Cambridge A Level Mathematics 9709 — 2016 May/June Paper 6 · Variant 3

1 In a group of 30 adults, 25 are right-handed and 8 wear spectacles. The number who are right-handed and do not wear spectacles is 19. (i) Copy and complete the following table to show the number of adults in each category. [2] Wears spectacles Does not wear spectacles Total Right-handed Not right-handed Total 30 An adult is chosen at random from the group. Event X is ‘the adult chosen is right-handed’; event Y is ‘the adult chosen wears spectacles’. (ii) Determine whether X and Y are independent events, justifying your answer. [3]

Mark scheme: Qu Answer Marks Guidance 1 (i) Wears Not Total specs wears specs RH 6 19 25 B1 One correct row or col including total Not other than the Total row/column 2 3 5 RH B1 [2] All correct Total 8 22 (ii) P(X) = 25/30, P(Y) = 8/30 M1 P(X) or P(Y) from their table or correct from question (denom 30) oe P(X) × P(Y) = 25/30 × 8/30 = 200/900 = 2/9 M1 Comparing their P(X) × P(Y) (values P(X∩Y) = 6/30 = 1/5 ≠ P(X) × P(Y) substituted) with their evaluated P(X∩Y) – not P(X)×P(Y) Not independent A1 [3]

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2 A group of children played a computer game which measured their time in seconds to perform a certain task. A summary of the times taken by girls and boys in the group is shown below. Minimum Lower quartile Median Upper quartile Maximum Girls 5 5.5 7 9 13 Boys 4 6 8.5 11 16 (i) On graph paper, draw two box-and-whisker plots in a single diagram to illustrate the times taken by girls and boys to perform this task. [3] (ii) State two comparisons of the times taken by girls and boys. [2]

Mark scheme: 2 (i) B1 Labels ‘time’ and ‘seconds’, ‘boys’ and girls ‘girls’ on correct plots and scaled line boys B1 One box and whisker all correct on graph paper – ignore boy or girl label B1 [3] Second box and whisker all correct (on 4 6 8 10 12 14 16 graph paper and ignore boy/girl label) on Time in seconds SAME scaled line. (ii) girls smaller range or IQ range than boys /girls B1 Any 2 comments – MUST be a less spread out oe comparison girls generally quicker than boys or girls B1 [2] median<boys median (not mean) oe boys almost symmetrical, girls +vely skewed oe

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3 Two ordinary fair dice are thrown. The resulting score is found as follows. • If the two dice show different numbers, the score is the smaller of the two numbers. • If the two dice show equal numbers, the score is 0. (i) Draw up the probability distribution table for the score. [4] (ii) Calculate the expected score. [2]

Mark scheme: 3 (i) P(0) = 6/36, P(1) = 10/36, P(2) = 8/36 B1 Table oe seen with 0, 1, 2, 3, 4, 5 (6 if P(6) = 0) B1 Any three probs correct M1 Σ p = 1 and at least 3 outcomes P(3) = 6/36, P(4) = 4/36, P(5) = 2/36 A1 [4] All probs correct (ii) mean score = (0×6+1×10 +16 +18 +16+10)/36 M1 Using Σxp (unsimplified) on its own – condone Σ p not =1 = 70/36 (35/18, 1.94) A1 [2]

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4 The monthly rental prices, $x, for 9 apartments in a certain city are listed and are summarised as follows. Σ x −c = 1845 Σ x −c 2 = 477 450 The mean monthly rental price is $2205. (i) Find the value of the constant c. [2] (ii) Find the variance of these values of x. [2] (iii) Another apartment is added to the list. The mean monthly rental price is now $2120.50. Find the rental price of this additional apartment. [2]

Mark scheme: 4 (i) 1845/9 (= 205) M1 Accept (1845± anything)/ 9 c = 2205 - 205 = 2000 A1 OR Σx = 2205× 9 (= 19845) M1 For 2205× 9 seen Σ x −Σ c = 1845 Σc = 19845 -1845 = 18000 A1 [2] c = 2000 477450 2 477450 2 (ii) var = − 205 M1 For − (their coded mean) 9 A1 9 = 11025 For their Σx2/9 – 22052 where Σx2 is 43857450 − 2205 2 M1 2 OR var = obtained from expanding Σ ( x − c ) with 9 = 11025 A1 [2] 2cΣx seen (iii) new total = 2120.5×10 = 21205 M1 Attempt at new total new price = 21205 – 19845 = 1360 A1 [2]

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5 The heights of school desks have a normal distribution with mean 69 cm and standard deviation 3 cm. It is known that 15.5% of these desks have a height greater than 70 cm. (i) Find the value of 3. [3] When Jodu sits at a desk, his knees are at a height of 58 cm above the floor. A desk is comfortable for Jodu if his knees are at least 9 cm below the top of the desk. Jodu’s school has 300 desks. (ii) Calculate an estimate of the number of these desks that are comfortable for Jodu. [5]

Mark scheme: 5 (i) z = 1.015 B1 Accept z between ±1.01 and 1.02 70 − 69 1.015 = M1 Standardising σ σ = 0.985 (200/203) A1 [3] (ii) 58 + 9 = 67 M1 58 + 9 seen or implied (or 69-58 or 69-9)  67 − 69  P ( > 67) = P  z >  M1 Standardising ± z no cc allow their sd  0.9852  (must be +ve)  9 − 11  Alt. 1 69-58 =11, P( >9)=P  z >   0.9852   58 − 60  Alt.2 69-9 =60, P( >58) =P  z >   0.9852  = P(z > – 2.03) M1 Correct prob area = 0.9788 Multiply their prob (from use of tables) by M1 300 300 × 0.9788 = 293.6 so 293 A1 [5] – accept 293 or 294 from fully correct working

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6 Find the number of ways all 9 letters of the word EVERGREEN can be arranged if (i) there are no restrictions, [1] (ii) the first letter is R and the last letter is G, [2] (iii) the Es are all together. [2] Three letters from the 9 letters of the word EVERGREEN are selected. (iv) Find the number of selections which contain no Es and exactly 1 R. [1] (v) Find the number of selections which contain no Es. [3]

Mark scheme: 6 (i) 7560 ways B1 [1] 7! (ii) RxxxxxxxG in B1 7! alone seen in num or 4! alone in denom 4! 7!× 2 Must be in a fraction. gets full 4!× 2 marks B1 [2] = 210 ways 6! (iii) eg EEEExxxxx in B1 6! or 5! × 6 seen in numerator or on own 2! Can be 6! × k but not 6! ± k B1 [2] = 360 ways (iv) 1 R eg RVG or RVN or RGN = 3 B1 [1] (v) no Rs eg VGN or 3C3 ways = 1 M1 Summing at least 2 options for R 2 Rs eg RRV or 3C1 ways = 3 A1 Correct outcome for no Rs or 2 Rs – Total = 7 A1 [3] evaluated

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7 Passengers are travelling to Picton by minibus. The probability that each passenger carries a backpack is 0.65, independently of other passengers. Each minibus has seats for 12 passengers. (i) Find the probability that, in a full minibus travelling to Picton, between 8 passengers and 10 passengers inclusive carry a backpack. [3] (ii) Passengers get on to an empty minibus. Find the probability that the fourth passenger who gets on to the minibus will be the first to be carrying a backpack. [2] (iii) Find the probability that, of a random sample of 250 full minibuses travelling to Picton, more than 54 will contain exactly 7 passengers carrying backpacks. [6]

Mark scheme: 7 (i) 12C8 ( 0.65)8(0.35)4 + 12C9 (0.65)9(0.35)3 + 12C10 M1 Bin term with 12Cr pr (1 – p)12-r seen r≠0 (0.65)10(0.35)2 any p<1 M1 Summing 2 or 3 bin probs p = 0.65 or 0.35, n = 12 = 0.541 A1 [3] (ii) P( RRRR ) = 0.35× 0.35 × 0.35 × 0.65 M1 Mult 4 probs either (0.35)3(0.65) or (0.65)3(0.35) A1 [2] = 0.0279 (iii) P(7) = 0.2039 (unsimplified) B1 12C7 (0.65)7(0.35)5 Mean = 250×’0.2039’ (= 50.9798) Correct unsimplified np and npq using Var = 250×’0.2039’ × ‘(1 – 0.2039)’ B1 ‘their 0.2039’ but not 0.65 or 0.35 ( = 40.5851)  54.5 − 50.9798  M1 Standardising need sq rt – must be from P(> 54) = P    40.5851  working with 54 M1 cc either 53.5 or 54.5 = P(z > 0.5526) = 1 – Φ(0.5526) = 1 – 0.7098 M1 correct area < 0.5 i.e. 1 – Φ - must be from working with 54 A1 [6] = 0.290

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