Questions FS1 AS (62 questions)

1 The probability distribution for the discrete random variable \(W\) is given in the table.

1. Show that \(\operatorname { Var } ( W ) = 0.8571\).
2. Find \(\operatorname { Var } ( 3 W + 6 )\).

2 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by \(X\).

1. State a further assumption needed for \(X\) to be well modelled by a Poisson distribution. Assume now that \(X\) can be well modelled by the distribution \(\operatorname { Po } ( 0.7 )\).
2. Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws. The number of flaws in 100 m of FOC of a larger diameter has the distribution \(\mathrm { Po } ( 1.6 )\).
3. Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4 . Judith believes that mathematical ability and chess-playing ability are related. She asks 20 randomly chosen chess players, with known British Chess Federation (BCF) ratings \(X\), to take a mathematics aptitude test, with scores \(Y\). The results are summarised as follows. $$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$
4. Calculate the value of Pearson's product-moment correlation coefficient \(r\).
5. State an assumption needed to be able to carry out a significance test on the value of \(r\).
6. Assume now that the assumption in part (ii) is valid. Test at the \(5 \%\) significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
7. There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\text { ELO rating } = 8 \times \text { BCF rating } + 650$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.
8. Calculate the value of Pearson's product-moment correlation coefficient \(r\).
9. State an assumption needed to be able to carry out a significance test on the value of \(r\).
10. Assume now that the assumption in part (b) is valid. Test at the \(5 \%\) significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
11. There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by $$\mathrm { ELO } \text { rating } = 8 \# \mathrm { BCF } \text { rating } + 650 .$$ Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion. An environmentalist measures the mean concentration, \(c\) milligrams per litre, of a particular chemical in a group of rivers, and the mean mass, \(m\) pounds, of fish of a certain species found in those rivers. The results are given in the table. \end{table}

    Question			Answer	Marks	AO	Guidance
    1	(a)		\(\begin{aligned}	0.25 + 0.36 + x + x ^ { 2 } = 1
    	x ^ { 2 } + x - 0.39 = 0
    	x = 0.3 \text { (or } - 1.3 \text { ) }
    	x \text { cannot be negative }
    	\mathrm { E } ( W ) = 2.23
    	\mathrm { E } \left( W ^ { 2 } \right) = \Sigma w ^ { 2 } \mathrm { p } ( w ) \quad [ = 5.83 ]
    	\text { Subtract } [ \mathrm { E } ( W ) ] ^ { 2 } \text { to get } \mathbf { 0 . 8 5 7 1 } \end{aligned}\)	\(\begin{gathered} \text { M1 }
    \text { A1 }
    \text { A1 }
    \text { B1ft }
    \text { B1 }
    \text { M1 }
    \text { A1 }
    { [ 7 ] } \end{gathered}\)	3.1a 1.1b 1.1b 2.3 1.1b 1.1 2.1	Equation using \(\Sigma p = 1\) Correct simplified quadratic Correctly obtain \(x = 0.3\) Explicitly reject other solution 2.23 or exact equivalent only Use \(\Sigma w ^ { 2 } \mathrm { p } ( w )\) Correctly obtain given answer, www	Can be implied Method needed ft on their quadratic Allow for \(\mathrm { E } ( W ) ^ { 2 } = 4.9729\) Need 2.23 or 4.9729 and 5.83 or full numerical \(\Sigma w ^ { 2 } \mathrm { p } ( w )\)
    1	(b)		\(9 \times 0.8571 = 7.7139\)	B1 [1]	1.1b	Allow 7.71 or 7.714
    2	(a)		Flaws must occur at constant average rate (uniform rate)	B1 [1]	1.2	Context (e.g. "flaws") needed Extra answers, e.g. "singly": B0	Not "constant rate" or "average constant rate".
    2	(b)		\(\operatorname { Po(2.1)~or~ } e ^ { - \lambda } \frac { \lambda ^ { 3 } } { 3 ! }\)	M1 A1 [2]	1.1 1.1b	Po(2.1) stated or implied, or formula with \(\lambda = 2.1\) stated Awrt 0.189
    2	(c)		Po(3) \(1 - \mathrm { P } ( \leq 3 )\)	M1 M1 A1 [3]	1.1 1.1 1.1b	\(\operatorname { Po } ( 2 \times 0.7 + 1.6 )\) stated or implied Allow \(1 - \mathrm { P } ( \leq 4 ) = 0.1847\), or from wrong \(\lambda\) Awrt 0.353	Or all combinations \(\leq 3\) \(1 -\) above, not just \(= 3\)

    Question			Answer	Marks	AO	Guidance
    3	(a)		0.4(00)	B2 [2]	1.1 1.1b	SC: if B0, give SC B1 for two of \(S _ { x x } = 12500 , S _ { y y } = 1600 , S _ { x y } = 1790\) and \(S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)\)	Also allow SC B1 for equivalent methods using Covariance \	SDs
    3	(b)		Data needs to have a bivariate normal distribution	B1 [1]	1.2	Needs "bivariate normal" or clear equivalent. Not just "both normally distributed"	Allow "scatter diagram forms ellipse"
    3	(c)		\(\mathrm { H } _ { 0 }\) : higher maths scores are not associated with higher BCF grading; \(\mathrm { H } _ { 1 }\) : positively associated CV 0.3783 \(0.400 > 0.3783\) so reject \(\mathrm { H } _ { 0 }\) Significant evidence that higher maths scores are associated with higher BCF grading	B1 B1 M1ft A1ft [4]	2.5 1.1b 2.2b 3.5a	Needs context and clearly onetailed \(O R \rho\) used and defined Not "evidence that ..." Allow 0.378 Reject/do not reject \(\mathrm { H } _ { 0 }\) Contextualised, not too definite Needn't say "positive" if \(\mathrm { H } _ { 1 } \mathrm { OK }\) SC 2-tail: B0; 0.4438, or 0.3783 B1; then M1A0	\(\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0\) where \(\rho\) is population pmcc (not \(r\) ) FT on their \(r\), but not CV Not "scores are associated ...". FT on their \(r\) only
    3	(d)		It makes no difference as this is a linear transformation	B1 [1]	2.2a	Need both "unchanged" oe and reason, need "linear" or exact equivalent	"oe" includes "their 0.4"
    4	(a)		Neither	B1 [1]	2.5	OE	Not "neither is independent of the other"
    4	(b)		\(c = 2.848 - 0.1567 m \quad \mathbf { B C }\)	B1 B1 B1 [3]	1.1 1.1 1.1	Correct \(a\), awrt 2.85 Correct \(b\), awrt 0.157 Letters correct from correct method (If both wrongly rounded, e.g. \(c = 2.84 - 0.156 m\), give B2)	\(\mathrm { SC } : m\) on \(c\) : \(m = 15.65 - 4.832 c\) : B2 \(y = 15.65 - 4.832 x\) : B1 \(c = 15.65 - 4.832 m : \mathrm { B } 1\) If B0B0, give B1 for correct letters from valid working

    Question		Answer	Marks	AO	Guidance
    4	(c)	\(a\) unchanged, \(b\) multiplied by 2.2 (allow " \(a\) unchanged, \(b\) increases", etc)	B1 [1]	2.2a	oe, e.g. \(c = 2.848 - 0.345 m\); \(m = 7.114 - 2.196 c\)	SC: \(m\) on \(c\) in (b): Both divided by 2.2 B1
    4	(d)	Draw approximate line of best fit Draw at least one vertical from line to point Say that "Best fit" line minimises the sum of squares of these distances	M1 M1 A1 [3]	1.1 2.4 2.4	Needs M2 and "minimises" and "sums of squares" oe	SC: Horizontal(s): full marks (indept of (b))

1 On any day, the number of orders received in one randomly chosen hour by an online supplier can be modelled by the distribution \(\mathrm { Po } ( 120 )\).

1. Find the probability that at least 28 orders are received in a randomly chosen 10 -minute period.
2. Find the probability that in a randomly chosen 10-minute period on one day and a randomly chosen 10-minute period on the next day a total of at least 56 orders are received.
3. State a necessary assumption for the validity of your calculation in part (b).

2 The members of a team stand in a random order in a straight line for a photograph. There are four men and six women.

1. Find the probability that all the men are next to each other.
2. Find the probability that no two men are next to one another.

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

1. 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.
2. 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 \(p\).
3. 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.
4. Assume that \(p = 0.3\). Find
   1. \(\mathrm { P } ( X \geqslant 5 )\),
   2. \(\operatorname { Var } ( X )\).
5. 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	AO	Guidance
   1	(a)		\(1 - \mathrm { P } ( \leq 27 )\) \(= 0.0525 \quad \mathbf { B C }\)	M1 A1 [2]	3.4 1.1	Allow M1 for \(1 - 0.9657 = 0.0343\) In range [0.0524, 0.0525] BC
   1	(b)		Po(40) \(1 - \mathrm { P } ( \leq 55 )\) \(= 0.00968\)	M1* depM1 A1 [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
   1	(c)		Orders on one day are independent of orders on the other	B1 [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}{*}{}	\(7 ! \times 4\) ! \(\div 10\) ! \(= \frac { 120960 } { 3628800 } = \frac { 1 } { 30 }\)	M1 M1 A1	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: \(7 \times \frac { 6 } { 10 } \times \frac { 4 } { 9 } \times \frac { 5 } { 8 } \times \frac { 3 } { 7 } \times \frac { 4 } { 6 } \times \frac { 2 } { 5 } \times \frac { 3 } { 4 } \times \frac { 1 } { 3 } \times \frac { 2 } { 2 } \times \frac { 1 } { 1 }\)	M1 M1 A1		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 \(m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m = { } ^ { 7 } C _ { 4 }\) \({ } ^ { 7 } C _ { 4 } \times \frac { 6 ! \times 4 ! } { 10 ! }\) \(= \frac { 1 } { 6 }\)	B1 М1 М1 A1	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 \(( 10 ! - 12 \times 9 ! + ( 3 \times 4 \times 8 ! + 12 \times 2 \times 8 ! ) - 24 \times 7 ! ) / 10 !\) \([ = ( 3628800 - 4354560 + 1451520 - 120960 ) / 10 ! ]\)	M2 A1 A1		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	AO	Guidance
   3	(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	B1 B1 M1ft A1 [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 ..."
   3	(b)	(i)	-0.534	B2 [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
   3	(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	B1 [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
   4	(a)		\(10 p ( 1 - p )\)	B1 [1]	1.2	Allow \(10 p q\) oe, e.g. \(10 p - 10 p ^ { 2 }\)	Not just \(n p ( 1 - p )\)
   4	(b)	(i)	\(0.7 ^ { 4 }\) = 0.240(1)	M1 A1 [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]
   4	(b)	(ii)	\(q / p ^ { 2 } = \frac { 70 } { 9 }\) or \(7.777 \ldots\)	B1 [1]	1.1	Allow 7.78, 7.778, etc	Allow 8 only if evidence, e.g. ( \(1 - 0.3\) )/ \(0.3 ^ { 2 }\)
   4	(c)		\(\begin{aligned}	( 1 - p ) ^ { 2 } p = \frac { 4 } { 25 } p
   	p = 0 \text { or } ( 1 - p ) ^ { 2 } = \frac { 4 } { 25 } \quad ( p \neq 0 )
   	( 1 - p ) = \pm \frac { 2 } { 5 }
   	p \neq \frac { 7 } { 5 }
   	p = \frac { 3 } { 5 } \end{aligned}\)	B1 M1 M1 B1ft A1 [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

1 Five observations of bivariate data \(( x , y )\) are given in the table.

1. Find the value of Pearson's product-moment correlation coefficient.
2. State what your answer to part (a) tells you about a scatter diagram representing the data.
3. A new variable \(a\) is defined by \(a = 3 x + 4\). Dee says "The value of Pearson's product-moment correlation coefficient between \(a\) and \(y\) will not be the same as the answer to part (a)." State with a reason whether you agree with Dee. An investor obtains data about the profits of 8 randomly chosen investment accounts over two one-year periods. The profit in the first year for each account is \(p \%\) and the profit in the second year for each account is \(q \%\). The results are shown in the table and in the scatter diagram.

   Account	A	B	C	D	E	F	G	H
   \(p\)	1.6	2.1	2.4	2.7	2.8	3.3	5.2	8.4
   \(q\)	1.6	2.3	2.2	2.2	3.1	2.9	7.6	4.8

   \(n = 8 \quad \Sigma p = 28.5 \quad \Sigma q = 26.7 \quad \Sigma p ^ { 2 } = 136.35 \quad \Sigma q ^ { 2 } = 116.35 \quad \Sigma p q = 116.70\)
   \includegraphics[max width=\textwidth, alt={}, center]{4c7546b9-03ee-47a1-915f-41e2b4ca19c0-03_762_1248_906_260}
4. State which, if either, of the variables \(p\) and \(q\) is independent.
5. Calculate the equation of the regression line of \(q\) on \(p\).
6. 1. Use the regression line to estimate the value of \(q\) for an investment account for which \(p = 2.5\).
   2. Give two reasons why this estimate could be considered reliable.
7. Comment on the reliability of using the regression line to predict the value of \(q\) when \(p = 7.0\).

3 At a cinema there are three film sessions each Saturday, "early", "middle" and "late". The numbers of the audience, in different age groups, at the three showings on a randomly chosen Saturday are given in Table 1. \begin{table}[h] \end{table}

1 Every time a spinner is spun, the probability that it shows the number 4 is 0.2 , independently of all other spins.

1. A pupil spins the spinner repeatedly until it shows the number 4 . Find the mean of the number of spins required.
2. Calculate the probability that the number of spins required is between 3 and 10 inclusive.
3. Each pupil in a class of 30 spins the spinner until it shows the number 4 . Out of the 30 pupils, the number of pupils who require at least 10 spins is denoted by \(X\). Determine the variance of \(X\).

2 After a holiday organised for a group, the company organising the holiday obtained scores out of 10 for six different aspects of the holiday. The company obtained responses from 100 couples and 100 single travellers. The total scores for each of the aspects are given in the following table. \end{table}

4

1. Four men and four women stand in a random order in a straight line. Determine the probability that no one is standing next to a person of the same gender.
2. \(x\) men, including Mr Adam, and \(x\) women, including Mrs Adam, are arranged at random in a straight line. Show that the probability that Mr Adam is standing next to Mrs Adam is \(\frac { 1 } { X }\).
3. The random variable \(X\) has the distribution \(\operatorname { Geo } ( 0.6 )\).
   (a) Find \(\mathrm { P } ( X \geq 8 )\).
   (b) Find the value of \(\mathrm { E } ( X )\).
   (c) Find the value of \(\operatorname { Var } ( X )\).
4. The random variable \(Y\) has the distribution \(\operatorname { Geo } ( p )\). It is given that \(\mathrm { P } ( Y < 4 ) = 0.986\) correct to 3 significant figures. Use an algebraic method to find the value of \(p\). Sabrina counts the number of cars passing her house during randomly chosen one minute intervals. Two assumptions are needed for the number of cars passing her house in a fixed time interval to be well modelled by a Poisson distribution.
5. State these two assumptions.
6. For each assumption in part (i) give a reason why it might not be a reasonable assumption for this context. Assume now that the number of cars that pass Sabrina's house in one minute can be well modelled by the distribution \(\operatorname { Po } ( 0.8 )\).
7. (a) Write down an expression for the probability that, in a given one minute period, exactly \(r\) cars pass Sabrina's house.
   (b) Hence find the probability that, in a given one minute period, exactly 2 cars pass Sabrina's house.
8. Find the probability that, in a given 30 minute period, at least 28 cars pass Sabrina's house.
9. The number of bicycles that pass Sabrina's house in a 5 minute period is a random variable with the distribution \(\operatorname { Po } ( 1.5 )\). Find the probability that, in a given 10 minute period, the total number of cars and bicycles that pass Sabrina's house is between 12 and 15 inclusive. State a necessary condition. 7 The discrete random variable \(X\) is equally likely to take values 0,1 and 2 . \(3 N\) observations of \(X\) are obtained, and the observed frequencies corresponding to \(X = 0 , X = 1\) and \(X = 2\) are given in the following table. \section*{2. Subject-specific Marking Instructions for AS Level Further Mathematics A} Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. 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 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. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner.
   If you are in any doubt whatsoever you should contact your Team Leader.
   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. \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} Mark for explaining a result or establishing a given result. 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.
   d 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.
   e 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. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner.
   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'. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.
   f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km , when this would be assumed to be the unspecified unit.) 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. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for \(g\). E marks will be lost except when results agree to the accuracy required in the question. Rules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.
   h 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 papers. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. 'Fresh starts' will not affect an earlier decision about a misread. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. If in any case the scheme operates with considerable unfairness consult your Team Leader. \end{table} PS = Problem Solving
   M = Modelling \section*{Summary of Updates}

   5(i)(a)
   5(i)(b)
   5(i)(c)
   5(ii)

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7 The discrete random variable \(X\) is equally likely to take values 0,1 and 2 . \(3 N\) observations of \(X\) are obtained, and the observed frequencies corresponding to \(X = 0 , X = 1\) and \(X = 2\) are given in the following table. The test statistic for a chi-squared goodness of fit test for the data is 0.3 . Find the value of \(N\).

8 The following table gives the mean per capita consumption of mozzarella cheese per annum, \(x\) pounds, and the number of civil engineering doctorates awarded, \(y\), in the United States in each of 10 years. \section*{2. Subject-specific Marking Instructions for AS Level Further Mathematics A} Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded. 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 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. When a candidate adopts a method which does not correspond to the mark scheme, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner.
If you are in any doubt whatsoever you should contact your Team Leader.
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. \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} Mark for explaining a result or establishing a given result. 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.
d 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.
e 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. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner.
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'. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.
f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km , when this would be assumed to be the unspecified unit.) 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. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for \(g\). E marks will be lost except when results agree to the accuracy required in the question. Rules for replaced work: if a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests; if there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.
h 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 papers. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. 'Fresh starts' will not affect an earlier decision about a misread. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. If in any case the scheme operates with considerable unfairness consult your Team Leader. \end{table} PS = Problem Solving
M = Modelling \section*{Summary of Updates} \section*{PLEASE DO NOT WRITE ON THIS PAGE}