OCR Further Statistics (Further Statistics) 2021 June

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.

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

1. 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.
2. 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.

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 \(X\).

1. 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 \(X\).
2. 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 \(\operatorname { Po } ( 7.2 )\).
3. 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.
4. State an assumption needed for your answer to part (c) to be valid.
5. Give a reason why the assumption in part (d) may not be valid in practice.

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 .

1. Find \(k\) in terms of \(n\).
2. 1. When \(n = 4\), find the cumulative distribution function of \(X\).
   2. Hence determine \(\mathrm { P } ( X > 7 \mid X > 5 )\) when \(n = 4\).
3. 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 Answer	Mark	AO	Guidance Guidance
   \multirow{2}{*}{}	\multirow{2}{*}{}	\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{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 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 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 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 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 1.1 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 1.1 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"