _ { 1 } ^ { \infty }

= \frac { k } { n - 1 } = 1 \text { so } k = n - 1 \end{aligned}\)

M1

B1

A1

[3]

1.1

Integral attempted, correct limits

Correct indefinite integral Correctly obtain $k = n - 1$, www

Don't need full details of $\lim ( a \rightarrow \infty )$

4

(b)

(i)

\(\begin{aligned}

\int 3 x ^ { - 4 } \mathrm {~d} x = - \frac { 1 } { x ^ { 3 } } + c

x = 1 , \mathrm {~F} ( x ) = 0 \text { so } c = 1 . \text { Hence } 1 - x ^ { - 3 }

\mathrm {~F} ( x ) = \begin{cases} 0

x < 1

1 - \frac { 1 } { x ^ { 3 } }

x \geq 1 \end{cases} \end{aligned}\)

M1

A1

B1

[3]

1.1

Needs $+ c$ or definite integral between 1 and $x$, oe

Fully correct active part of CDF

" 0 for $x < 1$ " stated and no wrong ranges (doesn't need M1 or A1)

Allow $\leq$ for $<$, and/or $>$ for $\geq$

Wrong $k$ : can get M1A0B1

Ignore ranges here

Or "0 otherwise" if " $x \geq 1$ " stated in active part

4

(ii)

\(\begin{aligned}

\frac { \mathrm { P } [ ( X > 7 ) \cap ( X > 5 ) ] } { \mathrm { P } ( X > 5 ) } = \frac { \mathrm { P } ( X > 7 ) } { \mathrm { P } ( X > 5 ) }

= \frac { 1 - \mathrm { F } ( 7 ) } { 1 - \mathrm { F } ( 5 ) }

= \frac { 125 } { 343 } \text { or } 0.364 ( 431 \ldots ) \end{aligned}\)

M1* A1

*dep M1

A1ft [4]

3.1a 3.1a

3.3

1.1

Use cond ${ } ^ { 1 }$ prob method $\mathrm { P } [ ( X > 7 ) \cap ( X > 5 ) ] = \mathrm { P } ( X > 7 )$

Convert probabilities into $\mathrm { F } ( X )$, not using $\mathrm { P } ( X > 7 ) \times \mathrm { P } ( X > 5 )$

Any exact fraction or awrt 0.364 , ft on $1 - a / x ^ { 3 } , a \neq 0,1$

$\frac { [ 1 - \mathrm { F } ( 7 ) ] [ 1 - \mathrm { F } ( 5 ) ] } { 1 - \mathrm { F } ( 5 ) }$ : can get M1A0M0A0

Allow from $\mathrm { F } ( x ) = 1 - a / x ^ { 3 }$, otherwise www

Question

Answer

Mark

AO

Guidance

\multirow{5}{*}{4}

\multirow{5}{*}{(c)}

$\mathrm { E } \left( X ^ { 2 } \right) = \int _ { 1 } ^ { \infty } k x ^ { 2 - n } \mathrm {~d} x = \left[ \frac { k x ^ { 3 - n } } { ( 3 - n ) } \right] _ { 1 } ^ { \infty } ( n \neq 3 )$

If $n = 3 , \mathrm { E } \left( X ^ { 2 } \right) = \lim _ { x \rightarrow \infty } [ 2 \ln ( x ) ]$, not defined

M1* B1

2.1 1.1

Correct limits needed somewhere

Correct indefinite integral or $\frac { n - 1 } { n - 3 }$

SC: $\mathrm { E } \left( X ^ { 2 } \right) = \frac { n - 1 } { n - 3 }$, M1B1 $\mathrm { E } ( X ) = \frac { n - 1 } { n - 2 } \Rightarrow n \neq 2$ or 3 : (not valid, must consider ln if $n = 2$ or 3 ): B0

No marks just for this unless last 3 marks all zero, then if this (or for $n = 2$ ) is shown, award SC B1 Make deduction based on convergence, ft

Infinite integral does not converge if $3 - n \geq 0$

*dep M1

2.2a

No limits used: M0B1M0B0

If $n \geq 4$ then $\mathrm { E } ( X ) = \left[ \frac { k x ^ { 2 - n } } { ( 2 - n ) } \right] _ { 1 } ^ { \infty }$ converges

B1

2.3

Consider convergence of $\mathrm { E } ( X )$

SC: $\operatorname { Var } ( X ) < 0$ when $n < 3$ : M1B1M1 (B0) A0

Therefore $\operatorname { Var } ( X )$ is not defined if and only if $n = 2$ or 3 .

A1 [5]

2.2a

Shown not defined for $n = 2$ or 3 and only for those

But no need to state "if and only if"

OCR Further Statistics 2021 June Q1

8 marks

1 Jo can use either of two different routes, A or B, for her journey to school. She believes that route A has shorter journey times. She measures how long her journey takes for 17 journeys by route A and 12 journeys by route B. She ranks the 29 journeys in increasing order of time taken, and she finds that the sum of the ranks of the journeys by route B is 219 .

Test at the $10 \%$ significance level whether route A has shorter journey times than route B . [8]
State an assumption about the 29 journeys which is necessary for the conclusion of the test to be valid.

OCR Further Statistics 2021 June Q2

27 marks

2 The random variable $X$ is equally likely to take any of the $n$ integer values from $m + 1$ to $m + n$ inclusive. It is given that $\mathrm { E } ( 3 X ) = 30$ and $\operatorname { Var } ( 3 X ) = 36$. Determine the value of $m$ and the value of $n$. 326 cards are each labelled with a different letter of the alphabet, A to Z. The letters A, E, I, O and U are vowels.

Five cards are selected at random without replacement. Determine the probability that the letters on at least three of the cards are vowels.

All 26 cards are arranged in a line, in random order.

Show that the probability that all the vowels are next to one another is $\frac { 1 } { 2990 }$.

Determine the probability that three of the vowels are next to each other, and the other two vowels are next to each other, but the five vowels are not all next to each other. A biased spinner has five sides, numbered 1 to 5 . Elmer spins the spinner repeatedly and counts the number of spins, $X$, up to and including the first time that the number 2 appears. He carries out this experiment 100 times and records the frequency $f$ with which each value of $X$ is obtained. His results are shown in Table 1, together with the values of $x f$. \begin{table}[h] \end{table}

Question

Answer

Mark

AO

Guidance

\multirow[t]{8}{*}{1}

\multirow{8}{*}{(a)}

$\mathrm { H } _ { 0 } : m _ { A } = m _ { B } , \mathrm { H } _ { 1 } : m _ { A } < m _ { B }$ where $m _ { A }$ and $m _ { B }$

B1

1.1

OR: Median journey times equal, oe. Allow if $m$ s used but not defined

Allow "mean" or "average" only if "population" stated

М1

1.1

Find either $\mathrm { P } ( \geq 219 )$ (218.5) or $\mathrm { P } ( \leq 141 )$ (141.5)

Use of 0.9559 is M0 here. For CV method see below

Consider correct tail, either 219 or 141 ( $R _ { m } = 219 , m ( m + n + 1 ) - R _ { m } = 141$ ) $p = \Phi \left( \frac { 141.5 - 180 } { \sqrt { 510 } } \right) = 0.0441 \ldots$

BC

М1

1.1

0.0421, 0.0401, 0.470 (no/wrong cc, $\sqrt { }$ ): M1

$0.9559 > 0.9 :$ A1A1 (M1A1) $0.9559 > 0.1 :$ A1A0 M0A0

0.0441 < 0.1

A1ft

1.1

Explicit comparison. FT on

Alternative:

CV $180 - z \times \sqrt { 5 } 10$ 141 (141.5) used $z = 1.282 \quad ( \mathrm { CV } = 151.05,151.058 .$. $141.5 < 151.05 ( 85 ) \quad$ or $218.5 > 208.95$

М1 M1 A1 A1

Allow $\sqrt { }$ errors Stated or implied CV and cc correct e.g. 141 < 150.55

$180 + 1.282 \sqrt { } 510$ etc is M0 unless 219 (218.5) used, in which case give M2(A1A1) E.g. $219 > 209.45$

Reject $\mathrm { H } _ { 0 }$. Significant evidence that route B takes longer

M1ft A1ft [8]

1.1 2.2b

Correct first conclusion Contextualised, not too definite

Needs like-with-like, e.g. 0.9559 with 0.9

SC Sum of A's ranks $= 435 - 219 = 216$ used: B1B0 M0M1A0A1 M1A1 max 5/8

Exx:

$\alpha$ : $\quad \mathrm { H } _ { 0 }$ : Journey times are the same, $\mathrm { H } _ { 1 }$ : journey times for $B$ are higher:

$\beta$ : $\quad \mathrm { H } _ { 0 }$ : No evidence that median journey times are different, etc:

B0

1

(b)

Must be a random sample (of all journeys) Or distributions must be same shape (necessary assumption for Wilcoxon ranksum test!)

B1 [1]

3.5b

Or equivalent. Allow "(journeys) independent"

Not "representative".

2

\(\begin{aligned}

3 \mathrm { E } ( X ) = 30 \text { or } \mathrm { E } ( X ) = 10

9 \times \operatorname { Var } ( X ) = 36 \text { or } \operatorname { Var } ( X ) = 4 \end{aligned}\)

B1 B1

2.2a 2.2a

Used, stated or implied One of these, used, stated or implied

Question			Answer	Mark	AO	Guidance
\multirow{5}{*}{}	\multirow{5}{*}{}	\multirow{5}{*}{}	$\frac { 1 } { 1 } \left( n ^ { 2 } - 1 \right) = 4$	M1	2.2a	$n = 7$ only, no need for "reject -7"	\multirow[b]{2}{*}{Allow if $\mathrm { E } ( 3 X + m )$ used rather than $\mathrm { E } [ 3 ( X + m ) ]$}
			$\mathrm { E } ( X - m ) = \frac { 1 } { 2 } ( n + 1 )$	M1	3.1b	Use expectation of uniform, e.g. $2 m + n + 1 = 20$.
			Alternative: $\operatorname { Var } ( Y + m ) = \frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)$	М1 A1 М1		$n = 7$ only Use expectation of uniform, e.g. $2 m + n + 1 = 20$.	No need for "reject -7"
			$10 - m = 4$	М1	2.1	Validly derive single equation for $m$
			$m = 6$	A1 [7]	2.2a	$m = 6$ only	NB: $\operatorname { Var } = ( n - 1 ) ^ { 2 } / 12$ is from continuous uniform!
\multirow{4}{*}{3}	\multirow{4}{*}{(a)}	\multirow{4}{*}{}	${ } ^ { 5 } C _ { 3 } \times { } ^ { 21 } C _ { 2 } + { } ^ { 5 } C _ { 4 } \times { } ^ { 21 } C _ { 1 } + 1 \quad [ = 2100 + 105 + 1 ]$	M1dep	3.1b	Any correct pair of ${ } ^ { n } C _ { r }$ s multiplied	Or $1 - \mathrm { P } ( 0,1,2 ) = 1 - .9665$
				A1	1.1	All terms correct
			$\div { } ^ { 26 } C _ { 5 } [ = 65780 ]$	*M1	1.1
			$\frac { 1103 } { 32890 }$ or $0.0335 \ldots$	A1 [4]	3.2a	Awrt 0.0335 or any exact fraction	e.g. $\frac { 2206 } { 65780 }$ or $\frac { 264720 } { 7893600 }$
			Alternative: $\frac { 5 } { 26 } \times \frac { 4 } { 25 } \times \frac { 3 } { 24 } \times \frac { 2 } { 23 } \times \frac { 1 } { 22 }$	B1		Must have 5 oe, e.g. ${ } ^ { 5 } C _ { 1 }$

3

(b)

(i)

$\frac { 22 ! \times 5 ! } { 26 ! } \left( = \frac { 1 \times 2 \times 3 \times 4 \times 5 } { 23 \times 24 \times 25 \times 26 } = \frac { 120 } { 358800 } \right)$

M1 A1

1.1 2.1

Oe. Allow M1 for 21! instead of 22! Fully correct

$\frac { 1 \times 2 \times 3 \times 4 \times 5 } { 22 \times 23 \times 24 \times 25 \times 26 } :$ M1

Question

Answer

Mark

AO

Guidance

$= \frac { 1 } { 2990 } \quad$ AG

A1 [3]

2.2a

Correctly obtain AG using exact method

Allow even if no working after $22 ! \times 5 ! \div 26$ !

\multirow{8}{*}{3}

\multirow{8}{*}{(b)}

\multirow{8}{*}{(ii)}

22 fences: 22 for [VVV] $\times 21$ for [VV]

M1

3.1b

Correct strategy, allow ${ } ^ { 22 } C _ { 2 }$ for ${ } ^ { 22 } P _ { 2 }$

Consonants arranged in 21! ways

M1

1.1

At least one of these, no subtraction

$21 ! \times 3 ! \times 2 ! \div 26 !$ M0M1

$21 ! \times 3 ! \times 2 ! \times 22 \times 21$ : M2A0

Vowels arranged in 5! ways ( $= { } ^ { 5 } P _ { 3 } \times { } ^ { 2 } P _ { 2 }$ )

A1

2.1

Both correct

NB: ${ } ^ { 5 } C _ { 3 } \times 3 ! \times 2 ! = 5 !$

\(\begin{aligned}

\text { Product } \div 26 ! = \frac { 21 } { 2990 }

\left( = 2.832 \times 10 ^ { 24 } \div 4.0329 \times 10 ^ { 26 } \right) \end{aligned}\)

A1

[4]

3.2a

Allow from calculator but must be exact fraction

Alternative:

Treat 21 consonants, [VVV] and [VV] as 23

М1

3.1b

Correct strategy, allow $23 ! \times 2 ! \times 3 !$

(Must subtract $2 \times 1 / 2990$ as 23! method counts

A1

2.1

Correct $\left( 5 ! = { } ^ { 5 } P _ { 3 } \times { } ^ { 2 } P _ { 2 } = \right. \left. { } ^ { 5 } C _ { 3 } \times 2 ! \times 3 ! \right)$

M1 also for subtracting $1 \times$

[VVVVV] twice, once as [VVV][VV] and once as [VV][VVV])

(11/1495 is M1A1M1A0)

Answer is $\frac { 21 } { 2990 }$

A1

1.1

1/2990

Final answer, exact fraction

\begin{center} \begin{tabular}{ | l | l | l | l | l | l | }

OCR Further Statistics 2021 June Q4

7 marks

$\mathbf { 4 }$ & $\mathbf { ( a ) }$ & Geometric & M1 & 1.1 & Stated explicitly
\end{tabular} \end{center}

Question

Answer

Mark

AO

Guidance

Mean $= 400 \div 100 ( = 4 )$ and $p = 1 /$ mean

M1

2.4

Use mean (or P(1) etc) to deduce $p$ ("Determine", so justification needed for 0.25)

Needs to deduce $p$ in part (a), not defer it to (b)

\multirow{3}{*}{4}

\multirow[t]{3}{*}{(b)}

Probability is $0.75 ^ { 6 } ( = 0.1779785 \ldots )$

M1

3.3

SC Geo(0.2): $0.8 { } ^ { 6 }$ M1A0

Or: 0.177978 or 0.177979 or better seen, or $1 - [ \mathrm { P } ( 1 ) + \ldots + \mathrm { P } ( 6 ) ]$ with evidence, e.g. formula

M1

Allow ± 1 term

Expected frequency $=$ probability $\times 100 = 17.798$

A1 [2]

2.1

17.798 correctly obtained, with sufficient evidence, www

$100 - \Sigma$ (other frequencies): SC B1

\multirow{6}{*}{4}

\multirow{6}{*}{(c)}

Ho: data consistent with (geometric)

B1

1.1

Both, allow equivalents, but not "evidence that ...". 9.005 or 9.01

Compare their $\Sigma X ^ { 2 }$ with 11.07

E.g. $\mathrm { H } _ { 0 } : X \sim \operatorname { Geo } ( p )$ Allow Geo(0.25)

$\Sigma X ^ { 2 } = 9.005$

B1

1.1

$9.005 < 11.07 ( v = 5 )$

B1

1.1

Do not reject $\mathrm { H } _ { 0 }$.

M1ft

1.1

Correct first conclusion, ft on their 9.005 or on 12.59, needs like-with-like

Allow from comparison with 12.59 but nothing else

Insufficient evidence that a geometric distribution is not a good fit.

A1ft [5]

2.2b

Contextualised, not too definite (needs double negative) Don't penalise "Geo(0.25)"

Allow addition slip in $\Sigma X ^ { 2 }$ SC Geo(0.2): can get full marks if given data used, $\Sigma X ^ { 2 } = 4.54$ used gets B1B1B0M1A1

OCR Pure 1 2017 Specimen Q2

2 The points $\mathrm { A } , \mathrm { B }$ and C have position vectors $3 \mathbf { i } - 4 \mathbf { j } + 2 \mathbf { k } , - \mathbf { i } + 6 \mathbf { k }$ and $7 \mathbf { i } - 4 \mathbf { j } - 2 \mathbf { k }$ respectively. M is the midpoint of BC .

Show that the magnitude of $\overrightarrow { O M }$ is equal to $\sqrt { 17 }$. Point D is such that $\overrightarrow { B C } = \overrightarrow { A D }$.
Show that position vector of the point D is $11 \mathbf { i } - 8 \mathbf { j } - 6 \mathbf { k }$.

OCR Pure 1 2017 Specimen Q3

3 The diagram below shows the graph of $y = \mathrm { f } ( x )$.
\includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-05_755_1398_388_326}

On the diagram in the Printed Answer Booklet, draw the graph of $y = \mathrm { f } \left( \frac { 1 } { 2 } x \right)$.
On the diagram in the Printed Answer Booklet, draw the graph of $y = \mathrm { f } ( x - 2 ) + 1$.

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OCR Further Statistics 2021 June Q3

OCR Further Statistics 2021 June Q4

OCR Further Statistics 2021 June Q1

OCR Further Statistics 2021 June Q2

OCR Further Statistics 2021 June Q4

OCR Pure 1 2017 Specimen Q2

OCR Pure 1 2017 Specimen Q3

Book	A	B	C	D	E	F	G	H	I	J	K	L
\(x\)	0.95	0.65	0.70	0.90	0.55	1.40	1.50	0.50	1.15	0.35	0.20	0.35
\(y\)	6.06	7.00	2.00	5.87	4.00	5.36	7.19	2.50	3.00	8.29	1.37	2.00