Cambridge A Level Mathematics 9709 — 2017 May/June Paper 6 · Variant 1

1 Kadijat noted the weights, x grams, of 30 chocolate buns. Her results are summarised by Σ x −k = 315, Σ x −k 2 = 4022, where k is a constant. The mean weight of the buns is 50.5 grams. (i) Find the value of k. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the standard deviation of x. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) EITHER: 315 10.5 30 30 − = = ∑x k (M1 and no more 5.5 10.5 40 = − = k A1) Correct answer from correct working OR: 50.5 30 1515 = × = ∑x , 1515 30 315 − = k (M1 Mult by 50.5 by 30 and + or – 315 and dividing by ±30 need all these k = 40 A1) Correct answer from correct working. 1200 gets M0 Total: 2 1(ii) EITHER: var = 4022/30–10.52(=23.817) (M1 Subst in correct coded variance formula sd = 4.88 A1) OR: ( ) ( ) 2 2 2 40 30 40 4022 − + = ∑ ∑ x x , 2 77222 = ∑x Var = 77222/30 – 50.52 (= 23.817) (M1 Expanding with ± 40Σx and ± 30(40)2 seen sd = 4.88 A1) Total: 2

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2 Ashfaq throws two fair dice and notes the numbers obtained. R is the event ‘The product of the two numbers is 12’. T is the event ‘One of the numbers is odd and one of the numbers is even’. By finding appropriate probabilities, determine whether events R and T are independent. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 P(R) = 4/36 = 1/9 M1 Attempt at P(R) by probability space diag or listing more than half the options, must see a prob, just a list is not enough ( ) ( ) ( ) P P O, E P E, O 1/ 4 1/ 4 1/ 2 T = + = + = OR ( ) P | 1/ 9 R T = M1 Attempt at P(T) or P(R|T) involving more than half the options ( ) ( ) ( ) P P 3, 4 P 4, 3 2 / 36 1/18 = + = = I R T OR ( ) P | 1/ 9 R T = B1 Value stated, not from P(R) × P(T) e.g. from probability space diagram ( ) ( ) ( ) As P P P × = I R T R T OR ( ) ( ) as P | P R T R = M1 Comparing product values with ( ) P I R T , or comparing P (R|T) with P(R) The events are independent. A1 Correct conclusion must have all probs correct Total: 5

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3 Redbury United soccer team play a match every week. Each match can be won, drawn or lost. At the beginning of the soccer season the probability that Redbury United win their first match is 5,3 with equal probabilities of losing or drawing. If they win the first match, the probability that they win the second match is 7 and the probability that they lose the second match is 1 If they draw the first 10 10. match they are equally likely to win, draw or lose the second match. If they lose the first match, the probability that they win the second match is 3 and the probability that they draw the second match 10 is 20.1 (i) Draw a fully labelled tree diagram to represent the first two matches played by Redbury United in the soccer season. [2] (ii) Given that Redbury United win the second match, find the probability that they lose the first match. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) M1 Correct shape i.e. 3 branches then 3 by 3 branches, labelled and clear annotation Condone omission of lines for first match result providing the probabilities are there. A1 All correct probs with fully correct shape and probs either fractions or decimals not 1.5/5 etc. Total: 2 7/10 W 2/10 D 3/5 1/10 L 1/3 W 1/5 D 1/3 D 1/3 L 1/5 3/10 W L 1/20 D 13/20 L Question Answer Marks Guidance 3(ii) ( ) ( ) ( ) 1 2 1 2 2 P P given P ∩ = L W L W W M1 Attempt at P(L1∩W2) as a two-factor prod only as num or denom of a fraction 1/ 5 3 /10 3 / 5 7 /10 1/ 5 1/ 3 1/ 5 3 /10 × = × + × + × M1 Attempt at P(W2) as sum of appropriate 3 two-factor probs OE seen anywhere A1 Unsimplified correct P(W2) num or denom of a fraction ( ) 3 / 50 9 / 82 0.110 41/ 75 = = A1 Total: 4

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4 The times taken, t seconds, by 1140 people to solve a puzzle are summarised in the table. Time (t seconds) 0 ≤t < 20 20 ≤t < 40 40 ≤t < 60 60 ≤t < 100 100 ≤t < 140 Number of people 320 280 220 220 100 (i) On the grid, draw a histogram to illustrate this information. [4] (ii) Calculate an estimate of the mean of t. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) fd 16, 14, 11, 505, 2.5 M1 Attempt at fd (must be at least 3 freq/cw) – may be implied by graph A1 Correct heights seen on graph i.e. must see a gap for fd = 2.5 etc. B1 Correct end points of bars and correct widths B1 labels fd, sec. Time can be optional. Linear axes, condone 0 ⩽ t < 20 etc. Total: 4 fd 20 15 10 5 0 20 40 60 80 100 120 140 time sec Question Answer Marks Guidance 4(ii) (10 × 320 + 30 × 280 + 50 × 220 + 80 × 220 + 120 × 100) / 1140 M1 using Σ fx / n with mid-point attempt ±0.5, not ends not class widths = 45.8 A1 Total: 2

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5 Eggs are sold in boxes of 20. Cracked eggs occur independently and the mean number of cracked eggs in a box is 1.4. (i) Calculate the probability that a randomly chosen box contains exactly 2 cracked eggs. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Calculate the probability that a randomly chosen box contains at least 1 cracked egg. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) A shop sells n of these boxes of eggs. Find the smallest value of n such that the probability of there being at least 1 cracked egg in each box sold is less than 0.01. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) p = 0.07 B1 ( ) ( ) ( ) 2 18 20 2 P 2 C 0.07 0.93 = M1 Bin term ( ) 20 20C 1 − − x x x p p their p = 0.252 A1 Total: 3 5(ii) P(at least 1 cracked egg)=1–(0.93)20=1–0.2342 M1 Attempt to find P(at least1 cracked egg) with their p from (i) allow 1 – P(0, 1) OE = 0.766 A1 Rounding to 0.766 Total: 2 5(iii) (0.7658)n<0.01 M1 Eqn or inequal containing (their 0.766)n or (their 0.234)n, together with 0.01 or 0.99 n = 18 A1 Total: 2

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7 (a) Eight children of different ages stand in a random order in a line. Find the number of different ways this can be done if none of the three youngest children stand next to each other. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) David chooses 5 chocolates from 6 different dark chocolates, 4 different white chocolates and 1 milk chocolate. He must choose at least one of each type. Find the number of different selections he can make. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) A password for Chelsea’s computer consists of 4 characters in a particular order. The characters are chosen from the following. ³ The 26 capital letters A to Z ³ The 9 digits 1 to 9 ³ The 5 symbols # ~ * ? ! The password must include at least one capital letter, at least one digit and at least one symbol. No character can be repeated. Find the number of different passwords that Chelsea can make. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) EITHER: e.g. xxxxx =5! for the other children (B1 Put y in 6 ways, then 5 then 4 for the youngest children B1 Mult by 6P3 OE Answer 5! × 6P3 = 14400 B1) Correct answer OR: total – 3 tog – 2 tog = 8! – 6!3! – 6! × 2 × 5 × 3 = 14400 (B1 8! – 6! × k ⩾1seen B1 6!3! or 6! × 2 × 5 × 3 seen subtracted B1) Correct answer Total: 3 7(b) D W M 2 2 1 = 6C2 × 4C2 × 1 = 90 B1 One correct unsimplified option 3 1 1 = 6C3 × 4 × 1 = 80 M1 Summing 2 or more 3-factor options which can contain perms or 3 factors added. The 1 can be implied 1 3 1 = 6 × 4C3 × 1 = 24 M1 Summing the correct 3 unsimplified outcomes only Total=194 ways A1 Total: 4 Question Answer Marks Guidance 7(c) C D S 2 1 1 = 26C2 × 9 × 5 × 4! = 351 000 M1 summing 2 or more options of the form (2 1 1), (1 2 1), (1 1 2), can have perms, can be added 1 2 1 = 26 × 9C2 × 5 × 4! = 112 320 M1 4 relevant products seen excluding 4! e.g. 26 × 9 × 8 × 5 or 26 × 9P2 × 5 for 2nd outcome, condone 26 × 9 × 5 × 37 as being relevant 1 1 2 = 26 × 9 × 5C2 × 4! = 56 160 M1 mult all terms by 4! or 4!/2! Total = 519 480 A1 Total: 4

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