General Certificate of Education Advanced Subsidiary Examination June 2013

Transcription

1 General Certificate of Education Advanced Subsidiary Examination June 2013 Mathematics Unit Statistics 1B Statistics Unit Statistics 1B Friday 17 May am to am For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed 1 hour 30 minutes MS/SS1B Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Fill in the boxes at the top of this page. Answer all questions. Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. You must answer each question in the space provided for that question. If you require extra space, use an AQA supplementary answer book; do not use the space provided for a different question. Do not write outside the box around each page. Show all necessary working; otherwise marks for method may be lost. Do all rough work in this book. Cross through any work that you do not want to be marked. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Condensed Information The marks for questions are shown in brackets. The maximum mark for this paper is 75. Unit Statistics 1B has a written paper only. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. You do not necessarily need to use all the space provided. P61084/Jun13/MS/SS1B 6/6/ MS/SS1B

2 2 1 The average maximum monthly temperatures, u degrees Fahrenheit, and the average minimum monthly temperatures, v degrees Fahrenheit, in New York City are as follows. Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Maximum (u) Minimum (v) (a) (i) Calculate, to one decimal place, the mean and the standard deviation of the 12 values of the average maximum monthly temperature. (2 marks) For comparative purposes with a UK city, it was necessary to convert the temperatures from degrees Fahrenheit ( F) to degrees Celsius ( C). The formula used to convert f F toc C is: c ¼ 5 ð f 32Þ 9 Use this formula and your answers in part (a)(i) to calculate, in C, the mean and the standard deviation of the 12 values of the average maximum monthly temperature. (3 marks) (b) The value of the product moment correlation coefficient, r uv, between the above 12 values of u and v is 0.997, correct to three decimal places. State, giving a reason, the corresponding value of r xy, where x and y are the exact equivalent temperatures in C ofu and v respectively. (2 marks) 2 The weight, X grams, of the contents of a tin of baked beans can be modelled by a normal random variable with a mean of 421 and a standard deviation of 2.5. (a) Find: (i) PðX ¼ 421Þ ; PðX < 425Þ ; (iii) Pð418 < X < 424Þ. (6 marks) (b) Determine the value of x such that PðX < xþ ¼0:98. (3 marks) (c) The weight, Y grams, of the contents of a tin of ravioli can be modelled by a normal random variable with a mean of m and a standard deviation of 3.0. Find the value of m such that PðY < 410Þ ¼0:01. (4 marks) (02) P61084/Jun13/MS/SS1B

3 3 3 An auction house offers items of jewellery for sale at its public auctions. Each item has a reserve price which is less than the lower price estimate which, in turn, is less than the upper price estimate. The outcome for any item is independent of the outcomes for all other items. The auction house has found, from past records, the following probabilities for the outcomes of items of jewellery offered for sale. Outcome Probability Item does not achieve its reserve price 0.15 Item achieves at least its reserve price 0.85 Item achieves at least its lower price estimate 0.50 Item achieves at least its upper price estimate For example, the probability that an item achieves at least its lower price estimate but not its upper price estimate is A particular auction includes exactly 40 items of jewellery that may be assumed to be a random sample of such items. (a) Use binomial distributions to find the probability that: (i) at most 10 items do not achieve their reserve prices; (1 mark) 25 or more items achieve at least their lower price estimates; (2 marks) (iii) exactly 2 items achieve at least their upper price estimates; (2 marks) (iv) more than 10 items but fewer than 15 items achieve at least their reserve prices but not their lower price estimates. (4 marks) (b) How many of the 40 items of jewellery would you expect to achieve at least their reserve prices but not their upper price estimates? (2 marks) Turn over s (03) P61084/Jun13/MS/SS1B

4 4 4 The girth, g metres, the length, l metres, and the weight, y kilograms, of each of a sample of 20 pigs were measured. The data collected is summarised as follows. S gg ¼ 0:1196 S ll ¼ 0:0436 S yy ¼ 5880 S gy ¼ 24:15 S ly ¼ 10:25 (a) Calculate the value of the product moment correlation coefficient between: (i) girth and weight; length and weight. (3 marks) (b) Interpret, in context, each of the values that you obtained in part (a). (3 marks) (c) Weighing pigs requires expensive equipment, whereas measuring their girths and lengths simply requires a tape measure. With this in mind, the following formula is proposed to make an estimate of a pig s weight, x kilograms, from its girth and length. x ¼ 69:3 g 2 l Applying this formula to the relevant data on the 20 pigs resulted in S xx ¼ 5656:15 S xy ¼ 5662:97 (i) By calculating a third value of the product moment correlation coefficient, state which of g, l or x is the most strongly correlated with y, the weight. (2 marks) Estimate the weight of a pig that has a girth of 1.25 metres and a length of 1.15 metres. (2 marks) (iii) Given the additional information that x ¼ 115:4 and y ¼ 116:0, calculate the equation of the least squares regression line of y on x, in the form y ¼ a þ bx. (3 marks) (iv) Comment on the likely accuracy of the estimated weight found in part (c). Your answer should make reference to the value of the product moment correlation coefficient found in part (c)(i) and to the values of b and a found in part (c)(iii). (4 marks) 5 Alison is a member of a tenpin bowling club which meets at a bowling alley on Wednesday and Thursday evenings. The probability that she bowls on a Wednesday evening is Independently, the probability that she bowls on a Thursday evening is (a) Calculate the probability that, during a particular week, Alison bowls on: (i) two evenings; exactly one evening. (3 marks) (04) P61084/Jun13/MS/SS1B

5 5 (b) David, a friend of Alison, is a member of the same club. The probability that he bowls on a Wednesday evening, given that Alison bowls on that evening, is The probability that he bowls on a Wednesday evening, given that Alison does not bowl on that evening, is The probability that he bowls on a Thursday evening, given that Alison bowls on that evening, is 1. The probability that he bowls on a Thursday evening, given that Alison does not bowl on that evening, is 0. Calculate the probability that, during a particular week: (i) Alison and David bowl on a Wednesday evening; (2 marks) Alison and David bowl on both evenings; (2 marks) (iii) Alison, but not David, bowls on a Thursday evening; (iv) neither bowls on either evening. (1 mark) (3 marks) 6 The weight, X kilograms, of sand in a bag can be modelled by a normal random variable with unknown mean m and known standard deviation 0.4. (a) The sand in each of a random sample of 25 bags from a large batch is weighed. The total weight of sand in these 25 bags is found to be kg. (i) Construct a 98% confidence interval for the mean weight of sand in bags in the batch. (5 marks) Hence comment on the claim that bags in the batch contain an average of 20 kg of sand. (2 marks) (iii) State why use of the Central Limit Theorem is not required in answering part (a)(i). (1 mark) (b) The weight, Y kilograms, of cement in a bag can be modelled by a normal random variable with mean and standard deviation A firm purchases 10 such bags. These bags may be considered to be a random sample. (i) Determine the probability that the mean weight of cement in the 10 bags is less than 25 kg. (4 marks) Calculate the probability that the weight of cement in each of the 10 bags is more than 25 kg. (4 marks) Copyright ª 2013 AQA and its licensors. All rights reserved. (05) P61084/Jun13/MS/SS1B

6 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A2,1 2 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

7 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments 1(a)(i) Mean = 62.2 to 62.3 B1 AWFW (62.25) SD = 17.4 to 17.6 or 16.7 to 16.9 B1 2 AWFW ( or ) Mean = to BF1 AWFW (16.806) F on (a)(i) only providing 45 < mean < 65 SD = 9.66 to 9.78 or 9.27 to 9.39 BF2 3 AWFW (9.733 or 9.319) F on (a)(i) only providing 10 < SD < 20 (b) r xy = B1 CAO Award on value only; ignore any explanation or working r xy = r uv with no value stated B0 r not affected by change(s) in/different units Accept Formula or It for r and reference to linear is not necessary or r not affected by linear scaling or B1 2 Accept Formula or It for r but reference to linear is necessary Scaling/coding/transformation/change/ conversion to u and v is linear Total 7 OE; but reference to linear is necessary 4

8 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments Accept percentage equivalents in (a) Weight, X ~ N(421, ) 2(a)(i) P(X = 421) = 0 or zero or nought or 0% B1 CAO; accept nothing else but ignore additional words providing that they are not contradictory (eg impossible so = 0) P(X < 425) = P Z < 2.5 Standardising 425 with 421 and 2.5 but allow ( ) = P(Z < 1.6) = to A1 AWRT ( ) (iii) P(418 < X < 424) = P( a < Z < a) = P(Z < a) (1 P(Z < a)) or 2 P(Z < a) 1 = ( ) = or = A1 OE; a = 1.2 or correct standardising are not required May be implied by (AWRT) seen anywhere or by a correct answer AWRT ( / ) Implied by a correct answer = to 0.77 A1 6 AWFW ( ) (b) 0.98 z = 2.05 to 2.06 B1 AWFW (2.0537) x = 2(.0) to 2.4 x = 426 to A1 3 Standardising x with 421 and 2.5 but allow (421 x); and equating to a z-value (ignore sign) AWFW (426.13) Must be consistent signs throughout (c) 0.01 z = 2.33 to 2.32 B1 AWFW; (ignore sign) ( ) z = 410 µ 3.0 or 2.5 Standardising 410 with µ and (3.0 or 2.5) but allow (µ 410) 410 µ Equating to a z-value (ignore sign) 3.0 = 2.6 to 2.3 A1 µ = 417 Adep1 4 Total 13 AWRT (416.98) Dependent on previous A1 Must be consistent signs throughout 5

9 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments Accept percentage equivalents except for 27 3(a)(i) O ~ B(40, p) P(NS 10) = 0.97 B1 1 AWRT (0.9701) P(LPE 25) = 1 ( or ) Requires 1 Accept 3 dp rounding Can be implied by ( to 0.077) but not by (0.04 to ) = A1 2 AWRT (0.0769) (iii) P(UPE = 2) = ( ) ( ) Correct expression; may be implied by a correct answer Ignore extra terms = A1 2 AWRT (0.0160) (iv) p = = 0.35 B1 CAO; award on value only May be implied by any of four probabilities below or by a correct answer P(10 < X < 15) = or (p 1 ) MINUS or (p 2 ) Accept 3 dp rounding Accept 3 dp rounding = 0.45 to A1 4 AWFW (0.4506) (b) p = = or p = B1 CAO; may be implied by 27 Each can be found in several ways CAO; may be implied by 13 or 27 Number = = 27 B1 2 CAO; can be found in several ways Total 11 6

10 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments 4(a)(i) in r gy = = 0.91 to (a)(i) or (a) or (c)(i) A1 AWFW ( ) r ly = = 0.64 to A1 3 AWFW ( ) (b) (Very) Strong positive correlation Bdep1 Dependent on 0.9 r gy < 1 (Some) Moderate positive correlation between girth and weight and/or length and weight Bdep1 Dependent on 0.6 r ly 0.7 Bdep0 for any mention of strong B1 3 At least one interpretation in context (c)(i) r xy = = 0.98 to B1 AWFW ( ) Most strongly correlated with y is x Bdep1 2 CAO; dependent on 0.97 r xy < 1 x = = 124 to 125 A1 2 AWFW (124.52) (iii) b = = 1 to A1 116/115.4 (= 1.005) M0 A0 AWFW ( ) a = b = 0.3 to 0.6 B1 3 AWFW ( ) (iv) r xy /nearly/almost/close to (+)1 or very strong/almost exact (positive) correlation (Stating r xy =0.98 to Bdep0) b =/ /nearly/almost/close to (+)1 a /nearly/almost/close to 0 (Stating a = 0.4 to 0.6 Bdep0) Bdep1 Bdep1 Bdep1 OE Dependent on 0.97 r xy < 1 OE; strong is not sufficient OE; must reference value of 1 or unity Dependent on A1 in (c)(iii) OE; must reference value of 0 or origin Dependent on B1 in (c)(iii) Estimate (not 'it' or 'this' or 'value', etc) is (very/highly/likely to be) accurate/precise/ reliable or (almost) exact/correct Bdep1 4 Total 17 OE; dependent on scoring at least 2 of the previous 3 marks in (c)(iv) Fairly accurate, good approximation, (quite) likely, (very) close, reasonable, etc Bdep0 7

11 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments 5(a)(i) P(A = 2) = = 0.85 to 0.86 B1 AWFW (0.855 or 171/200 OE) P(A = 1) = ( ) + ( ) or = 1 [ ( )] Do not ignore extra terms = 0.14 A1 3 CAO (7/50 OE) (b)(i) P(A W D W ) = = 0.72 A1 2 CAO (18/25 OE) P(A B D B ) = (b)(i) 0.95 ( 1) or = ( 1) or = (a)(i) to A1 2 AWFW (0.684 or 171/250 OE) (iii) P(A T D T ) = = 0 B1 1 CAO; award on value only (iv) P(neither) = P([A W D W ] [A T D T ]) (1 0.90) (1 0.15) (1 0.95) (1 0) or P(neither) = P(A W A T ) P(D W A W ) P(D T A T ) (1 0.90) (1 0.95) (1 0.15) (1 0) m1 () (m1) Accept or 17/200 OE Award and m1 on value(s) only Accept 0.05 or 1/20 OE Accept or 1/200 OE Award and m1 on value(s) only Accept 0.85 or 17/20 OE = or OE = to A1 3 AWFW ( or 17/4000 OE) Total 11 8

12 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments 6(a)(i) x = = 19.9 B1 CAO 25 98% (0.98) z = 2.32 to 2.33 B1 AWFW (2.3263) σ CI for µ is x ± z n Used with z (2.05 to 2.58), x (497.5 or 19 to 21) and σ (0.4) and n with n > 1 Thus ± A1 25 z (2.05 to 2.06 or 2.32 to 2.33 or 2.57 to 2.58), x (19.9) and σ (0.4) and 25 or 24 Hence 19.9 ± 0.2 or (19.7, 20.1) A1 5 CAO/AWRT ( ) AWRT Clear correct comparison of 20 with CI eg 20 is within CI or LCL < 20 < UCL so Agree with claim or no reason to doubt claim F on CI providing it contains 20 BF1 Quoting values for CI is not required Bdep1 2 OE; dependent on previous BF1 (iii) Weight of sand in a bag or X/x or original distribution or parent population is normal B1 1 It/mean/data/sample/information/sand is normal B0 Reference only to sample size or standard deviation B0 9

13 Mark Scheme General Certificate of Education (A-level) Mathematics MS/SS1B June 2013 Q Solution Marks Total Comments Accept percentage equivalent probabilities 6(b)(i) Y ~ N(25.25, ) V(mean) = /10 or to or SD (mean) = 0.35/ 10 or 0.11 to P( Y 25) P < = Z < B1 CAO/AWFW ( ) CAO/AWFW ( ) Standardising 25 using and 0.35/ 10 OE but allow ( ) = P(Z < ) = 1 P(Z < ) m1 Correct area change or an answer < 0.5 = 1 ( to ) = to A1 4 AWFW ( ) (0.987 to 0.989) B1 m0 A P( Y > 25) = P Z > 0.35 Standardising 25 using and 0.35 but allow ( ) = P(Z > ) = P(Z < ) = to A1 AWFW ( ) (0.236 to 0.239) A0 P(Y > 25 in each of 10) = p 10 Any p 10 providing 0 < p < 1 = to A1 4 AWFW ( ) Total 8 TOTAL 75 10

14 Scaled mark unit grade boundaries - June 2013 exams A-level Maximum Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E LAW02 LAW UNIT LAW03 LAW UNIT LAW04 LAW UNIT MD01 MATHEMATICS UNIT MD MD02 MATHEMATICS UNIT MD MFP1 MATHEMATICS UNIT MFP MFP2 MATHEMATICS UNIT MFP MFP3 MATHEMATICS UNIT MFP MFP4 MATHEMATICS UNIT MFP MB MATHEMATICS UNIT MB MM2B MATHEMATICS UNIT MM2B MM03 MATHEMATICS UNIT MM MM04 MATHEMATICS UNIT MM MM05 MATHEMATICS UNIT MM MPC1 MATHEMATICS UNIT MPC MPC2 MATHEMATICS UNIT MPC MPC3 MATHEMATICS UNIT MPC MPC4 MATHEMATICS UNIT MPC MS1A MATHEMATICS UNIT MS1A MS/SS1A/W MATHEMATICS UNIT S1A - WRITTEN MS/SS1A/C MATHEMATICS UNIT S1A - COURSEWORK MS1B MATHEMATICS UNIT MS1B MS2B MATHEMATICS UNIT MS2B