OCR FS1 AS 2021 June - A-Level Maths

Question 3 28 marks

3 Sixteen candidates took an examination paper in mechanics and an examination paper in statistics.

For all sixteen candidates, the value of the product moment correlation coefficient \(r\) for the marks on the two papers was 0.701 correct to 3 significant figures. Test whether there is evidence, at the \(5 \%\) significance level, of association between the marks on the two papers.
A teacher decided to omit the marks of the candidates who were in the top three places in mechanics and the candidates who were in the bottom three places in mechanics. The marks for the remaining 10 candidates can be summarised by
\(n = 10 , \Sigma x = 750 , \Sigma y = 690 , \Sigma x ^ { 2 } = 57690 , \Sigma y ^ { 2 } = 49676 , \Sigma x y = 50829\).
1. Calculate the value of \(r\) for these 10 candidates.
2. What do the two values of \(r\), in parts (a) and (b)(i), tell you about the scores of the sixteen candidates? A bag contains a mixture of blue and green beads, in unknown proportions. The proportion of green beads in the bag is denoted by \(p\).
Sasha selects 10 beads at random, with replacement. Write down an expression, in terms of \(p\), for the variance of the number of green beads Sasha selects. Freda selects one bead at random from the bag, notes its colour, and replaces it in the bag. She continues to select beads in this way until a green bead is selected. The first green bead is the \(X\) th bead that Freda selects.
Assume that \(p = 0.3\). Find
1. \(\mathrm { P } ( X \geqslant 5 )\),
2. \(\operatorname { Var } ( X )\).

In fact, on the basis of a large number of observations of \(X\), it is found that \(\mathrm { P } ( X = 3 ) = \frac { 4 } { 25 } \times \mathrm { P } ( X = 1 )\). Estimate the value of \(p\). \section*{Total Marks for Question Set 4: 29} \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. 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. 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'. 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 \(\mathbf { 3 ~ s . f . }\) 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. 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
oe	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

Guidance

(a)

\(1 - \mathrm { P } ( \leq 27 )\)

\(= 0.0525 \quad \mathbf { B C }\)

[2]

3.4

1.1

Allow M1 for \(1 - 0.9657 = 0.0343\)

In range [0.0524, 0.0525] BC

(b)

Po(40)

\(1 - \mathrm { P } ( \leq 55 )\)

\(= 0.00968\)

M1*

depM1

[3]

3.1b

1.1a

1.1

\(\operatorname { Po( } 2 \times\) their 20) stated or implied

Allow M1 for \(1 - \mathrm { P } ( \leq 56 ) = 0.00658\) or \(1 - 0.990 = 0.01\) or 0.0097

Awrt 0.00968

(c)

Orders on one day are independent of orders on the other

[1]

3.2b

Use "orders independent", clearly referred to the two different days, needs context [not "events"], and nothing else

Not anything affecting given separate Poissons, such as "orders must be independent" or "constant average rate".

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

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

\multirow{3}{*}{}

\(7 ! \times 4\) !

\(\div 10\) ! \(= \frac { 120960 } { 3628800 } = \frac { 1 } { 30 }\)

2.1

1.1a

1.1

Allow for \(6 ! \times 4\) ! or \(6 ! \times 4 ! \times 2\)

Divide by 10!, needs at least one factorial in numerator

Answer, exact or awrt 0.0333

3/3 for \(\frac { 1 } { 30 }\) www

Alternative: \(7 \times \frac { 6 } { 10 } \times \frac { 4 } { 9 } \times \frac { 5 } { 8 } \times \frac { 3 } { 7 } \times \frac { 4 } { 6 } \times \frac { 2 } { 5 } \times \frac { 3 } { 4 } \times \frac { 1 } { 3 } \times \frac { 2 } { 2 } \times \frac { 1 } { 1 }\)

no 7, one other error

only one error

correct answer

[3]

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

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

\multirow{3}{*}{}

Women placed in 6 ! ways, men in 4 ! \([ = 720 \times 24 ]\)

4 slots \(m\) in \(m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m = { } ^ { 7 } C _ { 4 }\) \({ } ^ { 7 } C _ { 4 } \times \frac { 6 ! \times 4 ! } { 10 ! }\)

\(= \frac { 1 } { 6 }\)

М1

2.1

3.1b

1.1a

1.1

\(6 ! \times 4 !\) anywhere, or \(6 ! \times\) attempt at \({ } ^ { 7 } P _ { 4 }\)

Or \({ } ^ { 7 } P _ { 4 }\). Allow for \(m\) and \(W\) reversed

Needs attempt at both terms

Or 0.167 or 0.1667 etc

Or \(6 ! \times 7 \times 6 \times 5 \times 4 \quad\) B2

\({ } ^ { 7 } P _ { 4 } \times 6\) !: B1M1

\(4 \times ( 6 ! \times 4 ! ) / 10 ! = 2 / 105 :\) В 1 М 1

Alternative: PIE

\(( 10 ! - 12 \times 9 ! + ( 3 \times 4 \times 8 ! + 12 \times 2 \times 8 ! ) - 24 \times 7 ! ) / 10 !\)

\([ = ( 3628800 - 4354560 + 1451520 - 120960 ) / 10 ! ]\)

Signs alternating, at least one term \(\sqrt { }\) Allow one term omitted or wrong Correct answer

[3]

Three together: \(7 \times 6 \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 5 }\)

Two pairs: \(\frac { 7 \times 6 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! }\)

\(\frac { 1 } { 10 }\)

One pair: \(7 \times \frac { 6 \times 5 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 2 }\)

Question

Answer

Marks

Guidance

(a)

\(\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho \neq 0\), where \(\rho\) is population pmcc

0.701 > 0.4973

Reject \(\mathrm { H } _ { 0 }\). There is significant evidence of association between the marks on the two papers

M1ft

[4]

2.5

1.1a

1.1

2.2b

Must use symbols. Allow no definition of letter if \(\rho\) used

Correct CV stated, allow 0.497

FT on wrong CV

Not FT. Needs context, and not too definite.

Not " \(\mathrm { H } _ { 0 }\) : there is no assoc' n , \(\mathrm { H } _ { 1 }\) : there is association"

Not There is association ..."

(b)

(i)

-0.534

[2]

1.1a

1.1

SC: if B0, give B1 for two of 1440, 2066, -921 and \(S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)\)

-0.53: B1

(b)

(ii)

6 candidates did very well or very badly on both papers; middle 10 tended to do badly on one paper and well on the other

[1]

2.4

Correct inference about scores oe, not "correlation/association/value of \(r\) ". Not "outliers" or "anomalies".

Allow inference for one group only, provided it is clearly for only one group \

any ref to other group is not wrong

(a)

\(10 p ( 1 - p )\)

[1]

1.2

Allow \(10 p q\) oe, e.g. \(10 p - 10 p ^ { 2 }\)

Not just \(n p ( 1 - p )\)

(b)

(i)

\(0.7 ^ { 4 }\)

= 0.240(1)

[2]

1.1a

1.1

\(0.7 ^ { 5 } = 0.168\) or \(0.7 ^ { 6 } = 0.118\) : M1

Allow 0.24

Or \(1 - 0.3 \left( 1 + 0.7 + 0.7 ^ { 2 } + 0.7 ^ { 3 } \right)\) Allow M1 if also \(0.3 \times 0.7 ^ { 4 }\) [0.15 is from binomial]

(b)

(ii)

\(q / p ^ { 2 } = \frac { 70 } { 9 }\) or \(7.777 \ldots\)

[1]

1.1

Allow 7.78, 7.778, etc

Allow 8 only if evidence, e.g. ( \(1 - 0.3\) )/ \(0.3 ^ { 2 }\)

(c)

\(\begin{aligned}

( 1 - p ) ^ { 2 } p = \frac { 4 } { 25 } p

p = 0 \text { or } ( 1 - p ) ^ { 2 } = \frac { 4 } { 25 } \quad ( p \neq 0 )

( 1 - p ) = \pm \frac { 2 } { 5 }

p \neq \frac { 7 } { 5 }

p = \frac { 3 } { 5 } \end{aligned}\)

B1ft

[5]

1.1

1.1a

1.1

2.3

2.1

Correct equation

Reduce to quadratic/cubic and solve

Obtain two non-zero solutions

Explicitly discard one solution, either here or in line 2 (not enough to give 2 answers and then only 1 )

Exact final answer exact (0.6) no others left, allow from ± omitted

e.g. \(p \left( p ^ { 2 } - 2 p + \frac { 21 } { 25 } \right) = 0\) ± omitted: M0B0A1
Allow " \(p = 0 , \frac { 3 } { 5 } , \frac { 7 } { 5 }\) but \(p \leq 1\) "
SC binomial: B0 then \(75 p ^ { 2 } = ( 1 - p ) ^ { 2 } \	\) solve M1 \(0.104 [ 0.1035 ] \quad\) A1 Explicitly reject 0 or - 0.13 B1 SC Poisson: 0

OCR FS1 AS (Further Statistics 1 AS) 2021 June

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Question 3 28 marks