Cambridge A Level Mathematics 9709 — 2013 May/June Paper 6 · Variant 2

1 The random variable Y is normally distributed with mean equal to five times the standard deviation. It is given that P Y > 20 = 0.0732. Find the mean. [3]

Mark scheme: 1 z = 1.452 B1 Rounding to ± 1.45 20 − µ 20 − µ 20 − 5σ 1.452 = or seen oe B1 µ / 5 µ / 5 σ µ = 15.5 B1 [3] rounding to correct answer

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2 A summary of the speeds, x kilometres per hour, of 22 cars passing a certain point gave the following information: Σ x −50 = 81.4 and Σ x −50 2 = 671.0. Find the variance of the speeds and hence find the value of Σx2. [4]

Mark scheme: 2 x = 50 + 81.4/22 = 53.7 M1 Attempt to find variance using coding in both, correct formula var = 671/22 – 3.72 = 16.81(16.8) A1 Correct answer using their var and their mean with 16.81 = Σx2/22 – 53.72 M1 uncoded formula for both = 63811(63800) A1 [4] correct answer OR Σx - 22×50 = 81.4 (Σx = 1181.4) M1 expanded eqn with 22×50 seen Σx2 -100Σx + 22×502 = 671 M1 expanded eqn with 2 or 3 terms correct Σx2 = 671 + 118140- 55000 = 63811 A1 correct answer A1 correct answer Var = Σx2/22 – (Σx/22)2 = 16.81

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3 Cans of lemon juice are supposed to contain 440 ml of juice. It is found that the actual volume of juice in a can is normally distributed with mean 445 ml and standard deviation 3.6 ml. (i) Find the probability that a randomly chosen can contains less than 440 ml of juice. [3] It is found that 94% of the cans contain between 445 −c ml and 445 + c ml of juice. (ii) Find the value of c. [3]

Mark scheme: 3 (i) P( x < 440)  440 − 445  M1 Standardising no cc no sq or sq rt = P  z <  = 1 - Φ (1.389)  6.3  M1 Correct area (1 – Φ) oe (indep) = 1 – 0.9176 Ans = 0.0824 A1 Rounding to correct answer accept [3] 0.0825 (ii) z = 1.881 M1 ±1.88 or 1.881 or 1.882 or 1.555 seen± c = 1.881 M1 Equation with ± c/3.6 or 2c/3.6 only = 6.3 z or prob (can be implied) c = 6.77 A1 [3] Correct answer accept 6.78 5

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4 Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive. (i) Find the probability that at least 2 of the 5 integers are less than or equal to 4. [3] Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the number of these n integers which are less than or equal to a certain integer k between 1 and 9 inclusive. It is given that the mean of X is 96 and the variance of X is 32. (ii) Find the values of n and k. [4]

Mark scheme: 4 (i) p = 4/9 or 5/9 B1 Binomial term 5Cxpx(1 – p)5 – x seen P(at least 2) = 1 – P(0, 1) M1 = 1 – (5/9)5 – (4/9)(5/9)4 5C1 = 0.735 A1 [3] Correct answer (ii) np = 96 npq = 32 p = P ( ≤ k) M1 Using np = 96 npq = 32 to obtain eqn in 1 variable p = 2/3 q = 1/3 n = 144 A1 1/3 or 2/3 seen or implied k = 6 A1ft Correct k ft k = 9p n = 144 A1 [4] correct n GCE AS/A LEVEL – May/June 2013 9709 62 5 (i) Stem leaf B1 Correct stem condone a space under the 1

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7 Susan has a bag of sweets containing 7 chocolates and 5 toffees. Ahmad has a bag of sweets containing 3 chocolates, 4 toffees and 2 boiled sweets. A sweet is taken at random from Susan’s bag and put in Ahmad’s bag. A sweet is then taken at random from Ahmad’s bag. (i) Find the probability that the two sweets taken are a toffee from Susan’s bag and a boiled sweet from Ahmad’s bag. [2] (ii) Given that the sweet taken from Ahmad’s bag is a chocolate, find the probability that the sweet taken from Susan’s bag was also a chocolate. [4] (iii) The random variable X is the number of times a chocolate is taken. State the possible values of X and draw up a table to show the probability distribution of X. [5]

Mark scheme: 7 (i) 5 2 1 M1 Mult their P(T) by 2/9 or 2/10 only P(T,B) = × = (0.0833) A1 Correct answer 12 10 12 [2] (ii) 7 4 28 P (C S ∩ C A ) = × = (0.2333) M1 Mult their P(CS) by 3/9 or 4/10 seen as 12 10 120 num or denom of a fraction 7 4 5 3 43 P(CA) = × + × = (0.3583) M1 Summing 2 two-factor products to find 12 10 12 10 120 P(CA) seen anywhere P (C ∩ C ) 28 / 120 P(CS CA) = = A1 Correct unsimplified P(CA) seen as num P (C A ) 43 / 120 or denom of a fraction 28 = (0.651) 43 A1 [4] Correct answer (iii) x 0 1 2 B1 x = 0, 1, 2, can be implied from table or Prob 7/24 19/40 7/30 working P(X = 0) = P(T, B) + P(T, T) M1 1 or 2 two-factor products, denoms 12 and 10 or 12 and 9, implied if ans is correct 5 2 5 5 7 = × + × = (0.292) A1 One correct unsimplified 12 10 12 10 24 7 4 28 P(X = 2) = P(C, C) = × = (0.233) B1 One other correct unsimplified 12 10 120 19 P(X = 1) = 1 – 7/24 – 28/120 = (0.475) B1ft [5] Third correct ft 1 – P(2 of their probs)) 40

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