Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 5 · Variant 2

1 Each of the 180 students at a college plays exactly one of the piano, the guitar and the drums. The numbers of male and female students who play the piano, the guitar and the drums are given in the following table. Piano Guitar Drums Male 25 44 11 Female 42 38 20 A student at the college is chosen at random. (a) Find the probability that the student plays the guitar. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the student is male given that the student plays the drums. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Determine whether the events ‘the student plays the guitar’ and ‘the student is female’ are independent, justifying your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 82 41 , 180 90 , 0.456 1 1(b) ( ) ( ) ( ) P M|D P M D P D   ∩ =       = 11 0.6011 180 or 20 11 0.1722 180 180 + M1 Their identified ( ) ( ) P M D P D ∩ 11 from data table 20 11 or + , accept unsimplified, condone × 180. 11 31, 0.355 A1 Final answer. 2 Question Answer Marks Guidance 1(c) P(F) = 100 5 , , 0.5556 180 9 OE P(G) = 82 41 , 0.4556 180 90 OE P(F∩G) = 38 19 , , 0.2111 180 90 OE ( ) ( ) 100 82 41 38 P F P G , 0.2531 OE 180 180 162 180   × = × = ≠     Not independent M1 Their identified P(F) × their identified P(G) or correct seen, can be unsimplified. A1 ( ) ( ) ( ) 41 38 , ,P F G and P F P G 162 180 ∩ × seen with correct conclusion, WWW. Values and labels must be seen. Alternative method for question 1(c) P(F∩G) = 38 19 , ,0.2111 180 90 OE P(G) = 82 41 , ,0.4556 180 90 OE ( ) 38 19 180 P F|G , 0.4634 82 41 180 = = OE ( ) 100 5 P F , 180 9 ≠ = ,0.5556 OE Not independent M1 P(F|G) (OE) unsimplified with their identified probs or correct A1 ( ) ( ) 19 100 , , P F G and P F|G 41 180 ∩ seen with correct conclusion WWW. Values and labels must be seen. 2

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2 A group of 6 people is to be chosen from 4 men and 11 women. (a) In how many different ways can a group of 6 be chosen if it must contain exactly 1 man? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Two of the 11 women are sisters Jane and Kate. (b) In how many different ways can a group of 6 be chosen if Jane and Kate cannot both be in the group? [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) M1 1848 A1 CAO as exact. 2 2(b) Method 1 [Identifying scenarios] [Neither selected =] 13C6 [= 1716] [Only Jane selected =] 13C5 [= 1287] [Only Kate selected =] 13C5 [= 1287] M1 Either 13C6 seen alone or 13C5 seen alone or × 2 (condone 13Pn, n = 5,6). [Total =] 1716 + 1287 + 1287 M1 Three correct scenarios only added, accept unsimplified (values may be incorrect). 4290 A1 Method 2 [Total number of selections – selections with Jane and Kate both picked] 15C6 - 13C4 [= 5005 – 715] M1 15C6 – k, k a positive integer < 5005, condone 15P6. M1 m – 13C4, m integer > 715, condone n – 13P4, n > 17 160. 4290 A1 3 SC Where the condition of 2(a) is also applied in 2(b), the final answer is 1512 SC M1 M1 A0 max. The method marks can be earned for the equivalent stages in each method. Method 1 4C1 × 9C5 + 4C1 × 9C4 × 2 Method 2 4C1 × 11C5 – 4C1 × 9C3

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4 (a) In how many different ways can the 9 letters of the word TELESCOPE be arranged? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) In how many different ways can the 9 letters of the word TELESCOPE be arranged so that there are exactly two letters between the T and the C? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 9! 3! M1 9!,! e e = 2, 3 60 480 A1 2 Question Answer Marks Guidance 4(b) 7! 2 6 3!× × M1 7! 3! k × seen, k an integer > 0. M1 ! 2 ! m q n × × 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3, 1 ⩽ q ⩽ 8 all integers. M1 ! 6 ! m p n × × 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3, 1 ⩽ p ⩽ 2 all integers. (Accept 3P2 for 6) If M0 M0 M0 awarded, SC M1 for t × 12, t an integer ⩾ 20, 5! 3! . 10 080 A1 Exact value. Alternative method for question 4(b) 7 2P 6! 2 3! × × M1 6! 3! k × seen, k an integer > 0. M1 7 2 ! P ! m q n × × m = 6,9, 1 ⩽ n ⩽ 3, 1 ⩽ q ⩽ 2 all integers. M1 ! ! m n × 7Pr × 2 m = 6, 9, 1 ⩽ n ⩽ 3, 1 ⩽ r ⩽ 5 all integers. If M0 M0 M0 awarded, SC M1 for t × 84, t an integer ⩾ 20, 5! 3!. 10 080 A1 Exact value.

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5 In a certain region, the probability that any given day in October is wet is 0.16, independently of other days. (a) Find the probability that, in a 10-day period in October, fewer than 3 days will be wet. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the first wet day in October is 8 October. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) For 4 randomly chosen years, find the probability that in exactly 1 of these years the first wet day in October is 8 October. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) [= 0.17490 + 0.333145 + 0.28555] A1 Correct unsimplified expression, or better. 0.794 A1 0.7935 < p ⩽ 0.794, mark at most accurate. If M0 scored, SC B1 for final answer 0.794. 3 5(b) ( ) 7 0.84 0.16 M1 (1 – p)7p, 0 < p < 1 0.0472 A1 0.0472144 to at least 3sf. 2 Question Answer Marks Guidance 5(c) 4 × 0.0472 × (1 – 0.0472)3 M1 4 × q(1 – q)3, q = their (b) or correct. 0.163 A1 0.163 ⩽ p ⩽ 0.1634, mark at most accurate from their probability to at least 3sf. 2

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6 The times taken, in minutes, to complete a particular task by employees at a large company are normally distributed with mean 32.2 and standard deviation 9.6. (a) Find the probability that a randomly chosen employee takes more than 28.6 minutes to complete the task. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) 20% of employees take longer than t minutes to complete the task. Find the value of t. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that the time taken to complete the task by a randomly chosen employee differs from the mean by less than 15.0 minutes. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) ( ) 28.6 32.2 P 28.6 9.6 X P Z −    > = >       ( ) 0.375 P Z = > −    M1 28.6, 32.2 and 9.6 substituted appropriately in ± Standardisation formula once, allow continuity correction of ± 0.05, no σ2, √σ. ( ) [Φ 0.375 their = ] their 0.6462 M1 Appropriate numerical area, from final process, must be probability, expect > 0.5. 0.646 A1 AWRT 3 6(b) 0.842 z = ± B1 0.841 < z ⩽ 0.842 or -0.842 ⩽ z < –0.841 seen. 32.2 0.842 9.6 t − = M1 Substituting 32.2 and 9.6 into ± standardisation formula, no continuity correction, allow σ2, √σ, must be equated to a z-value. 40.3 t = A1 40.28 ⩽ t ⩽ 40.3 WWW 3 Question Answer Marks Guidance 6(c) 15 15 9.6 9.6 P Z   − < <     ( 1.5625 1.5625) P Z − < < M1 Identifying at least one of 15 9.6 and 15 9.6 − as the appropriate z-values or substituting their (32.2 ± 15) into ± Standardisation formula once, no continuity correction, σ2 nor √σ. Condone ±1.563 for M1. [2 15 Φ(9.6 ) – 1 ] = 2 × 0.9409 – 1 A1 p = 0.941 AWRT SOI M1 Appropriate area 2Φ – 1 oe, (eg 1 – 2 × 0.0591, 2 × (0.9409 – 0.5) or 0.9409 – 0.0591), from final process, must be probability > 0.5. 0.882 A1 4

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7 The distances, x m, travelled to school by 140 children were recorded. The results are summarised in the table below. Distance, x m x ≤200 x ≤300 x ≤500 x ≤900 x ≤1200 x ≤1600 Cumulative frequency 16 46 88 122 134 140 (a) On the grid, draw a cumulative frequency graph to represent these results. [2] (b) Use your graph to estimate the interquartile range of the distances. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate estimates of the mean and standard deviation of the distances. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Cumulative frequency graph drawn B1 Axes labelled ‘cumulative frequency’ (or cf) from 0 to at least 140 and ‘distance (or d) [in] m’ from 0 to at least 1600, linear scales with at least 3 values stated. B1 All plotted correctly at correct upper end points (200 etc.) curve drawn accurately joined to (0, 0) (straight line segments B0) but no daylight above 140. Cf scale no less than 2 cm = 20 children . 2 Question Answer Marks Guidance 7(b) [UQ at 75% of 140 = 105, LQ at 25% of 140 = 35] [IQR:] 700 – 260 M1 Accept 660 ⩽ UQ ⩽ 720 – 240 ⩽ LQ ⩽ 290. If values are outside our range, FT providing scales linear and increasing cf drawn. 440 A1 Accept correct evaluation of 660 ⩽ their UQ ⩽ 720 – 240 ⩽ their LQ ⩽ 290 with clear indication that graph has been used for at least one of 105 or 35. 2 Question Answer Marks Guidance 7(c) [Mean =] 16 100 30 250 42 400 34 700 12 1050 6 1400 140 × + × + × + × + × + × B1 Frequencies 16 30 42 34 12 6 Mid-points 100 250 400 700 1050 1400 5 or 6 correct frequency values seen. B1 5 or 6 correct midpoint values seen. M1 Values substituted into mean formula using their midpoints which must be in the class – condone 1 data error. Accept 1600 7500 16 800 23800 12 600 8400 70 700 or 140 140 + + + + + . Condone 70 770 140 for M1. 505 A1 WWW Variance = 2 2 2 2 2 2 16 100 30 250 42 400 34 700 12 1050 6 1400 140 × + × + × + × + × + × 2 505 − M1 Values substituted into variance formula using (their mean)2 and their midpoints and their frequencies (including for denominator). Accept unsimplified. Condone 1 data error. Accept: 160 000 1875 000 6 720 000 16 660 000 13 230 000 11760 000 [ 140 + + + + + 2 50 405 000 or or 360 035.7143] 505 or 255 025 140   −  If formula stated accept 105 010 or 105 011 WWW. S.d. = 105 010.7 324   =   A1 WWW 6

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