abc Mathematics Statistics General Certificate of Education SPECIMEN UNITS AND MARK SCHEMES

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1 abc General Certificate of Education Mathematics Statistics SPECIMEN UNITS AND MARK SCHEMES ADVANCED SUBSIDIARY MATHEMATICS (56) ADVANCED SUBSIDIARY PURE MATHEMATICS (566) ADVANCED SUBSIDIARY FURTHER MATHEMATICS (57) ADVANCED MATHEMATICS (66) ADVANCED PURE MATHEMATICS (666) ADVANCED FURTHER MATHEMATICS (67)

2 General Certificate of Education Specimen Unit Advanced Subsidiary Examination abc MATHEMATICS Unit Statistics A MSA In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 5 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MSA. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maximum mark for this paper is 60. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. []

3 Answer all questions. Ten per cent of coloured beads used in costume jewellery are orange. (a) Find the probability that in a string of 40 beads, 4 or fewer beads are orange. ( marks) (b) Calculate the probability that in a string of 5 beads, exactly beads are orange. ( marks) (c) State one assumption that you have made in answering parts (a) and (b). ( mark) The weights of bags of red gravel may be modelled by a normal distribution with mean 5.8 kg and standard deviation 0.5 kg. (a) Determine the probability that a randomly selected bag of red gravel will weigh less than 5 kg. ( marks) (b) Determine, to two decimal places, the weight exceeded by 0% of bags. (4 marks) []

4 4 A cricket team meets for fielding practice. One exercise consists of a cricket ball being thrown at different heights, speeds and angles to one side of a fielder who tries to catch it using one hand. Each member of the team attempts 5 catches with each hand. The number of successful catches are given in the following table. Fielder A B C D E F G H I J K Left hand Right hand (a) Calculate the value of the product moment correlation between the number of catches with the left hand and the number of catches with the right hand. ( marks) (b) Comment on the performance of fielders E and I. ( marks) (c) When fielders E and I are omitted from the calculation, the value of the product moment correlation coefficient between the number of left-handed catches and the number of righthanded catches is 0.8, correct to three decimal places. Comment on this value and the value you calculated in part (a) ( marks) 5 Pencils produced on a certain machine have lengths, in millimetres, which are distributed with a mean of µ and a standard deviation of. A random sample of 90 pencils was taken and the length of each pencil measured. The mean length was found to be 78.5 millimetres. (a) Construct a 99% confidence interval for µ. (5 marks) (b) State why, in answering part (a), it is not necessary to assume that the length of pencils are normally distributed. ( marks) TURN OVER FOR THE NEXT QUESTION Turn over [4]

5 4 6 Last year the employees of a firm either received no pay rise, a small pay rise or a large pay rise. The following table shows the number in each category, classified by whether they were weekly paid or monthly paid. No pay rise Small pay rise Large pay rise Weekly Paid Monthly paid 4 8 A tax inspector decides to investigate the tax affairs of an employee selected at random. D is the event that a weekly paid employee is selected. E is the event that an employee who received no pay rise is selected. E is the event not E. (a) Find the value of: (i) P (D ); (ii) P (D E ); (iii) P (D E ). (5 marks) (b) The tax inspector now decides to select three employees. Find the probability that they are all weekly paid if: (i) one is selected at random from those who had no pay rise, one from those who had a small pay rise and one from those who had a large pay rise; ( marks) (ii) they are selected at random (without replacement) from all the employees of the firm. ( marks) [5]

6 5 7 [A sheet of graph paper is provided for use in this question.] Andrew (A), Charles (C) and Edward (E) are employed by the Palace Hotel. Each is responsible for one floor of the building and their duties include cleaning the bedrooms. The number of bedrooms occupied on each floor varies from day to day. The following table shows 0 observations of the number, x, of bedrooms to be cleaned and the time taken, y minutes, to carry out the cleaning. The employee carrying out the cleaning is also indicated. Employee A C E E C A A E C C x y (a) Plot a scatter diagram of the data. Identify the employee by labelling each point. ( marks) (b) Calculate the equation of the regression line of y on x. Draw the line on your scatter diagram. (6 marks) (c) Calculate the residuals for the three observations when Andrew did the cleaning. ( marks) (d) Comment on the times taken by Andrew to carry out his cleaning. ( mark) END OF QUESTIONS [6]

7 MSA Specimen abc Question Solution Marks Total Comments (a) Binomial n = 40 p = 0. P( 4 or fewer ) = 0.69 (b) P() = (5 4/) = (c) Beads selected randomly/independently E Total 7 (a) z = (5 5.8)/0.5 =.6 Probability less than 5kg = = (b) z =.86 Weight exceeded by 0% of bags = 6.44 (a) Class mid-mark Frequency x = 6. s = 5. m 4 Total 7 A A 5 Allow m for mean and s.d. if method shown. 6. (6. 6.) 5. (5.0 5.) (b) Journeys from Surrey have similar duration, on average, but are less variable than those from Essex. (c) People asked may not be representative. Times are estimated not measured. E E E E Total 9 Or any other sensible comments e.g. journey time not defined, weather conditions may be extreme etc [7]

8 MSA (cont) Question Solution Marks Total Comments 4(a) B allow M if method shown (b) E and I held more catches with left than with right hand - all others held more with right than left. E E (c) Correlation coefficient of 0.8 suggests that those who caught a lot of catches with one hand also caught a lot of catches with the other. When E and I (possibly left handers) are included the correlation coefficient of suggests no association between the number of catches with each hand. 5(a) 99% confidence interval for mean 78.5 ±.5758 / ± (b) Sample is large. Sample mean may be assumed to be Normally distributed by Central Limit Theorem. 6(a)(i) 5/50 = E E Total 7 m 5 E E Total 7 acf (ii) 5/9 = 0.86 acf (iii) 90/50 = 0.6 acf (b)(i) 5/9 85/9 5/8 = (ii) 5/50 4/49 /48 = Total 0 [8]

9 MSA (cont) Question Solution Marks Total Comments 7(a) See graph on next page (b) y = x x = 8 y = 96. x = y = 4. B B Allow for a and b if method shown + line on graph (c) Residuals = = =. (d) Andrew appears to be slowest (all residuals positive / all times longer than predicted by regression line) E Total TOTAL 60 method - ignore sign, allow read from graph one correct - ignore sign.7 ( 4).6 ( 4). ( 4) all correct, including sign [9]

10 MSA (cont) Graph for Question 7 y 60 E A 40 0 C 00 C A E C 40 E 0 A 00 C x [0]

11 abc General Certificate of Education Specimen Unit Advanced Subsidiary Examination MATHEMATICS Unit Statistics B MSB In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables; a sheet of graph paper for use in Question 6; a ruler. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MSB. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maximum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. []

12 Jeremy sells a magazine which is produced in order to raise money for homeless people. The probability of making a sale is 0.09 for each person he approaches. (a) Given that he approaches 40 people, find the probability that he will make: (i) or fewer sales; (ii) more than 5 sales. ( marks) ( marks) (b) Find the probability that he will make two sales given that he approaches 6 people.( marks) (c) State one assumption you have made in answering parts (a) and (b). ( mark) (a) A sample of people, who commute regularly from a town in Surrey into London, was asked for an estimate of the time taken on their most recent journey. The replies are summarised below. Time Frequency (minutes) Calculate estimates of the mean and the standard deviation of these times. (5 marks) (b) A sample of people who commute regularly from a town in Essex into London was also asked for an estimate of the time taken on their most recent journey. Their answers had a mean of 64 minutes and a standard deviation of minutes. Compare, briefly, the journey times estimated by commuters from the two towns. ( marks) (c) Give two reasons why the data presented in parts (a) and (b) may not adequately represent typical commuting times from the two towns. ( marks) []

13 A cricket team meets for fielding practice. One exercise consists of a cricket ball being thrown at different heights, speeds and angles to one side of a fielder who tries to catch it one handed. Each member of the team attempts 5 catches with each hand. The number of successful catches are given in the following table. Fielder A B C D E G H I J K L Left hand Right hand (a) Calculate the value of the product moment correlation between the number of catches with the left hand and the number of catches with the right hand. ( marks) (b) Comment on the performance of fielders E and J. ( marks) (c) When fielders E and J are omitted from the calculation, the value of the product moment correlation coefficient between the number of left-handed and the number of right-handed catches is 0.8, correct to three decimal places. Comment on this value and the value you calculated in part (a) ( marks) 4 The weights of the contents of jars of honey may be assumed to be normally distributed with the standard deviation. grams. The weights of the contents, in grams, of a random sample of eight jars were as follows: (a) Calculate a 95% confidence interval for the mean weight of the contents of all jars. (6 marks) (b) On each jar it states Contents 454 grams. Comment on this statement using the given sample and your results in part (a). ( marks) TURN OVER FOR THE NEXT QUESTION [] Turn over

14 4 5 Last year the employees of a firm either received no pay rise, a small pay rise or a large pay rise. The following table shows the number in each category, classified by whether they were weekly paid or monthly paid. No pay rise Small pay rise Large pay rise Weekly Paid Monthly paid 4 8 A tax inspector decides to investigate the tax affairs of an employee selected at random. D is the event that a weekly paid employee is selected. E is the event that an employee who received no pay rise is selected. E is the event not E. (a) Find the value of: (i) P (D); (ii) P (D E ) ; (iii) P (D E ). (5 marks) (b) The tax inspector now decides to select three employees. Find the probability that they are all weekly paid if: (i) one is selected at random from those who had no pay rise, one from those who had a small pay rise and one from those who had a large pay rise; ( marks) (ii) they are selected at random (without replacement) from all the employees of the firm. ( marks) 6 [A sheet of graph paper is provided for use in this question.] Andrew (A), Charles (C) and Edward (E) are employed by the Palace Hotel. Each is responsible for one floor of the building and their duties include cleaning the bedrooms. The number of bedrooms occupied on each floor varies from day to day. The following table shows 0 observations of the number, x, of bedrooms to be cleaned and the time taken, y minutes, to carry out the cleaning. The employee carrying out the cleaning is also indicated. Employee A C E E C A A E C C x y [4]

15 5 (a) Plot a scatter diagram of the data. Identify the employee by labelling each point.( marks) (b) Calculate the equation of the regression line of y on x. Draw the line on your scatter diagram. (6 marks) (c) Use your regression equation to estimate the time which would be taken to clean 8 bedrooms. ( mark) (d) Calculate the residuals for the three observations when Andrew did the cleaning. ( marks) (e) Modify your estimate in part (c), given that the 8 bedrooms are to be cleaned by Andrew. ( marks) 7 A gas supplier maintains a team of engineers who are available to deal with leaks reported by customers. Most reported leaks can be dealt with fairly quickly but some require a long time. The time (excluding travelling time), X, taken to deal with reported leaks is found to have a mean of 65 minutes and a standard deviation of 60 minutes. (a) Assuming that the times may be modelled by a normal distribution, find the probability that it will take: (i) more than 85 minutes to deal with a reported leak; (ii) between 50 minutes and 5 minutes to deal with a reported leak. ( marks) (4 marks) (b) The mean of the times taken to deal with each of a random sample of 90 leaks is denoted by X. (i) State the distribution of X. (ii) Find the probability that X is less than 70 minutes. ( marks) ( marks) (c) A statistician consulted by the gas supplier stated that, as the times had a mean of 65 minutes and a standard deviation of 60 minutes, the normal distribution would not provide an adequate model. (i) Explain the reason for the statistician s statement. ( marks) (ii) Give a reason why, despite the statistician s statement, your answer to part (b)(ii) is still valid. ( marks) END OF QUESTIONS [5]

16 abc MSB Specimen Question Solution Marks Total Comments (a)(i) Binomial n = 40 p = 0.09 P( or fewer ) = (ii) P(>5) = P(5 or fewer) = = (b) P() = (6 5/) = (c) probabilities independent/people selected E at random/equivalent Total 9 (a) Class mid-mark Frequency x = 6. s = 5. (b) Journeys from Surrey have similar duration, on average, but are less variable than those from Essex. (c) People asked may not be representative. Times are estimated not measured. AA 5 E E E E Total 9 Allow m for mean and s.d. if method shown. 6. (6. 6.) 5. (5.0-5.) Or any other sensible comments e.g. journey time not defined, weather conditions may be extreme etc (a) B allow M if method shown (b) E and J held more catches with left than with right hand - all others held more with right than left. E E (c) Correlation coefficient of 0.8 suggests that those who caught a lot of catches with one hand also caught a lot of catches with the other. When E and J (possibly left handers) are included the correlation coefficient of suggests no association between the number of catches with each hand. E E Total 7 [6]

17 MSB (cont) Question Solution Marks Total Comments 4(a) x = % confidence interval for mean ±.96./ ± M 6 (b) The confidence interval provides evidence that the mean contents are greater than 454 grams. However the sample shows that some jars will contain less than 454 grams. E E E Total 9 5(a)(i) 5/50 = acf (ii) 5/9 = 0.86 acf (iii) 90/50 = 0.6 acf E confidence interval refers to mean contents E evidence mean >454 E some individual contents <454 (b)(i) 5/9 85/9 5/8 = 0.4 (ii) 5/50 4/49 /48 = Total 0 6(a) See graph on next page (b) y = x BB Allow for a and b if method shown x = 8 y = 96. x = y = line on graph (c) , allow 90 (d) Residuals = = =. method - ignore sign, allow read from graph one correct - ignore sign.7 ( 4).6 ( 4). ( 4) all correct, including sign (e) = 05 Total 5 Any sensible method 0 [7]

18 MSB (cont) Graph for Question 6 y 60 E A 40 0 C 00 C A E C 40 E 0 A 00 C x [8]

19 MSB (cont) Question Solution Marks Total Comments 7(a)(i) ( 85 65) z = = P(X > 85) = = (ii) ( 50 65) (b)(i) (ii) z = = 60 ( 5 65) z = = P( 50 < X < 5)= ( ) = Normal, mean 65, s.d. 60/ 90 = 6. ( 70 65) z = 60 = Probability mean of 90 less than 70 is (c)(i) Mean is only a little more than one standard deviation above zero. For normal this implies substantial proportion of times would be negative. This is impossible so model must be inadequate m 4 E normal may be implied in (b)(ii) E (ii) Mean of large sample will be approximately normally distributed even if parent distribution is not. E E Total 6 TOTAL 75 [9]

20 General Certificate of Education Specimen Unit Advanced Level Examination MATHEMATICS Unit Statistics A MSA abc In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 5 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MSA. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maximum mark for this paper is 60. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [0]

21 Answer all questions. For her st birthday present, Joanne wishes to have a course of driving lessons. In order to select the better driving school available in her area, she decides to compare the recent performances of participants taking lessons at two driving schools P and Q. These performances are tabulated below. School P School Q Pass 00 0 Fail 6 4 Use a χ test, at the 0% level of significance, to determine whether there is an association between the performance of participants and driving school. (9 marks) The number of vehicles arriving at a toll bridge during a 5-minute period can be modelled by a Poisson distribution with mean.6. (a) State the value for the standard deviation of the number of vehicles arriving at a toll bridge during a 5-minute period. ( mark) (b) Find: (i) the probability that at least vehicles arrive in a 5-minute period; ( marks) (ii) the probability that at least vehicles arrive in each of three successive 5-minute periods. ( marks) (c) Show that the probability that no vehicles arrive in a 0-minute period is , correct to four decimal places. ( marks) []

22 At a cinema, the time, T minutes, that customers have to wait in order to collect their tickets has the following probability density function. t 8 f( t ) = ( 5 t ) t < t 5 otherwise (a) Write down the value of P(T = 4). (b) Show that the median waiting time is minutes. ( mark) ( marks) (c) Find the probability that customers have to wait for less than 4 minutes in order to collect their tickets. (4 marks) (d) Calculate the mean time that customers have to wait in order to collect their tickets. (4 marks) 4 The random variable X has the following distribution. x P(X = x) (a) Calculate values for E(X) and Var(X). ( marks) (b) A rectangle has sides of length X and X (i) Find values for the mean and variance of the area of the rectangle. (4 marks) (ii) By tabulating the distribution for Y X + 64 X =, or otherwise, show that E ( ) = 4. 5 Y. ( marks) (iii) Hence find the mean value for the perimeter of the rectangle. ( marks) Turn over []

23 4 5 Charles is an athlete specialising in throwing the javelin. During practice, he throws a javelin the following distances, in metres (a) Calculate unbiased estimates of the mean and the variance of the distance thrown. ( marks) (b) (i) Hence calculate a 95% confidence interval for the mean distance thrown. (ii) State two assumptions that you make in order to do this calculation. (5 marks) ( marks) 6 A random variable X is normally distributed with mean µ and variance The null hypothesis H 0 : µ = 50 is to be tested against the alternative hypothesis H : µ 50, using the 5% level of significance. The mean, X, of a random sample of 40 observations of X is to be used as the test statistic. (a) Write down the distribution of X assuming H 0 is true. ( marks) (b) Explain what is meant by: (i) a Type I error; (ii) a Type II error. (c) Write down the probability of a Type I error. (d) Calculate the acceptance region for X, giving the limits to two decimal places. ( marks) ( mark) (5 marks) END OF QUESTIONS []

24 MSA Specimen abc Question Solution Marks Total Comments H : no association Calculate E i : 0 6 = = = = α = O i Ei () 0.5 ν = χ =.706 α α E i total =.4 Reject H 0 at the 0% level Any correct method Use of ( O E) E Attempted use of Yates correction awfw.4 to.5 Evidence at the 0% level to suggest an association between performance and the driving school selected. E Any correct interpretation Total 9 [4]

25 MSA (cont) Question Solution Marks Total Comments (a) Standard deviation =. 6 = (b)(i) λ =.6 for a 5-minute period Let Y = no. of vehicles arriving in a 5-minute period then Y ~ (. 6 ) P 0 and P(Y )=0.07 P(Y )= P (Y ) = 0.07 = = (ii) p = P(at least arrive in succ. 5 min) =( ) = = 0.9 for p() i (c) let X = no. vehicles arriving in a 0-minute period then X ~ ( 7. ) P 0 P X = 0 = e and ( ) 7. = Total 7 ag [5]

26 MSA (cont) Question Solution Marks Total Comments P T = 4 = 0 (a) ( ) (b) Median = P T = P T > = 0.5 ( ) ( ) (c) P ( T > ) = f ( ) ( 5 ) 0. 5 = 0.5 P( T < 4) = P( T = = 8 7 = 8 4) 4 4 or t t dt = = (d) E t = d 4 8 t ( T ) dt + t ( 5 t) 0 5 = 4 t 7 8 = 4 0 5t + 8 t [( ) ( )] = m 4 Total Integration attempted (.958) [6]

27 MSA (cont) Question Solution Marks Total Comments 4 (a) 7 E( X ) = = ( X ) = E = 0.5 ( X ) = ( X ) ( X) Var E E = = 4.5 (b)(i) Area = 64 A= X + = X + 64 X ( A) ( X ) ( X) E = E + 64 = E + 64 = = 7 ( A) = ( X + ) Var Var 64 = 4Var( X ) = = 8 4 (ii) Y = X + 64 X Y P(Y=y) E ( Y ) = = (iii) P = Y + 4 E ( P ) = E( Y ) + 4 = = 5 Total [7]

28 MSA (cont) Question Solution Marks Total Comments 5(a) = 4 and x = 5. 4 (b)(i) x 4 ˆ µ = x = = = n 8 s = =.9 Confidence Interval for µ t s = x ± n Degrees of freedom = v = 8 = 7 ( ) 65 t = 7. CI forµ is awfw.90 to.9 (s =.8) ± = ± = (9., 4.7) accept (9./4, 4.6/7) (ii) distances are normally distributed distances are independent accept random sample Total 0 [8]

29 MSA (cont) Question Solution Marks Total Comments (a) X ~ N 50, ~ N( 50, 0.06) (b)(i) Type I error: Reject H 0 when H 0 true for Normal and 50 idea of σ ; correct n or equivalent (ii) Type II error: Accepting H0 when H0 incorrect (c) P( Type I error) = 0.05 or equivalent (d) Acceptance region: z =±.96 σ = n ± = ( 49.75,50.5 ) 5 Total TOTAL 60 [9]

30 General Certificate of Education Specimen Unit Advanced Level Examination MATHEMATICS Unit Statistics B MSB abc In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MSB. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maximum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [0]

31 Answer all questions. The number of strikes per game, obtained by a tenpin bowler, can be modelled by a Poisson distribution with mean λ. For the first game played, λ = 0. and, for each subsequent game played, λ =.. A match consists of three consecutive games. (a) Write down the distribution of T, the total number of strikes obtained by a tenpin bowler in a match. ( marks) (b) Write down the value of Var(T ). ( mark) (c) Find the probability that, in a match, the tenpin bowler will obtain a total of between and 5 strikes, inclusive. ( marks) Charles is an athlete specialising in throwing the javelin. During practice, he throws a javelin the following distances, in metres (a) Calculate unbiased estimates of the mean and the variance of the distance thrown. ( marks) (b) (i) Hence calculate a 95% confidence interval for the mean distance thrown. (5 marks) (ii) State two assumptions that you need to make in order to do this calculation. ( marks) The results of a recent police survey of traffic travelling on motorways produced information about the genders of drivers and the speeds, S miles per hour, of their vehicles, as tabulated below. S < S 90 S > 90 Male Female Investigate, at the % level of significance, the claim that there is no association between the gender of the driver and the speed of the car. ( marks) []

32 4 The continuous random variable X has a rectangular distribution with the following probability density function, where k is a constant. f ( x ) 6 = 0 6 x k otherwise (a) State the value of k. ( mark) (b) Hence find values for the mean, µ, and the variance, σ, of X. ( marks) (c) Determine ( X > µ + σ ) P. ( marks) 5 At a cinema, the time, T minutes, that customers have to wait in order to collect their tickets has the following probability density function. t 8 f ( t ) = (5 t) t < t 5 otherwise (a) Write down the value of P ( T = 4). ( mark) (b) Show that the median waiting time is minutes. ( marks) (c) Find the probability that customers have to wait for less than 4 minutes in order to collect their tickets. (4 marks) (d) Calculate the mean time that customers have to wait in order to collect their tickets. (4 marks) TURN OVER FOR THE NEXT QUESTION Turn over []

33 4 6 The random variable X has the following probability distribution. x 4 8 ( X = x) P (a) Calculate values for E ( X ) and ( X ) 6 Var. ( marks) (b) A rectangle has sides of length X and X (i) Find values for the mean and variance of the area of the rectangle. (4 marks) (ii) By tabulating the distribution for Y X + 64, X = or otherwise, show that ( Y ) = 4. 5 E. ( marks) (iii) Hence find the mean value for the perimeter of the rectangle. ( marks) 7 An instrument for measuring the speed of passing motorists is tested by a police force. A car is driven at known speeds along a straight road. The error (speed recorded by the instrument actual speed of the car) is observed on eight occasions with the following results in metres per second Assuming these data to be a random sample from a normal distribution, investigate, at the 5% level of significance, whether the instrument is biased (i.e. whether the mean error differs from zero). (0 marks) []

34 5 8 A random variable X is normally distributed with mean µ and variance The null hypothesis, H 0 : µ = 50, is to be tested against the alternative hypothesis, H : µ 50, using the 5% level of significance. The mean, X, of a random sample of 40 observations of X is to be used as the test statistic. (a) Write down the distribution of X assuming H 0 is true. ( marks) (b) Explain what is meant by: (i) a Type I error; (ii) a Type II error. (c) Write down the probability of a Type I error. ( marks) ( mark) (d) Calculate the acceptance region for X, giving the limits to two decimal places. (5 marks) END OF QUESTIONS [4]

35 MSB Specimen abc Question Solution Marks Total Comments (a) X = number of strikes per game then: X ~ Po(0.) X ~ Po(.) X ~ Po(.) T = X + X + X ~ Po(.4) (b) Var (T ) =.4 (c) (a) (b)(i) P( T 5) = P( T = or 4 or 5) = P( T 5) P(T ) = = = 4 and x = 5. 4 x ˆ 4 µ = x = = = 40.5 n 8 s = =.9 Confidence Interval for µ t s = x ± n Degres of freedom = ν = 8 = 7 t 7 (0.975) =.65 awfw 0.94 to 0.95 Total 6 awfw.90 to.9 (s =.8) Use of (ii) CI for µ is = 40.5 ± 8 = 40.5 ±.54 = (9., 4.7) distances are normally distributed distances are independent 5 Total 9 Accept (9./4, 4.6/7) Accept random sample [5]

36 MSB (cont) Question Solution Marks Total Comments H 0 : Gender of driver and speed of car are independent H : Gender of driver and speed of car are associated ν = ( ) ( ) = O i E i ( O E ) (5.4980) (0.0409) (.5565) (9.556) (0.07) (6.95) i E i i A χ % () = 9.0 χ calc = 5.0 (4.89) awfw 4.5 to 5.5 reject H 0 There is (very strong) evidence at the % level to suggest an association between the gender of the driver and the speed of the car. E Total [6]

37 MSB (cont) Question Solution Marks Total Comments 4 (a) k = (b) µ = 9 σ = (c) P(X > µ + σ ) ( > 9 ) = P X = 6 = 0. Total 5 6 5(a) P(T = 4) = 0 (b) Median = P ( T ) = P( T > ) = 0.5 P ( T > ) = f () (5 ) = 0.5 = 0.5 or 0 t t dt = = (c) (d) P(T < 4) = P(T 4) = 4 = 8 7 = 8 E( T ) = 0 t dt t(5 t) dt 4 4 t = t + 8 t [( ) ( )] = + = m 4 Total Integration attempted (.958) [7]

38 MSB (cont) Question Solution Marks Total Comments 6 (a) 7 E ( X ) = = 4 6 E ( X ) = = Var (X ) = E(X ) E (X ) = = 4.5 (b)(i) (ii) 64 Area = A = X + = X + 64 X E ( A ) = E(X + 64) = E( X ) + 64 = = 7 Var ( A ) = Var(X + 64) = 4Var( X ) = = 8 Y = X + 64 X y P(Y = y) (iii) E ( Y ) = = 4.5 P = Y E ( P ) = E( Y ) + 4 = = 5 Total [8]

39 MSB (cont) Question Solution Marks Total Comments 7 x = x n = 8 = 8 s = s = (a) (b)(i) H H 0 : µ = 0 : µ 0 0 t = = t 7 = ±.65 Accept H 0 No significant evidence of bias 0 Total X ~ N 50, ~ N(50, 0.06) 40 Type I error: Reject H 0 when H 0 true For Normal & 50 Idea of oe σ n ; correct (ii) Type II error: Accepting H 0 when H 0 incorrect oe (c) P( Type I error ) = 0.05 (d) Acceptance region: z = ±. 96 σ = 0.65 n 50 ± = (49.75, 50.5) 5 Total TOTAL 75 [9]

40 General Certificate of Education Specimen Unit Advanced Level Examination abc MATHEMATICS Unit Statistics MS0 In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MS0. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maximum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [40]

41 Answer all questions. In a local government election, 40% of adults voted. The random variable V denotes the number who voted out of a random sample of 00 adults. (a) (i) Write down the probability distribution for V. ( mark) (ii) Write down an appropriate distributional approximation for V. ( mark) (iii) Find an approximate value for P ( V > 90).. ( marks) (b) Only 5% of those who voted in the election voted for candidate A. (i) Find the probability that an adult, chosen at random, voted for candidate A. ( mark) (ii) Using a suitable distributional approximation, find the probability that, out of a random sample of 00 adults, more than 5 voted for candidate A. ( marks) Over a period of one year, a greengrocer sells tomatoes at six different prices (x pence per kilogram). He calculates the average number of kilograms, y, sold per day at each of the six different prices. His results are as follows. x y (a) Calculate the value of the product moment correlation coefficient. ( marks) (b) (i) Assuming that these data are a random sample from a bivariate distribution with correlation coefficient ρ, investigate, at the % level of significance, the hypothesis that ρ < 0. (4 marks) (ii) Interpret your result in the context of this question. ( mark) [4]

42 To travel to school in the morning, Sam either catches the bus, cycles or gets a lift with her father. The probabilities for these three modes of travel are 0., 0. and 0.4, respectively. The probability that Sam is late for school is: 0.4 when she catches the bus; 0.5 when she cycles; 0. when she gets a lift with her father. (a) Find the probability that, on a particular morning, Sam cycles to school and arrives late. ( marks) (b) Sam is late for school one morning. Find the probability that she cycled to school that morning. (4 marks) 4 (a) The random variable X has a Poisson distribution with parameter λ. Prove that the mean of X is λ. ( marks) (b) The daily number of missed appointments at a dental surgery can be modelled by a Poisson distribution with mean.8. In an attempt to reduce this figure, the surgery publicised the introduction of fines for missed appointments. (i) During the first 5 days after the fines were introduced, there was a total of 6 missed appointments. Use a test with the 0% level of significance to show that there is insufficient evidence to conclude that the introduction of fines has succeeded in its aim. (5 marks) (ii) A dentist at the surgery suggested that the number of missed appointments should be monitored over a much longer period of time. Subsequently, there was a total of 97 missed appointments during 80 days. Test, again at the 0% level of significance, the claim that the mean daily number of missed appointments is now less than.8. (8 marks) TURN OVER FOR NEXT QUESTION Turn over [4]

43 4 5 The random variable L has mean 65 and variance 6. The random variable S has mean 50 and variance 9. Variables L and S are independent and the random variable D is given by D = 4L 5S (a) (i) Show that D has mean 0. ( marks) (ii) Find the value of the variance of D. ( marks) (b) The weights of large eggs are normally distributed with mean 65 grams and standard deviation 4 grams. The weights of standard eggs are normally distributed with mean 50 grams and standard deviation grams. One large egg and one standard egg are chosen at random. Find the probability that the 4 weight of the standard egg is more than of the weight of the large egg. (5 marks) 5 6 It is claimed that women are faster than men at solving anagrams. To investigate the claim, the same list of anagrams was given to random samples of 0 men and 5 women under identical experimental conditions. The time, in minutes, taken by each person to solve the anagrams was recorded. The results are summarised in the table below, together with known values for the population standard deviations. Sample Size Sample Mean Population Standard Deviation Men (M) minutes. minutes Women (F) minutes.8 minutes (a) A hypothesis test is to be applied to these data to determine whether, on average, men take longer than women to solve anagrams. Stating null and alternative hypotheses concerning the population means using the 5% level of significance, find the critical region of this test. µ M and µ F, and Hence determine whether the evidence supports the claim that women are faster than men at solving anagrams. (8 marks) (b) If the alternative hypothesis states that µ M µ F = 0., find the power of the test. (5 marks) [4]

44 5 7 A company is introducing a new brand of toothpaste. The company s marketing department is investigating people s opinions of the new brand by giving out free samples and asking people to compare the new brand with their usual brand. (a) From a random sample of 00 people, 9 preferred the new brand of toothpaste to their usual brand. (i) Calculate an approximate 98% confidence interval for the proportion of people who prefer the new brand to their usual brand. (6 marks) (ii) Comment on the evidence for claiming that the majority of people prefer the new brand. ( mark) (b) In a further survey, 00 people aged under 5 and 50 people aged 5 or over were asked their opinion. The results are summarised below. Age (years) Sample size Proportion who preferred new brand Under or over (i) Calculate an approximate 98% confidence interval for the difference between the proportions of people aged under 5 and those aged 5 or over who prefer the new brand of toothpaste. (6 marks) (ii) Hence determine whether people aged under 5 are more likely than those aged 5 or over to prefer the new brand of toothpaste. ( mark) END OF QUESTIONS [44]

45 abc MS0 Specimen Question Solution Marks Total Comments (a)(i) V ~ B(00, 0.4) (ii) ~ N( 80,48) V (iii) P(V > 90) =P z > 48 = P ( z >.5) = = ( s f) Condone no continuity correction (.44) to ; cao (b)(i) P(voted for A) = = 0. 0 cao (ii) X ~ B(00, 0.0) Poisson(4) P(X > 5) = P(X 5) = 0.5 (a) r = 0.96 Total 9 B Use of Poisson awfw 0.96 to 0.96 Allow M if method shown (b)(i) H 0 : ρ = 0 H : ρ < 0 n = 6, α = % Critical value of coefficient = < 0.88 Reject H 0 at % level. Evidence suggests that ρ < 0 4 Both Accept positive value (ii) Fewer tomatoes are sold when price is higher. E Total 8 [45]

46 MS0 (cont) Question Solution Marks Total Comments (a) P(cycles and late) = = 0.5 (b) Denoting events by: A: Sam catches the bus, B: Sam cycles, C: Sam gets a lift, L: Sam is late for school, P(B L) = P(L B) P(B) P(L A) P(A)+P(L B) P(B)+P(L C) P(C) = ( ) + ( ) + ( ) 0.5 = = = Total 6 4(a) E X = r P X = r ( ) ( ) = r= 0 r= = λe r e λ λ n= 0 λ r λ λ r! λ n! = λe e = λ (b)(i) H 0 : Mean rate unchanged (or µ = 9) H : Mean rate decreased (or µ < 9) Under H 0, X ~ Poisson(9) P(X 6) = > 0% so insufficient evidence of decrease. (ii) If rate is unchanged at.8 per day, Y ~ Poisson(4) N(4, 4) P ( Y 97) = P Z 97 4 n 4 = P( Z.5) = 0.99 = ( s.f.) < 0% Evidence suggests that the mean daily rate is now less than.8. E 5 8 Total 6 Use of Bayes theorem with correct numerator and sum of probabilities in denominator. All four terms correct awrt Attempt at summation of products of r and P(r). where n = r Convincing use of series to prove result. Both; or equivalent Awrt 0.07 Must show comparison Use of normal approximation with correct mean & variance or, with continuity correction P( Z.47) = 0.07 (awrt) awrt Accept calculation of sample z-value and comparison with.8 [46]

47 MS0 (cont) Question Solution Marks Total Comments 5(a)(i) E(D) = 4E(L) 5E(S) = = 0 (ii) Var(D) = 4 Var(L) + 5 Var(S) = = 48 (b) L ~ N(65, 4 ); S ~ N(50, ) P S > 4 L = 4 P S L > = P ( 5S 4L > 0) = P( D < 0) D ~ N(0, 48) P( D < 0) = P Z < = P( Z < 0.456) = 0. to Total 9 6(a) H 0 : µ M = µ F H : µ M > µ F Under H 0, using central limit theorem, X M X F ~ N..8 0, = N(0, ) One-tailed test at 5% level so z =.6449 Critical region is X M X F > ie. X M X F > 0. 4 on part (a) (ii) both may be implied awfw 0.4 to 0.4 From the sample data, x M x F = 0. 5 Reject H 0. The evidence supports the claim. (b) H : µ M µ F = 0. Under H, X M X F ~ N(0., ) Power of test = P ( X M X F > 0.4 H ) = P Z > = Φ(0.479) = on critical region may be implied on critical value awfw 0. to 0. Total [47]

48 MS0 (cont) Question Solution Marks Total Comments 7(a)(i) 9 p ˆ = = cao 98% confidence interval so z =.6 Standard error = % confidence limits are 0.4 ± giving (0.6 to 0.64, to 0.497) 6 on pˆ and se (ii) Confidence interval lies below 50% so the claim is not supported. (b)(i) Denoting proportions of younger and older people by p X and p Y respectively, p ˆ pˆ = 0.09 X Y Estimated standard error for p p is = % confidence limits for p X py : 0.09 ± giving (-0.088, 0.99) (ii) Zero lies within confidence interval so not enough evidence for larger proportion. X Y 6 E Total 4 TOTAL 75 cao [48]

49 General Certificate of Education Specimen Unit Advanced Level Examination MATHEMATICS Unit Statistics 4 MS04 abc In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MS04. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maximum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [49]

50 Answer all questions. When an automatic bottling machine is working properly, the volume of liquid delivered into each bottle can be assumed to be normally distributed with standard deviation 0.5 ml. On a certain day, the Quality Control Manager believes that the standard deviation is larger than this. He therefore measures the contents of six randomly selected bottles with the following results. Contents (ml) Investigate the Quality Control Manager s belief at the 0% level of significance. (9 marks) The times, T minutes, between successive telephone calls to a small business can be assumed to be exponentially distributed with mean 0. (a) (i) Write down, in full, the probability density function of T. ( marks) (ii) Using integration, obtain an expression, valid for t 0, for the distribution function of T. ( marks) (b) A call is received at.00 am. Find the probability that the next call is received: (i) before.5 am; (ii) between.0 am and.0 am. ( marks) ( marks) Independent observations X, X and X are made on the random variable X which has mean µ and variance σ. (a) Determine which of the following random variables are unbiased estimators for µ. R = X + X + 6 X S = 5 X + 0 X + 5 X T = 4 X + X (b) Calculate the variance of those estimators which are unbiased. 4 X (5 marks) ( marks) (b) Hence state, giving a reason, which of these estimators is the best unbiased estimator for µ. ( mark) [50]

51 4 Eight patients have their pulse rates, in beats per minute, measured before and after receiving a certain treatment. The results, including the decrease in pulse rate for each patient, are given in the table. Patient A B C D E F G H Before After Decrease Using a t-test, investigate, at the 5% significance level, whether this treatment decreases, on average, patients pulse rates. (9 marks) 5 A six-sided die has the numbers,,, 4, 5 and 6 on its faces. The random variable R denotes the number of independent throws of the die needed to first obtain a score of either 5 or 6. Assuming that the die is unbiased, the probability model for R is: P(R = r) = 0 ( )( ) r r =,,,... otherwise. The frequency table below shows the results of 4 observations of R. Number of throws, r Frequency, f Using a χ goodness of fit test and the 5% level of significance, test the hypothesis that the die is unbiased. ( marks) Turn over [5]

52 4 6 The random variable X follows the probability distribution where p + q =. (a) Prove that: P(X = r) = r q p for r =,. Ε Χ = ; p (i) ( ) ( marks) (ii) Var ( ) q X =. (5 marks) p (b) In the case when p = 0.4, find the value of n such that P (X > n) < 0.00 (4 marks) 7 Radha, a gardener, buys tomato plants: 6 of variety X and 6 of variety Y. She plants these in her greenhouse and she keeps a record of the total yield, in kilograms, from each plant. She obtains the following results. Variety X Variety Y You may assume that these are random samples from populations distributed N( µ X, σ X ) and N( µ Y, σ Y ) respectively. (a) Test, at the 5% significance level, the hypotheses H 0 : σ X = σ Y against H : σ X σ Y. (7 marks) (b) (i) Obtain a 95% confidence interval for µ X µ Y. (7 marks) (ii) State, giving a reason, whether your result in part (b)(i) indicates a difference between the mean yields of the two varieties. ( marks) END OF QUESTIONS [5]

53 MS04 Specimen abc Question Solution Marks Total Comments H0 : σ = 0.5 or σ = 0. 5 both H : σ > 0.5 or σ > 0. 5 Significance level, α =0.0 (0%) Degrees of freedom, v = 6 = 5 cao Critical value, χ = 9.6 awfw 9. to 9.4 (a) (i) f (t) = (ii) x =.4 and x = (n )s = ( x ) x = n 5.49 use of, or equivalent accept n rather than (n ) awfw 5.49 to 5.50 s =.0986 s =. 048 σ = σ = χ = ( n ) s use of, or equivalent 5.49 = = σ 0.5 on calculation.97 awfw.9 to.0 evidence, at 0% level, to support manager s belief 9 on χ and CV Total 9 0 e t 0 0 t > 0 t < 0 t > 0 f (t) = 0 t < 0 cao 0 andboth ranges t F(t) = e 0 dx = 0 e 0 x 0 t 0 x = e t 0 (b) (i) Require P (T<5) = F (5) = to (ii) Require P (0< T< 0) = F(0) F (0) = = = 0. to 0. Total 9 cao use of f ( x) dx condone use of t integral & limits cao Use of F(5), or integration awfw use of difference of Fs or equivalent awfw [5]

54 MS04 (cont) Question Solution Marks Total Comments (a) E 6 ( R ) E X + X + X = = E 6 ( X ) + E( X ) + ( X ) = µ + µ + µ 6 E E E ( S ) = µ + µ µ = µ ( T ) = µ + µ µ R and T are unbiased 5 (b) Var( R ) = Var( X ) + Var( X ) + ( X ) 9 Var( T ) = Var 4 6 = σ + σ + σ = 7 σ σ + σ σ = 7 σ 8 (c) R best (smaller variance) E Total 9 4 H µ 0 H : µ 0 Both 0 : 0 = D > ν = 8 = 7 cao [54] Used on any one of R, S, T Use of any answer from (a) t =.895 awfe.89 to.90 crit 7. d d = = d = 7. =.75 awfw. to.4 8 d = = 7.75 S awrt 7.7 ( S =.67 / 8) t calc = use of on d x and S =.5 to.6 awfw, accept.5 or.6 evidence, at 5% level, that mean pulse 9 on tcalc and tcrit rate decreases Total 9

55 MS04 (cont) Question Solution Marks Total Comments 5 H 0 : biased H : unbiased At least H 0 ν = 6 = 5 cao χ =.070 awrt. crit r o P( R = r) e ( o e) e Probs to Prob r = e = 4p A, for all 6 e awrt integer for 4 or 5 e awrt integer ( o e) use of χ calc = = 7.5 to e awfw no evidence, at 5% level, that die is on χ biased Total calc and χ crit [55]

56 MS04 (cont) Question Solution Marks Total Comments 6(a) Ε ( x ) =.p + pq +... ) p = ( + q + q +...) = p p = p ( q) = p Accept Methods Using M.G.F.or P.G. F., which may be quoted, without proof, if known. (ii) E ( x )... =. p +. pg + pg = p ( + 4g + 9q +...) = p {( + q + 6q +... ) + q ( + q + 6q +...) = p {( ) ( q q + q ) } Var () = + p p = p q (b) n 0.4( 0.6 ) ( 0.6) n 0.6 < 0.00 log 0.00 n > log trials required > ( =.5) use of M if written down immediately Total [56]

57 MS04 (cont) Question Solution Marks Total Comments 7 x = 4.6 x = y 7.9 y = = (a) H : σ = σ H σ σ 0 x y : x y ν = ν 5 cao both = F crit = 7.5 cao (accept ) = (b)(i) (ii) S x = = awfw to S y F calc = = = =.7 to No evidence, at 5% level, to reject equal variances ˆ σ = awfw.650 to.65 accept.65 use of awfw 7 on F calc and F crit use of =.05 to.5 awfw x y = 0.55 cao ν = = 0 cao ( 0.975). 8 t awfw. to. 0 = CI is 0.55 ± (-.0 to.0, 0.9 to 0.9) 7 awfw No, since the CI contains zero on CI E Total 6 TOTAL 75 [57]