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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
  1. \(\operatorname { Var } ( 3 D + 5 )\),
  2. \(\mathrm { E } ( 3 D + 5 )\),
  3. \(\mathrm { P } ( D > \mathrm { E } ( D ) )\).
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 schemeMeaning
dep*Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
caoCorrect answer only
оеOr equivalent
rotRounded or truncated
soiSeen or implied
wwwWithout wrong working
AGAnswer given
awrtAnything which rounds to
BCBy Calculator
DRThis question included the instruction: In this question you must show detailed reasoning.
\end{table}
QuestionAnswerMarkAOGuidance
1(a)0.8392...B1 [1]1.1Awrt 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
B1
[2]
1.1
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.
QuestionAnswerMarkAOGuidance
\multirow[t]{4}{*}{3}(a)\(\mathrm { H } _ { 0 } : \mu = 500 , \mathrm { H } _ { 1 } : \mu < 500\)B11.1One 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)
\(\begin{aligned}\bar { X } \sim \mathrm {~N} \left( 500 , \frac { 80 ^ { 2 } } { 40 } \right) = \mathrm { N } ( 500,160 ) \text { and } \bar { X }
\mathrm { P } ( \bar { X } < 473 ) = 0.01640 \text { or } z = - 2.13 ( 45 )
\text { or } \mathrm { CV } = 470.6 \end{aligned}\)М13.3\(p\) 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 \(473 > 470.6\)A11.1Compare \(p\) with 0.01 or \(z\) with -2.326, or 2.326 used in CVMust 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]
1.1Correct first conclusion, needs correct method and like-with-like, ft on test statistic if method correct Contextualised, not too definiteBut 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 randomB1 [1]1.2No 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]2.1Allow "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
QuestionAnswerMarkAOGuidance
4(a)\(9 \times \frac { 40 } { 9 } = 40\)B1 [1]1.140 or awrt 40.0 only
4(b)\(\frac { 1 - p } { p ^ { 2 } } = \frac { 40 } { 9 }\)М13.1bUse correct formula for varianceSC: insufficient working, \(\frac { 3 } { 8 }\) only: M0B1 for \(\frac { 3 } { 8 }\), then B0
\(\begin{aligned}\mathrm { E } ( D ) = 1 / p \quad \left[ = \frac { 8 } { 3 } \right]
\mathrm { E } ( 3 D + 5 ) = 3 \times \frac { 8 } { 3 } + 5 \quad [ = 13 ] \end{aligned}\)B1ft2.3- 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
QuestionAnswerMarkAOGuidance
\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 \(Q\) andB11.1Allow \(m\) 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 \(= 1 / 2 \times 54 \times 55 = 1485\)М11.1Find sum of ranks
\(R _ { m } = 1485 - 726 = 759 \quad\) [or 561]A11.1Correct value of \(R _ { m }\) seenAllow 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 A1Not 759 - or 726 - ...; not wrong tail for comparison, but allow ± Needs correct ccOr 561.5 > 547.4 Wrong \(z\)-value: M1A1ft B0
\(p > 0.025,2 p > 0.05 , z < 1.96\), or 1.96 used in CVB11.1Explicit correct comparisonNeeds 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 definiteft 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]
BookABCDEFGHIJKL
\(x\)0.950.650.700.900.551.401.500.501.150.350.200.35
\(y\)6.067.002.005.874.005.367.192.503.008.291.372.00
\captionsetup{labelformat=empty} \caption{Table 1}
\end{table} $$n = 12 , \Sigma x = 9.20 , \Sigma y = 54.64 , \Sigma x ^ { 2 } = 8.9950 , \Sigma y ^ { 2 } = 310.4572 , \Sigma x y = 46.0545$$
  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.
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 \(X\).
  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.
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 .
  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 \(n = 4\).
  2. 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 schemeMeaning
    dep*Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark
    caoCorrect answer only
    оеOr equivalent
    rotRounded or truncated
    soiSeen or implied
    wwwWithout wrong working
    AGAnswer given
    awrtAnything which rounds to
    BCBy Calculator
    DRThis question included the instruction: In this question you must show detailed reasoning.
    \end{table}
    Answer
    Answer
    MarkAO
    Guidance
    Guidance
    \multirow{2}{*}{}\multirow{2}{*}{}\multirow{2}{*}{}\multirow[t]{2}{*}{
    \(53.1 \pm 1.96 \sqrt { \frac { 30 } { 8 } }\)
    (49.30, 56.90)
    }
    M13.3Square root correct Awrt 1.96 used, can be implied\multirow{2}{*}{Allow e.g. (49.30, 56.9)}
    A1 [4]3.4Both, only these numbers (4 sf needed at least once)
    2(a)(i)The points do not lie very close to a straight lineB1 [1]1.1Or 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 2018B21.1 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 methodSC: if B0 allow B1 for any 3 of 8.85, 46.35, 8.8725, 241.7331, 43.153
    QuestionAnswerMarkAOGuidance
    \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]
    1.1Need "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 systemsCan 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
    QuestionAnswerMarkAOGuidance
    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)\(\begin{aligned}\int 3 x ^ { - 4 } \mathrm {~d} x = - \frac { 1 } { x ^ { 3 } } + c
    x = 1 , \mathrm {~F} ( x ) = 0 \text { so } c = 1 . \text { Hence } 1 - x ^ { - 3 }
    \mathrm {~F} ( x ) = \begin{cases} 0x < 1
    1 - \frac { 1 } { x ^ { 3 } }x \geq 1 \end{cases} \end{aligned}\)
    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
    QuestionAnswerMarkAOGuidance
    \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* B12.1 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 M12.2aNo limits used: M0B1M0B0
    If \(n \geq 4\) then \(\mathrm { E } ( X ) = \left[ \frac { k x ^ { 2 - n } } { ( 2 - n ) } \right] _ { 1 } ^ { \infty }\) convergesB12.3Consider 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]2.2aShown not defined for \(n = 2\) or 3 and only for thoseBut 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 .
  1. Test at the \(10 \%\) significance level whether route A has shorter journey times than route B . [8]
  2. 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.
  1. Five cards are selected at random without replacement. Determine the probability that the letters on at least three of the cards are vowels.
  2. All 26 cards are arranged in a line, in random order.
    1. Show that the probability that all the vowels are next to one another is \(\frac { 1 } { 2990 }\).
    2. 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}
      QuestionAnswerMarkAOGuidance
      \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 \(m _ { B }\)B11.1OR: Median journey times equal, oe. Allow if \(m\) s used but not definedAllow "mean" or "average" only if "population" stated
      М11.1Find 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
      М11.10.0421, 0.0401, 0.470 (no/wrong cc, \(\sqrt { }\) ): M1\(0.9559 > 0.9 :\) A1A1 (M1A1) \(0.9559 > 0.1 :\) A1A0 M0A0
      0.0441 < 0.1A1ft1.1Explicit 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 A1Allow \(\sqrt { }\) 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 longerM1ft A1ft [8]1.1 2.2bCorrect first conclusion Contextualised, not too definiteNeeds 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
      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]3.5bOr 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 B12.2a 2.2aUsed, stated or implied One of these, used, stated or implied
      QuestionAnswerMarkAOGuidance
      \multirow{5}{*}{}\multirow{5}{*}{}\multirow{5}{*}{}\(\frac { 1 } { 1 } \left( n ^ { 2 } - 1 \right) = 4\)M12.2a\(n = 7\) only, no need for "reject -7"\multirow[b]{2}{*}{Allow if \(\mathrm { E } ( 3 X + m )\) used rather than \(\mathrm { E } [ 3 ( X + m ) ]\)}
      \(\mathrm { E } ( X - m ) = \frac { 1 } { 2 } ( n + 1 )\)M13.1bUse 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"
      \(10 - m = 4\)М12.1Validly derive single equation for \(m\)
      \(m = 6\)A1 [7]2.2a\(m = 6\) onlyNB: \(\operatorname { Var } = ( n - 1 ) ^ { 2 } / 12\) is from continuous uniform!
      \multirow{4}{*}{3}\multirow{4}{*}{(a)}\multirow{4}{*}{}\({ } ^ { 5 } C _ { 3 } \times { } ^ { 21 } C _ { 2 } + { } ^ { 5 } C _ { 4 } \times { } ^ { 21 } C _ { 1 } + 1 \quad [ = 2100 + 105 + 1 ]\)M1dep3.1bAny correct pair of \({ } ^ { n } C _ { r }\) s multipliedOr \(1 - \mathrm { P } ( 0,1,2 ) = 1 - .9665\)
      A11.1All terms correct
      \(\div { } ^ { 26 } C _ { 5 } [ = 65780 ]\)*M11.1
      \(\frac { 1103 } { 32890 }\) or \(0.0335 \ldots\)A1 [4]3.2aAwrt 0.0335 or any exact fractione.g. \(\frac { 2206 } { 65780 }\) or \(\frac { 264720 } { 7893600 }\)
      Alternative: \(\frac { 5 } { 26 } \times \frac { 4 } { 25 } \times \frac { 3 } { 24 } \times \frac { 2 } { 23 } \times \frac { 1 } { 22 }\)B1Must 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 A11.1 2.1Oe. 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
      QuestionAnswerMarkAOGuidance
      \(= \frac { 1 } { 2990 } \quad\) AGA1 [3]2.2aCorrectly obtain AG using exact methodAllow 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]M13.1bCorrect strategy, allow \({ } ^ { 22 } C _ { 2 }\) for \({ } ^ { 22 } P _ { 2 }\)
      Consonants arranged in 21! waysM11.1At 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 ( \(= { } ^ { 5 } P _ { 3 } \times { } ^ { 2 } P _ { 2 }\) )A12.1Both correctNB: \({ } ^ { 5 } C _ { 3 } \times 3 ! \times 2 ! = 5 !\)
      \(\begin{aligned}\text { Product } \div 26 ! = \frac { 21 } { 2990 }
      \left( = 2.832 \times 10 ^ { 24 } \div 4.0329 \times 10 ^ { 26 } \right) \end{aligned}\)
      A1
      [4]
      3.2aAllow from calculator but must be exact fraction
      Alternative:
      Treat 21 consonants, [VVV] and [VV] as 23
      М13.1bCorrect strategy, allow \(23 ! \times 2 ! \times 3 !\)(Must subtract \(2 \times 1 / 2990\) as 23! method counts
      A12.1Correct \(\left( 5 ! = { } ^ { 5 } P _ { 3 } \times { } ^ { 2 } P _ { 2 } = \right. \left. { } ^ { 5 } C _ { 3 } \times 2 ! \times 3 ! \right)\)
      M1 also for subtracting \(1 \times\)
      [VVVVV] twice, once as [VVV][VV] and once as [VV][VVV])
      (11/1495 is M1A1M1A0)
      Answer is \(\frac { 21 } { 2990 }\)A11.1
      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}
QuestionAnswerMarkAOGuidance
Mean \(= 400 \div 100 ( = 4 )\) and \(p = 1 /\) meanM12.4Use 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 \(0.75 ^ { 6 } ( = 0.1779785 \ldots )\)M13.3SC 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. formulaM1Allow ± 1 term
Expected frequency \(=\) probability \(\times 100 = 17.798\)A1 [2]2.117.798 correctly obtained, with sufficient evidence, www\(100 - \Sigma\) (other frequencies): SC B1
\multirow{6}{*}{4}\multirow{6}{*}{(c)}Ho: data consistent with (geometric)B11.1
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)
\(\Sigma X ^ { 2 } = 9.005\)B11.1
\(9.005 < 11.07 ( v = 5 )\)B11.1
Do not reject \(\mathrm { H } _ { 0 }\).M1ft1.1Correct first conclusion, ft on their 9.005 or on 12.59, needs like-with-likeAllow from comparison with 12.59 but nothing else
Insufficient evidence that a geometric distribution is not a good fit.A1ft [5]2.2bContextualised, 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 .
  1. Show that the magnitude of \(\overrightarrow { O M }\) is equal to \(\sqrt { 17 }\). Point D is such that \(\overrightarrow { B C } = \overrightarrow { A D }\).
  2. 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}
  1. On the diagram in the Printed Answer Booklet, draw the graph of \(y = \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
  2. On the diagram in the Printed Answer Booklet, draw the graph of \(y = \mathrm { f } ( x - 2 ) + 1\).
OCR Pure 1 2017 Specimen Q4
4 The diagram shows a sector \(A O B\) of a circle with centre \(O\) and radius \(r \mathrm {~cm}\).
\includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-05_519_604_1617_335} The angle \(A O B\) is \(\theta\) radians. The arc length \(A B\) is 15 cm and the area of the sector is \(45 \mathrm {~cm} ^ { 2 }\).
  1. Find the values of \(r\) and \(\theta\).
  2. Find the area of the segment bounded by the arc \(A B\) and the chord \(A B\).
OCR Pure 1 2017 Specimen Q5
5 In this question you must show detailed reasoning.
Use logarithms to solve the equation \(3 ^ { 2 x + 1 } = 4 ^ { 100 }\), giving your answer correct to 3 significant figures.
OCR Pure 1 2017 Specimen Q7
7 Business A made a \(\pounds 5000\) profit during its first year.
In each subsequent year, the profit increased by \(\pounds 1500\) so that the profit was \(\pounds 6500\) during the second year, £8000 during the third year and so on. Business B made a \(\pounds 5000\) profit during its first year.
In each subsequent year, the profit was 90\% of the previous year's profit.
  1. Find an expression for the total profit made by business A during the first \(n\) years. Give your answer in its simplest form.
  2. Find an expression for the total profit made by business B during the first \(n\) years. Give your answer in its simplest form.
  3. Find how many years it will take for the total profit of business A to reach \(\pounds 385000\).
  4. Comment on the profits made by each business in the long term.
OCR Pure 1 2017 Specimen Q8
8
  1. Show that \(\frac { 2 \tan \theta } { 1 + \tan ^ { 2 } \theta } = \sin 2 \theta\).
  2. In this question you must show detailed reasoning. Solve \(\frac { 2 \tan \theta } { 1 + \tan ^ { 2 } \theta } = 3 \cos 2 \theta\) for \(0 \leq \theta \leq \pi\).
OCR Pure 1 2017 Specimen Q10
10 A curve has equation \(x = ( y + 5 ) \ln ( 2 y - 7 )\).
  1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of y .
  2. Find the gradient of the curve where it crosses the \(y\)-axis.
OCR Pure 1 2017 Specimen Q11
11 For all real values of \(x\), the functions f and g are defined by \(\mathrm { f } ( x ) = x ^ { 2 } + 8 a x + 4 a ^ { 2 }\) and \(\mathrm { g } ( x ) = 6 x - 2 a\), where \(a\) is a positive constant.
  1. Find \(\mathrm { fg } ( x )\). Determine the range of \(\mathrm { fg } ( x )\) in terms of \(a\).
  2. If \(\operatorname { fg } ( 2 ) = 144\), find the value of \(a\).
  3. Determine whether the function fg has an inverse.
OCR Pure 1 2017 Specimen Q14
85 marks
14 John wants to encourage more birds to come into the park near his house. Each day, starting on day 1 , he puts bird food out and then observes the birds for one hour. He records the maximum number of birds that he observes at any given moment in the park each day. He believes that his observations may be modelled by the following differential equation, where \(n\) is the maximum number of birds that he observed at any given moment on day \(t\). $$\frac { \mathrm { d } n } { \mathrm {~d} t } = 0.1 n \left( 1 - \frac { n } { 50 } \right)$$
  1. Show that the general solution to the differential equation can be written in the form $$n = \frac { 50 A } { \mathrm { e } ^ { - 0.1 t } + A } , \text { where } A \text { is an arbitrary positive constant. }$$
  2. Using his model, determine the maximum number of birds that John would expect to observe at any given moment in the long term.
  3. Write down one possible refinement of this model.
  4. Write down one way in which John's model is not appropriate. \section*{END OF QUESTION PAPER} {www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
    For queries or further information please contact the Copyright Team, The Triangle Building, Shaftesbury Road, Cambridge CB2 8EA.
    OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }\section*{...day June 20XX - Morning/Afternoon} A Level Mathematics A
    H240/01 Pure Mathematics \section*{SAMPLE MARK SCHEME}
    \includegraphics[max width=\textwidth, alt={}]{2c30c315-c666-47d6-b05e-da71cb4decdd-09_268_1241_1256_777}
    \section*{Text Instructions} \section*{1. Annotations and abbreviations} \section*{2. Subject-specific Marking Instructions for A Level 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 units. 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.
    j If in any case the scheme operates with considerable unfairness consult your Team Leader. PS = Problem Solving
    M = Modelling \section*{Summary of Updates} \section*{Accredited} \section*{A Level Mathematics A} \section*{H240/01 Pure Mathematics} Printed Answer Booklet \section*{Date - Morning/Afternoon} \section*{Time allowed: \(\mathbf { 2 }\) hours} \section*{You must have:}
    • Question Paper H240/01 (inserted)
    \section*{You may use:}
    • a scientific or graphical calculator
      \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-25_115_487_1052_1239}
      \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-25_302_1454_1293_328}
    \section*{INSTRUCTIONS}
    • The Question Paper will be found inside the Printed Answer Booklet.
    • Use black ink. HB pencil may be used for graphs and diagrams only.
    • Complete the boxes provided on the Printed Answer Booklet with your name, centre number and candidate number.
    • Answer all the questions.
    • Write your answer to each question in the space provided in the Printed Answer Booklet. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s).
    • Do not write in the bar codes.
    • You are permitted to use a scientific or graphical calculator in this paper.
    • Final answers should be given to a degree of accuracy appropriate to the context.
    • The acceleration due to gravity is denoted by \(g \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Unless otherwise instructed, when a numerical value is needed, use \(g = 9.8\).
    \section*{INFORMATION}
    • You are reminded of the need for clear presentation in your answers.
    • The Printed Answer Booklet consists of \(\mathbf { 1 6 }\) pages. The Question Paper consists of \(\mathbf { 8 }\) pages.
      \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-26_2287_1632_234_238}
      \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-27_2268_1582_182_238}
      \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-28_2283_1580_182_240}
      \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-29_2293_1589_176_236}
    7(a)
    7(b)
    7(c)
    7(d)
    8(a)
    8(b)
    \includegraphics[max width=\textwidth, alt={}]{2c30c315-c666-47d6-b05e-da71cb4decdd-32_2268_1632_180_238}
    9(d)
    10(a)
    10(b)
    11(c)
    12(a)
    12(b)
    \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-36_1552_1582_180_238}
    \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-37_2280_1582_180_238}
    \includegraphics[max width=\textwidth, alt={}, center]{2c30c315-c666-47d6-b05e-da71cb4decdd-38_2283_1586_180_236}
    14(b)
    \section*{DO NOT WRITE ON THIS PAGE} \section*{Copyright Information:} OCR is committed to seeking permission to reproduce all third-party content that it uses in the assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answerrelated information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements booklet. This is produced for each series of examinations and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, The Triangle Building, Shaftesbury Road, Cambridge CB2 8EA.
    OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge
OCR Stats 1 2017 Specimen Q1
1 Simplify fully.
  1. \(\sqrt { a ^ { 3 } } \times \sqrt { 16 a }\)
  2. \(\left( 4 b ^ { 6 } \right) ^ { \frac { 5 } { 2 } }\)
OCR Stats 1 2017 Specimen Q3
3 A publisher has to choose the price at which to sell a certain new book.
The total profit, \(\pounds t\), that the publisher will make depends on the price, \(\pounds p\).
He decides to use a model that includes the following assumptions.
  • If the price is low, many copies will be sold, but the profit on each copy sold will be small, and the total profit will be small.
  • If the price is high, the profit on each copy sold will be high, but few copies will be sold, and the total profit will be small.
The graphs below show two possible models. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cac31da4-f1ad-4c34-a47f-2bc68c2304f1-05_330_443_964_438} \captionsetup{labelformat=empty} \caption{Model A}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cac31da4-f1ad-4c34-a47f-2bc68c2304f1-05_330_450_962_1104} \captionsetup{labelformat=empty} \caption{Model B}
\end{figure}
  1. Explain how model A is inconsistent with one of the assumptions given above.
  2. Given that the equation of the curve in model B is quadratic, show that this equation is of the form \(t = k \left( 12 p - p ^ { 2 } \right)\), and find the value of the constant \(k\).
  3. The publisher needs to make a total profit of at least \(\pounds 6400\). Use the equation found in part (b) to find the range of values within which model B suggests that the price of the book must lie.
  4. Comment briefly on how realistic model B may be in the following cases.
    • \(p = 0\)
    • \(p = 12.1\)
OCR Stats 1 2017 Specimen Q4
4
  1. Express \(\frac { 1 } { ( x - 1 ) ( x + 2 ) }\) in partial fractions
  2. In this question you must show detailed reasoning. Hence find \(\int _ { 2 } ^ { 3 } \frac { 1 } { ( x - 1 ) ( x + 2 ) } \mathrm { d } x\).
    Give your answer in its simplest form.
OCR Stats 1 2017 Specimen Q5
5 The diagram shows the circle with centre O and radius 2 , and the parabola \(y = \frac { 1 } { \sqrt { 3 } } \left( 4 - x ^ { 2 } \right)\).
\includegraphics[max width=\textwidth, alt={}, center]{cac31da4-f1ad-4c34-a47f-2bc68c2304f1-06_850_916_996_352} The circle meets the parabola at points \(P\) and \(Q\), as shown in the diagram.
  1. Verify that the coordinates of \(Q\) are \(( 1 , \sqrt { 3 } )\).
  2. Find the exact area of the shaded region enclosed by the arc \(P Q\) of the circle and the parabola.
OCR Stats 1 2017 Specimen Q6
6 Helga invests \(\pounds 4000\) in a savings account.
After \(t\) days, her investment is worth \(\pounds y\).
The rate of increase of \(y\) is \(k y\), where \(k\) is a constant.
  1. Write down a differential equation in terms of \(t , y\) and \(k\).
  2. Solve your differential equation to find the value of Helga's investment after \(t\) days. Give your answer in terms of \(k\) and \(t\). It is given that \(k = \frac { 1 } { 365 } \ln \left( 1 + \frac { r } { 100 } \right)\) where \(r \%\) is the rate of interest per annum.
    During the first year the rate of interest is \(6 \%\) per annum.
  3. Find the value of Helga's investment after 90 days. After one year (365 days), the rate of interest drops to 5\% per annum.
  4. Find the total time that it will take for Helga's investment to double in value.
OCR Stats 1 2017 Specimen Q7
7
  1. The heights of English men aged 25 to 34 are normally distributed with mean 178 cm and standard deviation 8 cm .
    Three English men aged 25 to 34 are chosen at random.
    Find the probability that all three men have a height less than 194 cm .
  2. The diagram shows the distribution of heights of Scottish women aged 25 to 34 .
    \includegraphics[max width=\textwidth, alt={}, center]{cac31da4-f1ad-4c34-a47f-2bc68c2304f1-08_559_1389_855_412} The distribution is approximately normal. Use the diagram in the Printed Answer Booklet to estimate the standard deviation of these heights, explaining your method.
OCR Stats 1 2017 Specimen Q9
9 The diagram below shows some "Cycle to work" data taken from the 2001 and 2011 UK censuses. The diagram shows the percentages, by age group, of male and female workers in England and Wales, excluding London, who cycled to work in 2001 and 2011.
\includegraphics[max width=\textwidth, alt={}, center]{cac31da4-f1ad-4c34-a47f-2bc68c2304f1-10_899_1537_524_283} The following questions refer to the workers represented by the graphs in the diagram.
  1. A researcher is going to take a sample of men and a sample of women and ask them whether or not they cycle to work. Why would it be more important to stratify the sample of men? A research project followed a randomly chosen large sample of the group of male workers who were aged 30-34 in 2001.
  2. Does the diagram suggest that the proportion of this group who cycled to work has increased or decreased from 2001 to 2011?
    Justify your answer.
  3. Write down one assumption that you have to make about these workers in order to draw this conclusion.