Non-geometric distribution identification

4 Three fair 4-sided spinners each have sides labelled 1,2,3,4. The spinners are spun at the same time and the number on the side on which each spinner lands is recorded. The random variable \(X\) denotes the highest number recorded.

1. Show that \(\mathrm { P } ( X = 2 ) = \frac { 7 } { 64 }\).
2. Complete the probability distribution table for \(X\).

   \(x\)	1	2	3	4
   \(\mathrm { P } ( X = x )\)		\(\frac { 7 } { 64 }\)	\(\frac { 19 } { 64 }\)

   On another occasion, one of the fair 4 -sided spinners is spun repeatedly until a 3 is obtained. The random variable \(Y\) is the number of spins required to obtain a 3 .
3. Find \(\mathrm { P } ( Y = 6 )\).
4. Find \(\mathrm { P } ( Y > 4 )\).

7 Each of three students, \(\mathrm { X } , \mathrm { Y }\) and Z , was given an identical pack of 48 cards, of which 12 cards were red and 36 were blue. They were each told to carry out a different experiment, as follows: Student X: Choose a card from the pack, at random, 20 times altogether, with replacement. Record how many times you obtain a red card. Student Y: Choose a card from the pack, at random, 20 times altogether, without replacement. Record how many times you obtain a red card. Student Z: Choose single cards from the pack at random, with replacement, until you obtain the first red card. Record how many cards you have chosen, including the first red card.

1. Find the probability that student Z has to choose more than 8 cards in order to obtain the first red card. Each student carries out their experiment 30 times. The frequencies of the results recorded by each student are shown in the following table, but not necessarily with the rows in the order \(\mathrm { X } , \mathrm { Y } , \mathrm { Z }\) :

   	Number recorded	0	1	2	3	4	5	6	7	8	\(\geqslant 9\)	Observed Mean	Observed Variance
   \multirow{3}{*}{Observed Frequencies}	Student 1	0	0	1	3	7	8	6	4	1	0	5.03	1.97
   	Student 2	0	8	5	4	2	3	3	2	1	2	4.03	11.57
   	Student 3	0	1	2	5	4	6	5	3	4	0	4.97	3.70

   \section*{(b) In this question you must show detailed reasoning.} Two other students make the following statements about the results. For each of the statements, explain whether you agree with the statement. Do not carry out any hypothesis tests, but in each case you should give two justifications for your answer.
   1. "The second row is a good match with the expected results for student Z ."
   2. "The third row is definitely student X 's results."

4

1. Four men and four women stand in a random order in a straight line. Determine the probability that no one is standing next to a person of the same gender.
2. \(x\) men, including Mr Adam, and \(x\) women, including Mrs Adam, are arranged at random in a straight line. Show that the probability that Mr Adam is standing next to Mrs Adam is \(\frac { 1 } { X }\).

2. A bag contains a large number of counters. Each counter has a single digit number on it and the mean of all the numbers in the bag is the unknown parameter \(\mu\). The number 2 is on \(40 \%\) of the counters and the number 5 is on \(25 \%\) of the counters. All the remaining counters have numbers greater than 5 on them. A random sample of 10 counters is taken from the bag.

1. State whether or not each of the following is a statistic
   1. \(S =\) the sum of the numbers on the counters in the sample,
   2. \(D =\) the difference between the highest number in the sample and \(\mu\),
   3. \(F =\) the number of counters in the sample with a number 5 on them. The random variable \(T\) represents the number of counters in a random sample of 10 with the number 2 on them.
2. Specify the sampling distribution of \(T\). The counters are selected one by one.
3. Find the probability that the third counter selected is the first counter with the number 2 on it.

6 Tom has read in a newspaper that you can tell the air temperature by counting how often a cricket chirps in a period of 20 seconds. (A cricket is a type of insect.) He wants to know exactly how the temperature can be predicted. On 8 randomly selected days, when Tom can hear crickets chirping, he records the number of chirps, \(x\), made by a cricket in a 20-second interval, and also the temperature, \(y ^ { \circ } \mathrm { C }\), at that time. The data are summarised as follows.
\(n = 8 \quad \sum x = 268 \quad \sum y = 141.9 \quad \sum x ^ { 2 } = 9618 \quad \sum y ^ { 2 } = 2630.55 \quad \sum \mathrm { xy } = 5009.1\)
These data are illustrated below.
\includegraphics[max width=\textwidth, alt={}, center]{8f1e0c68-a334-4657-823e-386ab0994c02-5_661_1035_699_242}

1. Determine the equation of the regression line of \(y\) on \(x\). Give your answer in the form \(\mathrm { y } = \mathrm { ax } + \mathrm { b }\), giving the values of \(a\) and \(b\) correct to \(\mathbf { 3 }\) significant figures.
2. Use the equation of the regression line to predict the temperature for the following values of \(x\).
   • 35
   • 10
   • Comment on the reliability of your predictions in part (b).
   • State the coordinates of the point of intersection of the line whose equation you have calculated with the regression line of \(x\) on \(y\).

1. A biased coin has probability 0.4 of showing a head. In an experiment, the coin is spun until a head appears. If a head has not appeared after 4 spins, the coin is not spun again. The random variable \(X\) represents the number of times the coin is spun.

For example, \(X = 3\) if the first two spins do not show a head but the third spin does show a head. The coin would not then be spun a fourth time since the coin has already shown a head.

1. Show that \(\mathrm { P } ( X = 3 ) = 0.144\) The table gives some values for the probability distribution of \(X\)

   \(x\)	1	2	3	4
   \(\mathrm { P } ( X = x )\)		0.24	0.144

2. 1. Write down the value of \(\mathrm { P } ( X = 1 )\)
   2. Find \(\mathrm { P } ( X = 4 )\)
3. Find \(\mathrm { E } ( X )\)
4. Find \(\operatorname { Var } ( X )\) The random variable \(H\) represents the number of heads obtained when the coin is spun in the experiment.
5. Explain why \(H\) can only take the values 0 and 1 and find the probability distribution of \(H\).
6. Write down the value of
   1. \(\mathrm { P } ( \{ X = 3 \} \cap \{ H = 0 \} )\)
   2. \(\mathrm { P } ( \{ X = 4 \} \cap \{ H = 0 \} )\) The random variable \(S = X + H\)
7. Find the probability distribution of \(S\)

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

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

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

1. For all sixteen candidates, the value of the product moment correlation coefficient \(r\) for the marks on the two papers was 0.701 correct to 3 significant figures. Test whether there is evidence, at the \(5 \%\) significance level, of association between the marks on the two papers.
2. A teacher decided to omit the marks of the candidates who were in the top three places in mechanics and the candidates who were in the bottom three places in mechanics. The marks for the remaining 10 candidates can be summarised by
   \(n = 10 , \Sigma x = 750 , \Sigma y = 690 , \Sigma x ^ { 2 } = 57690 , \Sigma y ^ { 2 } = 49676 , \Sigma x y = 50829\).
   1. Calculate the value of \(r\) for these 10 candidates.
   2. What do the two values of \(r\), in parts (a) and (b)(i), tell you about the scores of the sixteen candidates? A bag contains a mixture of blue and green beads, in unknown proportions. The proportion of green beads in the bag is denoted by \(p\).
3. Sasha selects 10 beads at random, with replacement. Write down an expression, in terms of \(p\), for the variance of the number of green beads Sasha selects. Freda selects one bead at random from the bag, notes its colour, and replaces it in the bag. She continues to select beads in this way until a green bead is selected. The first green bead is the \(X\) th bead that Freda selects.
4. Assume that \(p = 0.3\). Find
   1. \(\mathrm { P } ( X \geqslant 5 )\),
   2. \(\operatorname { Var } ( X )\).
5. In fact, on the basis of a large number of observations of \(X\), it is found that \(\mathrm { P } ( X = 3 ) = \frac { 4 } { 25 } \times \mathrm { P } ( X = 1 )\). Estimate the value of \(p\). \section*{Total Marks for Question Set 4: 29} \section*{Mark scheme} \section*{Marking Instructions} a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
   b The following types of marks are available. \section*{M} A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
   A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly. \section*{A} Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. \section*{B} Mark for a correct result or statement independent of Method marks. \section*{E} A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given. The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'. We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
   • When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
   • When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s . f . }\) unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
   Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
   Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. Rules for replaced work and multiple attempts:
   • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
   • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
   • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
   For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
   If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
   h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Abbreviations}

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

   \end{table}

   Question			Answer	Marks	AO	Guidance
   1	(a)		\(1 - \mathrm { P } ( \leq 27 )\) \(= 0.0525 \quad \mathbf { B C }\)	M1 A1 [2]	3.4 1.1	Allow M1 for \(1 - 0.9657 = 0.0343\) In range [0.0524, 0.0525] BC
   1	(b)		Po(40) \(1 - \mathrm { P } ( \leq 55 )\) \(= 0.00968\)	M1* depM1 A1 [3]	3.1b 1.1a 1.1	\(\operatorname { Po( } 2 \times\) their 20) stated or implied Allow M1 for \(1 - \mathrm { P } ( \leq 56 ) = 0.00658\) or \(1 - 0.990 = 0.01\) or 0.0097 Awrt 0.00968
   1	(c)		Orders on one day are independent of orders on the other	B1 [1]	3.2b	Use "orders independent", clearly referred to the two different days, needs context [not "events"], and nothing else	Not anything affecting given separate Poissons, such as "orders must be independent" or "constant average rate".
   \multirow[t]{3}{*}{2}	\multirow[t]{3}{*}{(a)}	\multirow{3}{*}{}	\(7 ! \times 4\) ! \(\div 10\) ! \(= \frac { 120960 } { 3628800 } = \frac { 1 } { 30 }\)	M1 M1 A1	2.1 1.1a 1.1	Allow for \(6 ! \times 4\) ! or \(6 ! \times 4 ! \times 2\) Divide by 10!, needs at least one factorial in numerator Answer, exact or awrt 0.0333	3/3 for \(\frac { 1 } { 30 }\) www
   			Alternative: \(7 \times \frac { 6 } { 10 } \times \frac { 4 } { 9 } \times \frac { 5 } { 8 } \times \frac { 3 } { 7 } \times \frac { 4 } { 6 } \times \frac { 2 } { 5 } \times \frac { 3 } { 4 } \times \frac { 1 } { 3 } \times \frac { 2 } { 2 } \times \frac { 1 } { 1 }\)	M1 M1 A1		no 7, one other error only one error correct answer
   				[3]
   \multirow[t]{3}{*}{2}	\multirow[t]{3}{*}{(b)}	\multirow{3}{*}{}	Women placed in 6 ! ways, men in 4 ! \([ = 720 \times 24 ]\) 4 slots \(m\) in \(m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m \mathrm {~W} m = { } ^ { 7 } C _ { 4 }\) \({ } ^ { 7 } C _ { 4 } \times \frac { 6 ! \times 4 ! } { 10 ! }\) \(= \frac { 1 } { 6 }\)	B1 М1 М1 A1	2.1 3.1b 1.1a 1.1	\(6 ! \times 4 !\) anywhere, or \(6 ! \times\) attempt at \({ } ^ { 7 } P _ { 4 }\) Or \({ } ^ { 7 } P _ { 4 }\). Allow for \(m\) and \(W\) reversed Needs attempt at both terms Or 0.167 or 0.1667 etc	Or \(6 ! \times 7 \times 6 \times 5 \times 4 \quad\) B2 \({ } ^ { 7 } P _ { 4 } \times 6\) !: B1M1 \(4 \times ( 6 ! \times 4 ! ) / 10 ! = 2 / 105 :\) В 1 М 1
   			Alternative: PIE \(( 10 ! - 12 \times 9 ! + ( 3 \times 4 \times 8 ! + 12 \times 2 \times 8 ! ) - 24 \times 7 ! ) / 10 !\) \([ = ( 3628800 - 4354560 + 1451520 - 120960 ) / 10 ! ]\)	M2 A1 A1		Signs alternating, at least one term \(\sqrt { }\) Allow one term omitted or wrong Correct answer
   				[3]
   			Three together: \(7 \times 6 \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 5 }\)	Two pairs: \(\frac { 7 \times 6 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! }\)		\(\frac { 1 } { 10 }\)	One pair: \(7 \times \frac { 6 \times 5 } { 2 } \times \frac { 6 ! 4 ! } { 10 ! } = \frac { 1 } { 2 }\)

   Question			Answer	Marks	AO	Guidance
   3	(a)		\(\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho \neq 0\), where \(\rho\) is population pmcc 0.701 > 0.4973 Reject \(\mathrm { H } _ { 0 }\). There is significant evidence of association between the marks on the two papers	B1 B1 M1ft A1 [4]	2.5 1.1a 1.1 2.2b	Must use symbols. Allow no definition of letter if \(\rho\) used Correct CV stated, allow 0.497 FT on wrong CV Not FT. Needs context, and not too definite.	Not " \(\mathrm { H } _ { 0 }\) : there is no assoc' n , \(\mathrm { H } _ { 1 }\) : there is association" Not There is association ..."
   3	(b)	(i)	-0.534	B2 [2]	1.1a 1.1	SC: if B0, give B1 for two of 1440, 2066, -921 and \(S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)\)	-0.53: B1
   3	(b)	(ii)	6 candidates did very well or very badly on both papers; middle 10 tended to do badly on one paper and well on the other	B1 [1]	2.4	Correct inference about scores oe, not "correlation/association/value of \(r\) ". Not "outliers" or "anomalies".	Allow inference for one group only, provided it is clearly for only one group \	any ref to other group is not wrong
   4	(a)		\(10 p ( 1 - p )\)	B1 [1]	1.2	Allow \(10 p q\) oe, e.g. \(10 p - 10 p ^ { 2 }\)	Not just \(n p ( 1 - p )\)
   4	(b)	(i)	\(0.7 ^ { 4 }\) = 0.240(1)	M1 A1 [2]	1.1a 1.1	\(0.7 ^ { 5 } = 0.168\) or \(0.7 ^ { 6 } = 0.118\) : M1 Allow 0.24	Or \(1 - 0.3 \left( 1 + 0.7 + 0.7 ^ { 2 } + 0.7 ^ { 3 } \right)\) Allow M1 if also \(0.3 \times 0.7 ^ { 4 }\) [0.15 is from binomial]
   4	(b)	(ii)	\(q / p ^ { 2 } = \frac { 70 } { 9 }\) or \(7.777 \ldots\)	B1 [1]	1.1	Allow 7.78, 7.778, etc	Allow 8 only if evidence, e.g. ( \(1 - 0.3\) )/ \(0.3 ^ { 2 }\)
   4	(c)		\(\begin{aligned}	( 1 - p ) ^ { 2 } p = \frac { 4 } { 25 } p
   	p = 0 \text { or } ( 1 - p ) ^ { 2 } = \frac { 4 } { 25 } \quad ( p \neq 0 )
   	( 1 - p ) = \pm \frac { 2 } { 5 }
   	p \neq \frac { 7 } { 5 }
   	p = \frac { 3 } { 5 } \end{aligned}\)	B1 M1 M1 B1ft A1 [5]	1.1 1.1a 1.1 2.3 2.1	Correct equation Reduce to quadratic/cubic and solve Obtain two non-zero solutions Explicitly discard one solution, either here or in line 2 (not enough to give 2 answers and then only 1 ) Exact final answer exact (0.6) no others left, allow from ± omitted	e.g. \(p \left( p ^ { 2 } - 2 p + \frac { 21 } { 25 } \right) = 0\) ± omitted: M0B0A1 Allow " \(p = 0 , \frac { 3 } { 5 } , \frac { 7 } { 5 }\) but \(p \leq 1\) " SC binomial: B0 then \(75 p ^ { 2 } = ( 1 - p ) ^ { 2 } \ \) solve M1 \(0.104 [ 0.1035 ] \quad\) A1 Explicitly reject 0 or - 0.13 B1 SC Poisson: 0