ࡱ>  bjbjQFQF 3,3,6i77!!!$!!!P7"4k#!VG'`+:+++:::UUUUUUUjZ ]lU!WCj8 :WCWCU77++ tVKJKJKJWC7+!+UKJWCUKJKJ:cO, O+@%q!DO qUV0VORx]Fx]OOx]!}P:=*KJ?xA:::UUH:::VWCWCWCWCx]::::::::: : Paper Reference(s) 6684/01 Edexcel GCE Statistics S2 Gold Level G2 Time: 1 hour 30 minutes Materials required for examination Items included with question papers Mathematical Formulae (Green) Nil Candidates may use any calculator allowed by the regulations of the Joint Council for Qualifications. Calculators must not have the facility for symbolic algebra manipulation, differentiation and integration, or have retrievable mathematical formulas stored in them. Instructions to Candidates Write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Statistics S2), the paper reference (6684), your surname, initials and signature. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks may be obtained for answers to ALL questions. There are 8 questions in this question paper. The total mark for this paper is 75. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. Suggested grade boundaries for this paper: A*ABCDE655547362617 1. A bag contains a large number of counters. A third of the counters have a number 5 on them and the remainder have a number 1. A random sample of 3 counters is selected. (a) List all possible samples. (2) (b) Find the sampling distribution for the range. (3)  2. A test statistic has a distribution B(25, p). Given that H0 : p = 0.5, H1 : p ( 0.5, (a) find the critical region for the test statistic such that the probability in each tail is as close as possible to 2.5%. (3) (b) State the probability of incorrectly rejecting H0 using this critical region. (2)  3. Bacteria are randomly distributed in a river at a rate of 5 per litre of water. A new factory opens and a scientist claims it is polluting the river with bacteria. He takes a sample of 0.5 litres of water from the river near the factory and finds that it contains 7 bacteria. Stating your hypotheses clearly test, at the 5% level of significance, the claim of the scientist. (7) ___________________________________________________________________________ 4. Past records suggest that 30% of customers who buy baked beans from a large supermarket buy them in single tins. A new manager questions whether or not there has been a change in the proportion of customers who buy baked beans in single tins. A random sample of 20 customers who had bought baked beans was taken. (a) Using a 10% level of significance, find the critical region for a two-tailed test to answer the managers question. You should state the probability of rejection in each tail which should be less than 0.05. (5) (b) Write down the actual significance level of a test based on your critical region from part (a). (1) The manager found that 11 customers from the sample of 20 had bought baked beans in single tins. (c) Comment on this finding in the light of your critical region found in part (a). (2)  5. In a game, players select sticks at random from a box containing a large number of sticks of different lengths. The length, in cm, of a randomly chosen stick has a continuous uniform distribution over the interval [7, 10]. A stick is selected at random from the box. (a) Find the probability that the stick is shorter than 9.5 cm. (2) To win a bag of sweets, a player must select 3 sticks and wins if the length of the longest stick is more than 9.5 cm. (b) Find the probability of winning a bag of sweets. (2) To win a soft toy, a player must select 6 sticks and wins the toy if more than four of the sticks are shorter than 7.6 cm. (c) Find the probability of winning a soft toy. (4) ___________________________________________________________________________ 6. A company has a large number of regular users logging onto its website. On average 4 users every hour fail to connect to the companys website at their first attempt. (a) Explain why the Poisson distribution may be a suitable model in this case. (1) Find the probability that, in a randomly chosen 2 hour period, (b) (i) all users connect at their first attempt, (ii) at least 4 users fail to connect at their first attempt. (5) The company suffered from a virus infecting its computer system. During this infection it was found that the number of users failing to connect at their first attempt, over a 12 hour period, was60. (c) Using a suitable approximation, test whether or not the mean number of users per hour who failed to connect at their first attempt had increased. Use a 5% level of significance and state your hypotheses clearly. (9)  7. (a) Define the critical region of a test statistic. (2) A discrete random variable X has a Binomial distribution B(30, p). A single observation is used to test H0 : p = 0.3 against H1 : p `" 0.3 (b) Using a 1% level of significance find the critical region of this test. You should state the probability of rejection in each tail which should be as close as possible to 0.005. (5) (c) Write down the actual significance level of the test. (1) The value of the observation was found to be 15. (d) Comment on this finding in light of your critical region. (2) ___________________________________________________________________________ 8. The continuous random variable X has probability density function given by f(x) =  EMBED Equation.3  (a) Sketch f(x) showing clearly the points where it meets the x-axis. (2) (b) Write down the value of the mean, (, of X. (1) (c) Show that E(X 2) = 9.8. (4) (d) Find the standard deviation, (, of X. (2) The cumulative distribution function of X is given by F(x) =  EMBED Equation.3  where a is a constant. (e) Find the value of a. (2) (f) Show that the lower quartile of X,  EMBED Equation.3 , lies between 2.29 and 2.31. (3) (g) Hence find the upper quartile of X, giving your answer to 1 decimal place. (1) (h) Find, to 2 decimal places, the value of k so that P(( k( < X < ( + k( ) = 0.5. (2) ___________________________________________________________________________  TOTAL FOR PAPER: 75 MARKS END Question NumberSchemeMarks 1.(a)(1, 1, 1), (5, 5, 5), (1, 5, 5), (1, 5, 1) B1(1,1,1); (5,5,5); (1, 5, 5); (5, 1, 5); (5, 5, 1) (5, 1, 1); (1, 5, 1); (1, 1, 5) B1(2)(b)r : 0 and 4 B1P(R = 0) = EMBED Equation.DSMT4 P(R = 4) = EMBED Equation.DSMT4M1d A1(3)[5] 2. (a)X ~ B(25,0.5) may be implied by calculations in part a or b M1P( X EMBED Equation.DSMT4 7) = 0.0216 P(X EMBED Equation.DSMT4 18) = 0.0216 CR X EMBED Equation.DSMT4 7; EMBED Equation.DSMT4 X EMBED Equation.DSMT4 18 A1,A1 (3)(b)P(rejecting H0) = 0.0216 + 0.0216 M1 = 0.0432 awrt 0.0432/0.0433A1 (2)Total 5 Question NumberSchemeMarks  3One tail test Method 1 Ho : l = 5 (l = 2.5) may use l or  H1 : l > 5 (l > 2.5) X ~ Po (2.5) may be implied  P(X e" 7) = 1 " P(X d" 6) [ P(X e" 5) = 1" 0.8912 = 0.1088 ] att P(X e" 7) P(X e" 6) = 1 " 0.9858 P(X e" 6) = 1" 0.9580 = 0.0420 = 0.0142 CR X e" 6 awrt 0.0142 0.0142 < 0.05 7 e" 6 or 7 is in critical region or 7 is significant (Reject H0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified Method 2 Ho : l = 5 (l = 2.5) may use l or  H1 : l > 5 (l > 2.5) X ~ Po (2.5) may be implied  P(X < 7) [P(X < 5) = 0.8912] att P(X < 7) P(X < 6) P(X < 6) = 0.9580 = 0.9858 CR X e" 6 wrt 0.986 0.9858 > 0.95 7 e" 6 or 7 is in critical region or 7 is significant (Reject H0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified Two tail test Method 1 Ho : l = 5 (l = 2.5) may use l or  H1 : l  EMBED Equation.DSMT4 5 (l  EMBED Equation.DSMT4 2.5) X ~ Po (2.5)  P(X e" 7) = 1 " P(X d" 6) [P(X e" 6) = 1" 0.9580 = 0.0420] att P(X e" 7) P(X e" 7) = 1 " 0.9858 P(X e" 7) = 1" 0.9858 = 0.0142 = 0.0142 CR X e" 7 awrt 0.0142 0.0142 < 0.025 7 e" 7 or 7 is in critical region or 7 is significant (Reject H0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified  Method 2 Ho : l = 5 (l = 2.5) may use l or  H1 : l  EMBED Equation.DSMT4 5 (l  EMBED Equation.DSMT4 2.5) X ~ Po (2.5)  P(X < 7) [P(X < 6) = 0.9580] att P(X < 7) P(X < 7) P(X < 7) = 0.9858 = 0.9858 CR X e" 7 awrt 0.986 0.9858 > 0.975 7 e" 7 or 7 is in critical region or 7 is significant (Reject H0.) There is significant evidence at the 5% significance level that the factory is polluting the river with bacteria. or The scientists claim is justified  B1 B1 M1 M1 A1 M1 B1 (7) Total 7 B1 B1 M1 M1 A1 M1 B1 (7) B1 B0 M1 M1 A1 M1 B1 (7) B1 B0 M1 M1A1 M1 B1 (7)  Question NumberSchemeMarks 4. (a) (b) (c)X~ B(20, 0.3) P(X < 2) = 0.0355 P(X < 9) = 0.9520 so P(X > 10) = 0.0480 Therefore the critical region is  EMBED Equation.DSMT4  0.0355 + 0.0480 = 0.0835 awrt (0.083 or 0.084) 11 is in the critical region there is evidence of a change/ increase in the proportion/number of customers buying single tins M1 A1 A1 A1 A1 (5) B1 (1) B1ft B1ft (2) (8 marks) 5. (a) EMBED Equation.DSMT4 M1 EMBED Equation.DSMT4  awrt 0.833A1(2) (b)P(Longest > 9.5) = 1 - P(all < 9.5) =  EMBED Equation.DSMT4 M1  EMBED Equation.DSMT4 A1(2) (c)P(a stick < 7.6) =  EMBED Equation.DSMT4 B1Let Y = number of sticks (out of 6) <7.6 then Y~B(6, 0.2)M1P(Y > 4) = 1 - P(Y < 4)M1 = 1 - 0.9984 = 0.0016 or  EMBED Equation.3  A1 (4)8 Question NumberSchemeMarks 6. (a)Connecting occurs at random/independently, singly or at a constant rateB1 (1)(b)Po (8)B1(i)P(X = 0) = 0.0003M1A1(ii)P(X ( 4) = 1 P(X ( 3)M1 = 1 0.0424A1 (5) = 0.9576(c)H0 :  EMBED Equation.DSMT4 = 4 (48) H1 :  EMBED Equation.DSMT4 > 4 (48)B1N(48,48)M1 A1Method 1 Method 2 P(X > 59.5) =  EMBED Equation.DSMT4   EMBED Equation.3 =1.6449 = P ( Z > 1.66) = 1 0.9515 = 0.0485 x = 59.9 M1 M1 A1 A10.0485 < 0.05Reject H0. Significant. 60 lies in the Critical regionM1The number of failed connections at the first attempt has increased.A1 ft (9)[15] Question NumberSchemeMarks 7. (a) The set of values of the test statistic for which B1the null hypothesis is rejected in a hypothesis test.B1 (2) (b)X~B(30,0.3)M1 EMBED Equation.DSMT4  EMBED Equation.DSMT4 A1 EMBED Equation.DSMT4  EMBED Equation.DSMT4 A1Critical region is  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4  A1A1 (5) (c)Actual significance level 0.0021+0.0064=0.0085 or 0.85%B1 (1) (d)15 (it) is not in the critical region not significantBft 2, 1, 0No significant evidence of a change in  EMBED Equation.DSMT4  accept H0, (reject H1)  EMBED Equation.3  (2)Total [10] Question NumberSchemeMarks 8.  EMBED Equation.DSMT4  shape which does not go below the x-axis [condone missing patios]B1(a)Graph must end at the points (1,0) and (5,0) and the points labelled at 1 and 5 B1(2)(b)E(X) = 3 (by symmetry) B1(1)(c) EMBED Equation.3  EMBED Equation.3 M1=  EMBED Equation.DSMT4 A1=  EMBED Equation.DSMT4  = 9.8 (*)M1 A1 cso (4)(d)s.d. =  EMBED Equation.DSMT4 , M1  = 0.8944 awrt 0.894A1(2)(e)F(1) = 0  EMBED Equation.DSMT4 , leading to a = 7M1 A1(2)(f)F(2.29) = 0.2449, F(2.31) = 0.2515M1 A1Since F( EMBED Equation.DSMT4 ) = 0.25 and these values are either side of 0.25 then 2.29<  EMBED Equation.DSMT4  < 2.31A1(3)(g)Since the distribution is symmetric  EMBED Equation.DSMT4 =5-1.3 = 3.7 caoB1(1)(h)We know P( EMBED Equation.DSMT4 =2.3 < X < 3.7= EMBED Equation.DSMT4 ) = 0.5 so  EMBED Equation.DSMT4  M1so k =  EMBED Equation.DSMT4  = awrt 0.78A1 (2)17 Examiner reports Question 1 This question was answered well by many candidates. In part (a) the vast majority of candidates were able to score one mark and many scored both marks as they could list all 8 combinations. Part (b) caused some candidates problems as they failed to recognise that the range was the difference between the largest and smallest values. Some candidates found the sampling distribution of the mean or the sum rather than the range. Others thought that all samples were equally likely and hence obtained the wrong probabilities. Those candidates who could identify the range usually went on to calculate the correct corresponding probabilities. A common error was to give the range as 0 and 1. Question 2 Part (a) was a routine question with many fully correct answers being seen. However, a minority of candidates were unable to write down the critical regions correctly. The two most common errors were to write down the regions in terms of probabilities, e.g.  EMBED Equation.3 or to simply write X = 7 and X = 18. In part (b) many candidates were distracted by the word incorrectly. We reject the null hypothesis incorrectly when it is in fact true. In this case, this means that p really is 0.5, so that we can use the distribution B(25, 0.5) to work out probabilities. This is the distribution that had already been used in part (a), so that all that needed to be done was to simply add the two probabilities that were used to identify the two parts of the critical region in part (a): 0.0216 + 0.0216 = 0.0432. It was all too common to see correct working followed by incorrect and irrelevant work that invalidated the whole response. The most common incorrect postscripts were 0.05 0.0432 = 0.0068 and 1 0.0432 = 0.9568. Question 3 The majority of candidates found this question straightforward. They were most successful if they used the probability method and compared it with 0.05. Those who attempted to use 95% were less successful and this is not a recommended route for these tests. Most candidates knew how to specify the hypotheses with most candidates using 2.5 rather than 5. Some candidates used p, or did not use a letter at all, in stating their hypotheses, but most of the time they used . A minority found P(X=7) and some worked with Po(5). If using the critical region method, not all candidates showed clearly, either their working, or a comparison with the value of 7 and the CR X > 7. A sizeable minority of candidates failed to put their conclusion back into the given context. Reject H0 is not sufficient. Question 4 This was a very well answered question. Candidates were able to use binomial tables and gave the answer to the required number of decimal places. As in previous years there were some candidates who confused the critical region with the probability of the test statistic being in that region but this error has decreased. Candidates were able to describe the acceptance of the hypothesis in context although sometimes it would be better if they just repeated the wording from the question which would help them avoid some of the mistakes seen. There were still a few candidates who did not give a reason in context at all. In part (a) many candidates failed to read this question carefully assuming it was identical to similar ones set previously. Most candidates correctly identified B(20,0.3) to earn the method mark and many had 0.0355 written down to earn the first A mark, although in light of their subsequent work, this may often have been accidental. A majority of candidates did not gain the second A mark as they failed to respond to the instruction state the probability of rejection in each case. In the more serious cases, candidates had shown no probabilities from the tables, doing all their work mentally, only writing their general strategy:  EMBED Equation.DSMT4 . Whilst many candidates were able to write down the critical region using the correct notation there are still some candidates who are losing marks they should have earned, by writing  EMBED Equation.DSMT4  for the critical region  EMBED Equation.3  Part (b) was usually correct. Part (c) provided yet more evidence of candidates who had failed to read the question: in the light of your critical region. Some candidates chose not to mention the critical region and a number of those candidates who identified that 11 was in the critical region did not refer to the managers question. Question 5 Part (a) was routine and the vast majority of candidates demonstrated familiarity with the probability density function of a rectangular distribution. Part (b) required some careful initial thought that eluded a large majority of the candidates. The successful solutions fell into two camps. On the one hand were those candidates who identified the Binomial distribution  EMBED Equation.3  and then evaluated the probability  EMBED Equation.3  by conventional means. On the other hand there were candidates who worked from first principles and produced either an elaborate tree diagram or a list of all possible outcomes. This latter approach (from first principles) is of course valid, but takes longer and is more susceptible to error. There were many complete and correct solutions to part (c). Finding the length of a stick shorter than 7.6 cm was straightforward for most but many attempts reflected candidates' lack of understanding of what the question required of them in this part. Successful candidates realised it was a binomial situation using B(6, p) and used tables to find  EMBED Equation.3 or calculated P(X = 5) + P(X = 6). Common errors included misinterpretation of P(X > 4) as  EMBED Equation.3 or even P(X = 4) and a small minority of candidates gave the answer as EMBED Equation.3 which gained no marks as it is an incorrect method. Question 6 The majority of candidates were familiar with the technical terms in part (a), but failed to establish any context. Part (b) was a useful source of marks for a large proportion of the candidates. The only problems were occasional errors in detail. In part (i) a few did not spot the change in time scale and used Po(4) rather than Po(8). Some were confused by the wording and calculated P(X = 8) rather than P(X = 0). The main source of error for (ii) was to find 1 P(X ( 4) instead of 1 P(X ( 3). In part (c) the Normal distribution was a well-rehearsed routine for many candidates with many candidates concluding the question with a clear statement in context. The main errors were Some other letter (or none) in place of  EMBED Equation.3  or  EMBED Equation.3  Incorrect Normal distribution: e.g. N(60, 60) Omission of (or an incorrect) continuity correction Using 48 instead of 60 Calculation errors A minority of candidates who used the wrong distribution (usually Poisson) were still able to earn the final two marks in the many cases when clear working was shown. This question was generally well done with many candidates scoring full marks. Question 7 Part (a) tested candidates understanding of the critical region of a test statistic and responses were very varied, with many giving answers in terms of a region or area and making no reference to the null hypothesis or the test being significant. Many candidates lost at least one mark in part (b), either through not showing the working to get the probability for the upper critical value, i.e. 1 P(X d" 15) = P(X e" 16) = 0.0064, or by not showing any results that indicated that they had used B(30, 0.3) and just writing down the critical regions, often incorrectly. 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