Cambridge A Level Mathematics 9709 — 2010 May/June Paper 6 · Variant 3

1 A bottle of sweets contains 13 red sweets, 13 blue sweets, 13 green sweets and 13 yellow sweets. 7 sweets are selected at random. Find the probability that exactly 3 of them are red. [3]

Mark scheme: C 3 × C 41 M1 Using combinations with attempt to evaluate 52 C 7 product of 2 in num and only 1 in denom M1 Correct numerator or denominator = 0.176 A1 Correct answer OR P(RRR) = M1 OR Multiplying 3 unequal red probs with 4 13 12 11 39 38 37 36 7 unequal non-red probs × × × × × × × C 3 M1 Multiplying a probability by 7C3 52 51 50 49 48 47 46 A1 Correct answer = 0.176 [3]

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2 The heights, x cm, of a group of 82 children are summarised as follows. Σ(x −130) = −287, standard deviation of x = 6.9. (i) Find the mean height. [2] (ii) Find Σ(x −130)2. [2]

Mark scheme: 2 (i) x = 130 − 287 / 82 M1 287/82 seen added or subt to 130 OR 287 seen added or subt to 82 × 130 = 126.5 (126, 127) cm A1 Correct answer [2] Σ ( x − 130) 2 2 2 (ii) − ( −5.3 ) = 9.6 M1 6.92 + (±their coded mean)2 seen or implied 82 Σ(x – 130)2 = 4908.5 cm (4910) A1 correct answer [2] 7 6 7

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3 Christa takes her dog for a walk every day. The probability that they go to the park on any day is 0.6. If they go to the park there is a probability of 0.35 that the dog will bark. If they do not go to the park there is a probability of 0.75 that the dog will bark. (i) Find the probability that they go to the park on more than 5 of the next 7 days. [2] (ii) Find the probability that the dog barks on any particular day. [2] (iii) Find the variance of the number of times they go to the park in 30 days. [1]

Mark scheme: 3 (i) P(> 5) = 7C6(0.6)6(0.4) + (0.6)7 M1 Summing 2 or 3 binomial probs of the form = 0.1306 + 0.02799 7Cr(0.6)r(0.4)7–r = 0.159 A1 Correct answer [2] (ii) P(bark) = P(park, bark) + P(not park, bark) M1 Summing two appropriate 2-factor = 0.6 × 0.35 + 0.4 × 0.75 probabilities = 0.51 A1 Correct answer [2] (iii) Variance (number of times) = 7.2 B1 Correct final answer [1] GCE AS/A LEVEL – May/June 2010 9709 63

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4 Three identical cans of cola, 2 identical cans of green tea and 2 identical cans of orange juice are arranged in a row. Calculate the number of arrangements if (i) the first and last cans in the row are the same type of drink, [3] (ii) the 3 cans of cola are all next to each other and the 2 cans of green tea are not next to each other. [5]

Mark scheme: 4 (i) ends cola, 5!/2!2! = 30 M1 Considering all three options ends green tea, 5!/3!2! = 10 ends orange juice, 5!/3!2! = 10 A1 Any one option correct total = 50 ways A1 Correct answer 3 2 2 1 2 1 OR P(ends same) = × + × + × M1 OR Considering all three options 7 6 7 6 7 6 5 = A1 Correct fraction 21 5 !7 × = 50 ways A1 Correct answer 21 !2!2!3 [3] (ii) colas together, no restrictions, 5!/2!2! M1 Considering all colas together, or 5! seen = 30 ways A1 Correct answer colas together and green tea together, 4!/2! M1 Considering all colas tog and all green tea tog, or 4! seen = 12 ways A1 Correct answer 30 – 12 = 18 ways. A1 Correct final answer OR1 Attempt to list M1A1 OR1 10 or more, 12 or more correct M1A1 14 or more, 16 or more correct A1 18 correct 4 × 3 OR2 3 × = 18 M1 OR2 Considering all colas together, or 3! seen 2 A1 3 ways for colas and orange juice M1 Considering green teas not together A1 4 × 3 or (4 × 3)/2 A1 Correct final answer [5]

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5 Set A consists of the ten digits 0, 0, 0, 0, 0, 0, 2, 2, 2, 4. Set B consists of the seven digits 0, 0, 0, 0, 2, 2, 2. One digit is chosen at random from each set. The random variable X is defined as the sum of these two digits. (i) Show that P(X = 2) = 37. [2] (ii) Tabulate the probability distribution of X. [2] (iii) Find E(X) and Var(X). [3] (iv) Given that X = 2, find the probability that the digit chosen from set A was 2. [2]

Mark scheme: 5 (i) P(2) = P(0,2) + P(2,0) M1 Summing two 2-factor probabilities = 6/10 × 3/7 + 3/10 × 4/7 = 30/70 = 3/7 AG A1 Correct answer legit obtained [2] (ii) x 0 2 4 6 B1 Correct values for rv X P(X = x) 24/70 30/70 13/70 3/70 B1 Correct probs [2] (iii) E(X) = 13/7 B1ft Var(X) = 120/70 + 208/70 + 108/70 – (13/7)2 M1 Using variance formula correctly with mean2 subtracted numerically, no extra division = 2.78 A1 Correct final answer [3] 3 / 10 × 4 / 7 (iv) P(A2│Sum 2) = M1 Correct numerator with a 0 < denom < 1 30 / 70 = 0.4 A1 Correct answer [2] GCE AS/A LEVEL – May/June 2010 9709 63

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6 The lengths of some insects of the same type from two countries, X and Y, were measured. The stem-and-leaf diagram shows the results. Country X Country Y (10) 9 7 6 6 6 4 4 4 3 2 80 (18) 8 8 8 7 7 6 6 5 5 5 4 4 3 3 3 2 2 0 81 1 1 2 2 3 3 3 5 5 6 7 8 9 (13) (16) 9 9 9 8 8 7 7 6 5 5 3 2 2 1 0 0 82 0 0 1 2 3 3 3 q 4 5 6 6 7 8 8 (15) (16) 8 7 6 5 5 5 3 3 2 2 2 1 1 1 0 0 83 0 1 2 2 4 4 4 4 5 5 6 6 7 7 7 8 9 (17) (11) 8 7 6 5 5 4 4 3 3 1 1 84 0 0 1 2 4 4 5 5 6 6 7 7 7 8 9 (15) 85 1 2 r 3 3 5 5 6 6 7 8 8 (12) 86 0 1 2 2 3 5 5 5 8 9 9 (11) Key: 5 | 81 | 3 means an insect from country X has length 0.815 cm and an insect from country Y has length 0.813 cm. (i) Find the median and interquartile range of the lengths of the insects from country X. [2] (ii) The interquartile range of the lengths of the insects from country Y is 0.028 cm. Find the values of q and r. [2] (iii) Represent the data by means of a pair of box-and-whisker plots in a single diagram on graph paper. [4] (iv) Compare the lengths of the insects from the two countries. [2]

Mark scheme: 6 (i) for X: Median = 0.825 cm B1 Correct median IQ range = 0.019 cm (0.833 – 0.814) B1 Correct IQ range [2] (ii) q = 4 B1 r = 2 B1 Must be 4 and 2 not 3 and 1 [2] SR q = 0.824 and r = 0.852 B1 (iii) B1 Labels X, Y and length/cm, linear scale Y from 0.80 to 0.87 and both on one diagram X B1ft Correct median and quartiles for X ft theirs must be a box B1ft Correct median and quartiles for Y ft theirs must be a box 0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 B1 Whiskers correct no line through middle length in cm [4] (iv) Y has longer insects on average B1 Correct statement about lengths Y has larger range B1 Correct statement about spreads [2] 135 − µ

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7 The heights that children of a particular age can jump have a normal distribution. On average, 8 children out of 10 can jump a height of more than 127 cm, and 1 child out of 3 can jump a height of more than 135 cm. (i) Find the mean and standard deviation of the heights the children can jump. [5] (ii) Find the probability that a randomly chosen child will not be able to jump a height of 145 cm. [3] (iii) Find the probability that, of 8 randomly chosen children, at least 2 will be able to jump a height of more than 135 cm. [3]

Mark scheme: 135 µ 7 (i) 0.431 = B1 One ±z-value correct, accept 0.430 σ B1 A second ±z-value correct 127 − µ –0.842 = M1 Solving two equations relating µ, σ, 135, σ 127 and their z-values (must be z-values) σ = 6.29 A1 Correct answer accept 6.28 µ = 132 A1 Correct answer [5]  145 − 1323.  (ii) P(X < 145) = P  z <  M1 Standardising no sq rt no cc  .6 284  =P(z < 2.023) M1 Correct use of normal tables = 0.978 A1 Answer rounding to 0.978 or 0.979 [3] (iii) p = 1/3 P(at least 2) = 1 – P(0, 1) M1 Binomial expression with powers summing to 8 and 8Csomething. (any p) = 1 – [ ( 2 / 3) 8 + 8 C1 × 1( / 3)1 ( 2 / 3) 7 ] A1 Correct unsimplified expression = 0.805 A1 Answer rounding to 0.805 [3]

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