Questions - OCR FS1 AS - A-Level Maths

OCR FS1 AS 2021 June Q1

1 A book reviewer estimates that the probability that he receives a delivery of books to review on any one weekday is 0.1 . The first weekday in September on which he receives a delivery of books to review is the $X$ th weekday of September.

State an assumption needed for $X$ to be well modelled by a geometric distribution.
Find $\mathrm { P } ( X = 11 )$.
Find $\mathrm { P } ( X \leqslant 8 )$.
Find $\operatorname { Var } ( X )$.
Give a reason why a geometric distribution might not be an appropriate model for the first weekday in a calendar year on which the reviewer receives a delivery of books to review.

OCR FS1 AS 2021 June Q2

2 The probability distribution for the discrete random variable $W$ is given in the table.

$w$	1	2	3	4
$\mathrm { P } ( W = w )$	0.25	0.36	$x$	$x ^ { 2 }$

Show that $\operatorname { Var } ( W ) = 0.8571$.
Find $\operatorname { Var } ( 3 W + 6 )$. In this question you must show detailed reasoning.
The random variable $T$ has a binomial distribution. It is known that $\mathrm { E } ( T ) = 5.625$ and the standard deviation of $T$ is 1.875 . Find the values of the parameters of the distribution.

OCR FS1 AS 2021 June Q4

30 marks

4 The table shows the results of a random sample drawn from a population which is thought to have the distribution $\mathrm { U } ( 20 )$. \end{table}

OCR FS1 AS 2017 December Q1

1 Bill and Gill send letters to potential sponsors of a show. On past experience, they know that $5 \%$ of letters receive a favourable reply.

Bill sends a letter to each of 40 potential sponsors. Assuming that the number $N$ of favourable responses can be modelled by a binomial distribution, find the mean and variance of $N$.
Gill sends one letter at a time to potential sponsors. $L$ is the number of letters she sends, up to and including the first letter that receives a favourable response.
(a) State two assumptions needed for $L$ to be well modelled by a geometric distribution.
(b) Using the assumptions in part (ii)(a), find the smallest number of letters that Gill has to send in order to have at least a $90 \%$ chance of receiving at least one favourable reply.

OCR FS1 AS 2017 December Q2

2 Each letter of the words NEW COURSE is written on a card (including one blank card, representing the space between the words), so that there are 10 cards altogether.

All 10 cards are arranged in a random order in a straight line. Find the probability that the two cards containing an E are next to each other.
4 cards are chosen at random. Find the probability that at least three consonants ( $\mathrm { N } , \mathrm { W } , \mathrm { C } , \mathrm { R } , \mathrm { S }$ ) are on the cards chosen.

OCR FS1 AS 2017 December Q3

3 Over a long period Jenny counts the number of trolleys used at her local supermarket between 10 am and 10.20 am each day. She finds that the mean number of trolleys used between these times on a weekday is 40.00. You should assume that the use of trolleys occurs randomly, independently of one another, and at a constant average rate.

Calculate the probability that, on a randomly chosen weekday, the number of trolleys used between these times is between 32 and 50 inclusive.
Write down an expression for the probability that, on a randomly chosen weekday, exactly 5 trolleys are used during a time period of $t$ minutes between 10 am and 10.20 am. Jenny carries out this process for seven consecutive days. She finds that the mean number of trolleys used between 10 am and 10.20 am is 35.14 and the variance is 91.55 .
Explain why this suggests that the distribution of the number of trolleys used between these times on these seven consecutive days is not well modelled by a Poisson distribution.
Give a reason why it might not be appropriate to apply the Poisson model to the total number of trolleys used between these times on seven consecutive days.

OCR FS1 AS 2017 December Q4

4 The discrete random variable $X$ has the distribution $\mathrm { U } ( n )$.

Use the results $\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$ and $\mathrm { E } ( X ) = \frac { n + 1 } { 2 }$ to show that $\operatorname { Var } ( X ) = \frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)$. It is given that $\mathrm { E } ( X ) = 13$.
Find the value of $n$.
Find $\mathrm { P } ( X < 7.5 )$. It is given that $\mathrm { E } ( a X + b ) = 10$ and $\operatorname { Var } ( a X + b ) = 117$, where $a$ and $b$ are positive.
Calculate the value of $a$ and the value of $b$.

OCR FS1 AS 2017 December Q5

5 A shop manager recorded the maximum daytime temperature $T ^ { \circ } \mathrm { C }$ and the number $C$ of ice creams sold on 9 summer days. The results are given in the table and illustrated in the scatter diagram.

$T$	17	21	25	26	27	27	29	30	30
$C$	21	16	20	38	32	37	35	39	42

\includegraphics[max width=\textwidth, alt={}]{64d7ed6d-fadd-4c59-afb0-97d1788ba369-3_661_1189_1320_431}

$$n = 9 , \Sigma t = 232 , \Sigma c = 280 , \Sigma t ^ { 2 } = 6130 , \Sigma c ^ { 2 } = 9444 , \Sigma t c = 7489$$

State, with a reason, whether one of the variables $C$ or $T$ is likely to be dependent upon the other.
Calculate Pearson's product-moment correlation coefficient $r$ for the data.
State with a reason what the value of $r$ would have been if the temperature had been measured in ${ } ^ { \circ } \mathrm { F }$ rather than ${ } ^ { \circ } \mathrm { C }$.
Calculate the equation of the least squares regression line of $c$ on $t$.
The regression line is drawn on the copy of the scatter diagram in the Printed Answer Booklet. Use this diagram to explain what is meant by "least squares".

OCR FS1 AS 2017 December Q6

6 Arlosh, Sarah and Desi are investigating the ratings given to six different films by two critics.

Arlosh calculates Spearman's rank correlation coefficient $r _ { s }$ for the critics' ratings. He calculates that $\Sigma d ^ { 2 } = 72$. Show that this value must be incorrect.
Arlosh checks his working with Sarah, whose answer $r _ { s } = \frac { 29 } { 35 }$ is correct. Find the correct value of $\Sigma d ^ { 2 }$.
Carry out an appropriate two-tailed significance test of the value of $r _ { s }$ at the $5 \%$ significance level, stating your hypotheses clearly. Each critic gives a score out of 100 to each film. Desi uses these scores to calculate Pearson's product-moment correlation coefficient. She carries out a two-tailed significance test of this value at the $5 \%$ significance level.
Explain with a reason whether you would expect the conclusion of Desi's test to be the same as the result of the test in part (iii).

OCR FS1 AS 2017 December Q7

7 Josh is investigating whether sticking pins into a map at random, while blindfolded, provides a random sample of regions of the map. Josh divides the map into 49 squares of equal size and asks each of 98 friends to stick a pin into the map at random, while blindfolded. He then notes the number of pins in each square. To analyse the results he groups the squares as shown in the diagram.

D	D	D	D	D	D	D
D	C	C	C	C	C	D
D	C	B	B	B	C	D
D	C	B	A	B	C	D
D	C	B	B	B	C	D
D	C	C	C	C	C	D
D	D	D	D	D	D	D

The results are summarised in the table.

Region	A	B	C	D
Number of squares	1	8	16	24
Number of pins	6	21	33	38

Test at the 10\% significance level whether the use of pins in this way provides a random sample of regions of the map.
What can be deduced from considering the different contributions to the test statistic? \section*{OCR} \section*{Oxford Cambridge and RSA}

OCR FS1 AS 2018 March Q1

1 A learner driver keeps taking the driving test until she passes. The number of attempts taken, up to and including the pass, is denoted by $X$.

State two assumptions needed for $X$ to be well modelled by a geometric distribution. Assume now that $X \sim \operatorname { Geo } ( 0.4 )$.
Find $\mathrm { P } ( X < 6 )$.
Find $\mathrm { E } ( X )$.
Find $\operatorname { Var } ( X )$.

OCR FS1 AS 2018 March Q2

2 The number of calls received by a customer service department in 30 minutes is denoted by $W$. It is known that $\mathrm { E } ( W ) = 6.5$.

It is given that $W$ has a Poisson distribution.
(a) Write down the standard deviation of $W$.
(b) Find the probability that the total number of calls received in a randomly chosen period of 2 hours is less than 30 .
It is given instead that $W$ has a uniform distribution on $[ 1 , N ]$. Calculate the value of $\mathrm { P } ( W > 3 )$.

OCR FS1 AS 2018 March Q3

3 A pack of 40 cards consists of 10 cards in each of four colours: red, yellow, blue and green. The pack is dealt at random into four "hands", each of 10 cards. The hands are labelled North, South, East and West.

Find the probability that West has exactly 3 red cards.
Find the probability that West has exactly 3 red cards, given that East and West have between them a total of exactly 5 red cards.
South has 5 red cards and 5 blue cards. These cards are placed in a row in a random order. Find the probability that the colour of each card is different from the colour of the preceding card.

OCR FS1 AS 2018 March Q4

4 A spinner has edges numbered $1,2,3,4$ and 5 . When the spinner is spun, the number of the edge on which it lands is the score. The probability distribution of the score, $N$, is given in the table.

Score, $N$	1	2	3	4	5
Probability	0.3	0.2	0.2	$x$	$y$

It is known that $\mathrm { E } ( N ) = 2.55$.

Find $\operatorname { Var } ( N )$.
Find $\mathrm { E } ( 3 N + 2 )$.
Find $\operatorname { Var } ( 3 N + 2 )$.

OCR FS1 AS 2018 March Q5

5 The speed $v \mathrm {~ms} ^ { - 1 }$ of a car at time $t$ seconds after it starts to accelerate was measured at 1 -second intervals. The results are shown in the following diagram.
\includegraphics[max width=\textwidth, alt={}, center]{d5843350-52f9-4fed-adf4-86ceb958033f-3_661_1186_1078_443}

State whether $t$ or $v$ or neither is a controlled variable. The value of the product moment correlation coefficient $r$ for the data is 0.987 correct to 3 significant figures.
The speed of the car is converted to miles per hour and the time to minutes. State the value of $r$ for the converted data.
State the value of Spearman's rank correlation coefficient $r _ { s }$ for the data.
What information does $r$ give about the data that is not given by $r _ { s }$ ?

OCR FS1 AS 2018 March Q6

7 marks

6 The discrete random variable $R$ has the distribution $\operatorname { Po } ( \lambda )$.
Use an algebraic method to find the range of values of $\lambda$ for which the single most likely value of $R$ is 7. [7]

OCR FS1 AS 2018 March Q7

7 The numbers of students taking A levels in three subjects at a school were classified by the year in which they entered the school as follows.

\cline { 2 - 5 } \multicolumn{1}{c|}{}

Subject

Mathematics

English

Physics

\multirow{3}{*}{

Year of

Entry

}

Year 7

17

16

7

\cline { 2 - 5 }

Year 12

13

2

5

The Head of the school carries out a significance test at the $10 \%$ level to test whether subjects taken are independent of year of entry.

Show that in carrying out the test it is necessary to combine columns.
Suggest a reason why it is more sensible to combine the columns for Mathematics and Physics than the columns for Physics and English.
Carry out the test.
State which cell gives the largest contribution to the test statistic.
Interpret your answer to part (iv).

OCR FS1 AS 2018 March Q8

8 In a competition, entrants have to give ranks from 1 to 7 to each of seven resorts. The correct ranks for the resorts are decided by an expert.

One competitor chooses his ranks randomly. By considering all the possible rankings, find the probability that the value of Spearman's rank correlation coefficient $r _ { s }$ between the competitor's ranks and the expert's ranks is at least $\frac { 27 } { 28 }$.
Another competitor ranks the seven resorts. A significance test is carried out to test whether there is evidence that this competitor is merely guessing the rank order of the seven resorts. The critical region is $r _ { s } \geqslant \frac { 27 } { 28 }$. State the significance level of the test. \section*{END OF QUESTION PAPER}

OCR FS1 AS 2021 June Q1

1 The probability distribution for the discrete random variable $W$ is given in the table.

$w$	1	2	3	4
$\mathrm { P } ( W = w )$	0.25	0.36	$x$	$x ^ { 2 }$

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

OCR FS1 AS 2021 June Q2

23 marks

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 $X$.

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 $\operatorname { Po } ( 0.7 )$.
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 $\mathrm { Po } ( 1.6 )$.
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. $$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$
Calculate the value of Pearson's product-moment correlation coefficient $r$.
State an assumption needed to be able to carry out a significance test on the value of $r$.
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.
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.
Calculate the value of Pearson's product-moment correlation coefficient $r$.
State an assumption needed to be able to carry out a significance test on the value of $r$.
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.

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 { B1ft }

\text { B1 }

\text { M1 }

\text { A1 }

{ [ 7 ] } \end{gathered}\)

3.1a

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

A1

[3]

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

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

[3]

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

A1

[3]

1.1

2.4

Needs M2 and "minimises" and "sums of squares" oe

SC: Horizontal(s):

full marks (indept of (b))

OCR FS1 AS 2021 June Q1

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

Find the probability that at least 28 orders are received in a randomly chosen 10 -minute period.
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.
State a necessary assumption for the validity of your calculation in part (b).

OCR FS1 AS 2021 June Q2

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.

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

OCR FS1 AS 2021 June Q3

28 marks

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

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

In fact, on the basis of a large number of observations of $X$, it is found that $\mathrm { P } ( X = 3 ) = \frac { 4 } { 25 } \times \mathrm { P } ( X = 1 )$. Estimate the value of $p$. \section*{Total Marks for Question Set 4: 29} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'. We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.

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

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

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

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

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

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

\end{table}

Question

Answer

Marks

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

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

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

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

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

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

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

OCR FS1 AS 2021 June Q1

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

$x$	7	8	12	6	4
$y$	20	16	7	17	23

Find the value of Pearson's product-moment correlation coefficient.
State what your answer to part (a) tells you about a scatter diagram representing the data.
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}
State which, if either, of the variables $p$ and $q$ is independent.
Calculate the equation of the regression line of $q$ on $p$.
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.
Comment on the reliability of using the regression line to predict the value of $q$ when $p = 7.0$.

OCR FS1 AS 2021 June Q3

38 marks

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}

Question

Solution

Marks

AOs

Guidance

1

(a)

-0.954 BC

B2 [2]

1.1 1.1

SC: If B0, give B1 if two of 7.04, 29.0[4], -13.6[4] (or 35.2, 145[.2], -68.2) seen

1

(b)

Points lie close to a straight line Line has negative gradient

B1 B1 [2]

2.2b 1.1

Must refer to line, not just "negative correlation"

1

(c)

No, it will be the same as $x \rightarrow a$ is a linear transformation

B1 [1]

2.2a

OE. Either "same" with correct reason, or "disagree" with correct reason. Allow any clear valid technical term

2

(a)

Neither

B1 [1]

1.2

2

(b)

$q = 1.13 + 0.620 p$

B1B1 B1 [3]

1.1,1.1 1.1

0.62(0) correct; both numbers correct Fully correct answer including letters

2

(c)

(i)

2.68

B1ft [1]

1.1

awrt 2.68, ft on their (b) if letters correct

2

(c)

(ii)

2.5 is within data range, and points (here) are close to line/well correlated

B1 B1 [2]

2.2b 2.2b

At least one reason, allow "no because points not close to line" Full argument, two reasons needed

2

(d)

Not much data here/points scattered/ possible outliers

So not very reliable

M1 A1 [2]

2.3 1.1

Reason for not very reliable (not "extrapolation") Full argument and conclusion, not too assertive (not wholly unreliable!)

3

(a)

Expected frequency for Middle/25 to 60 is 4.4 which is < 5 so must combine cells

B1*ft depB1 [2]

2.4 3.5b

Correctly obtain this $F _ { E }$, ft on addition errors " < 5" explicit and correct deduction

3

(b)

Early	Middle	Late
29.4	23.1	31.5
26.6	20.9	28.5

Early	Middle	Late
0.9918	0.4160	2.2937
1.0962	0.4598	2.5351

B1

1.1

Both, allow 28.4 for 28.5

awrt 2.29, but allow 2.3 In range [2.53, 2.54]

Question

Solution

Marks

AOs

Guidance

3

(c)

$\mathrm { H } _ { 0 }$ : no association between session and age group. $\mathrm { H } _ { 1 }$ : some association

$\Sigma X ^ { 2 } = 7.793$

$v = 2 , \chi ^ { 2 } ( 2 ) _ { \text {crit } } = 5.991$

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

Significant evidence of association between session attended and age group.

B1

M1ft

A1ft [5]

1.1

2.2b

Both. Allow "independent" etc

Correct value of $X ^ { 2 }$, awrt 7.79 (allow even if wrong in (b))

Correct CV and comparison

Correct first conclusion, FT on their TS only

Contextualised, not too assertive

3

(d)

The two biggest contributions to $\chi ^ { 2 }$ are both for the late session ... ... when the proportion of younger people is higher, and of older people is lower, than the null hypothesis would suggest.

M1ft

A1ft

[2]

1.1

2.4

Refer to biggest contribution(s), FT on their answers to (b), needs "reject $\mathrm { H } _ { 0 }$ "

Full answer, referring to at least one cell (ignore comments on next highest cells)

\multirow[t]{2}{*}{4}

\multirow{2}{*}{}

\multirow{2}{*}{OR:}

$\frac { { } ^ { 2 m } C _ { 2 } \times m } { { } ^ { 3 m } C _ { 3 } }$

$= \frac { 2 m ( 2 m - 1 ) } { 2 } \times m \div \frac { 3 m ( 3 m - 1 ) ( 3 m - 2 ) } { 6 }$

$= \frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) }$ $\frac { 2 m ( 2 m - 1 ) } { ( 3 m - 1 ) ( 3 m - 2 ) } = \frac { 28 } { 55 }$

$\Rightarrow 16 m ^ { 2 } - 71 m + 28 = 0$

$m = 4$ BC

Reject $m = \frac { 7 } { 16 }$ as $m$ is an integer

M1

A1

M1

A1

M1

A1

[7]

3.1b

2.1

3.1a

2.1

1.1

3.2a

Use ${ } ^ { 2 m } C _ { 2 }$ and $m$
Divide by ${ } ^ { 3 m } C _ { 3 }$
Correct expression in terms of $m$ (allow with $m$ not cancelled yet)
Equate to $\frac { 28 } { 55 }$ \	simplify to three-term quadratic
Correct simplified quadratic, or (quadratic) $\times m , = 0$, aef Solve to get both 4 and $\frac { 7 } { 16 }$
Explicitly reject $m = \frac { 7 } { 16 }$

$\frac { 2 m ( 2 m - 1 ) \times m \times 3 ! } { 3 m ( 3 m - 1 ) ( 3 m - 2 ) \times 2 }$ then as above

Multiplication method can get full marks, but if no 3 or 3 !, max

M1M0A0 M1A0M0A0

\(w\)	1	2	3	4
\(\mathrm { P } ( W = w )\)	0.25	0.36	\(x\)	\(x ^ { 2 }\)

\(w\)	1	2	3	4
\(\mathrm { P } ( W = w )\)	0.25	0.36	\(x\)	\(x ^ { 2 }\)

Questions — OCR FS1 AS (30 questions)

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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

\(T\)	17	21	25	26	27	27	29	30	30
\(C\)	21	16	20	38	32	37	35	39	42