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Mathematics 9709/52 October/November 2021: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_w21_qp_529709_w21_ms_529709/52/O/N/21

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/52/O/N/21
9709_w21_ms_52.pdf

9709/52 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 15 marks

Question

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 B1 , , 0.456 180 90 1 1(b) 11 M1 P ( M ∩ D) P(M ∩ D) 0.6011 Their identified P ( M|D ) = 180 = 20 11 or P ( D) P ( D ) + 0.1722 11 180 180 or from data table , accept unsimplified, condone × 180. 20 + 11 11 A1 Final answer. , 0.355 31 2 1(c) 100 5 82 41 M1 Their identified P(F) × their identified P(G) or correct seen, can be P(F) = , , 0.5556 OE P(G) = , 0.4556 OE 180 9 180 90 unsimplified. 38 19 P(F∩G) = , , 0.2111 OE A1 41 38 180 90 , , P ( F ∩ G ) and P ( F ) × P ( G ) seen with correct 162 180 100 82 41 38 P ( F) × P ( G ) = × = , 0.2531 OE ≠ conclusion, WWW. Values and labels must be seen. 180 180 162 180 Not independent Alternative method for question 1(c) 38 19 82 41 M1 P(F|G) (OE) unsimplified with their identified probs or correct P(F∩G) = , ,0.2111 OE P(G) = , , 0.4556 OE 180 90 180 90 A1 19 100 38 , , P ( F ∩ G ) and P ( F|G ) seen with correct conclusion WWW. 19 41 180 P ( F|G ) = 180 = , 0.4634 OE Values and labels must be seen. 82 41 180 100 5 ≠ P ( F) = , ,0.5556 OE 180 9 Not independent 2 11

Question 25 marks

Question

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) C5 × 4C1 M1 11 C5 × 4C1 condone 11P5 × 4P1 no +, –, × or ÷. 1848 A1 CAO as exact. 2 2(b) Method 1 [Identifying scenarios] [Neither selected =] 13C6 [= 1716] M1 Either 13C6 seen alone or 13C5 seen alone or × 2 [Only Jane selected =] 13C5 [= 1287] (condone 13Pn, n = 5,6). [Only Kate selected =] 13C5 [= 1287] [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] 15 C6 - 13C4 [= 5005 – 715] M1 15 C6 – 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

Question 37 marks

Question

A bag contains 5 yellow and 4 green marbles. Three marbles are selected at random from the bag, without replacement. 5. (a) Show that the probability that exactly one of the marbles is yellow is 14 [3] The random variable X is the number of yellow marbles selected. (b) Draw up the probability distribution table for X . [3] (c) Find E X . [1]

Mark scheme

3(a) For one yellow: YGG + GYG +GGY M1 a b c × × , 0 < a,b,c integers ⩽ 5, for one arrangement. 5 4 3 9 8 7 × × ×3 9 8 7 M1 Their three-factor probability × 3, 3C1, 3C2 or 3P1, (or repeated adding) no additional terms. 180 5 A1 AG. Convincingly shown, including identifying possible scenarios, 504 = 14 may be on tree diagram WWW. 3 Alternative method for question 3(a) 5 C1 × 4 C2 M1 5 C1 × 4 C2 9 9 , r = 2, 3, 4 C3 Cr M1 5 Cs × 4 Ct 9 ,s+t=3 C3 30 5 A1 AG. Convincingly shown, WWW. 84 = 14 3 3(b) B1 Table with correct X values and one correct probability inserted X 0 1 2 3 appropriately. P(X) Condone any additional X values if probability stated as 0. 24 180 240 60 504 504 504 504 B1 Second identified correct probability, may not be in table. 1 5 10 5 = 21 , = 14 , = 21 , = 42 , B1 All probabilities identified and correct . 0.0476 0.357 0.476 0.119 SC if less than 2 correct probabilities or X value(s) omitted: SC B1 3 or 4 probabilities summing to one. 3 3(c) 840 5 B1 OE Must be evaluated. [E(X) =] , , 1.67 SC B1 FT correct unsimplified expression from incorrect 3(b) using at 504 3 least 3 probabilities, 0 < p < 1. 1

Question 46 marks

Question

(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! M1 9! , e = 2, 3 3! e! 60 480 A1 2 4(b) 7! M1 7! × 2×6 × k seen, k an integer > 0. 3! 3! M1 m! ×2 × q 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3, 1 ⩽ q ⩽ 8 all integers. n! M1 m! × p × 6 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3, 1 ⩽ p ⩽ 2 all integers. n! (Accept 3P2 for 6) 5! If M0 M0 M0 awarded, SC M1 for t × 12, t an integer ⩾ 20, . 3! 10 080 A1 Exact value. Alternative method for question 4(b) 7 P2 × 6! × 2 M1 6! × k seen, k an integer > 0. 3! 3! M1 m! 7 × P2 × q m = 6,9, 1 ⩽ n ⩽ 3, 1 ⩽ q ⩽ 2 all integers. n! M1 m! 7 × Pr × 2 m = 6, 9, 1 ⩽ n ⩽ 3, 1 ⩽ r ⩽ 5 all integers. n! 5! If M0 M0 M0 awarded, SC M1 for t × 84, t an integer ⩾ 20, . 3! 10 080 A1 Exact value. 4(b) Alternative method for question 4(b) 7! M1 7! × 4P2 × k seen, k an integer > 0. 3! 3! M1 5! t × 4P2 or 12, t an integer ⩾ 20, . 3! M1 m! × 4P2 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3 all integers. n! 10 008 A1 Exact value. 4

Question 57 marks

Question

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) [P(0, 1, 2) =] 10C0 0.160 0.8410 + 10C1 0.161 0.849 + 10C2 0.162 0.848 M1 One term: 10Cx px (1 – p)10–x for 0 < x < 10, any p. [= 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) ( 0.84 ) 0.16 7 M1 (1 – p)7p, 0 < p < 1 0.0472 A1 0.0472144 to at least 3sf. 2 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

Question 610 marks

Question

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

Question 710 marks

Question

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] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

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 7(b) [UQ at 75% of 140 = 105, LQ at 25% of 140 = 35] M1 Accept 660 ⩽ UQ ⩽ 720 – 240 ⩽ LQ ⩽ 290. [IQR:] 700 – 260 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 7(c) [Mean =] B1 Frequencies 16 30 42 34 12 6 16 × 100 + 30 × 250 + 42 × 400 + 34 × 700 + 12 × 1050 + 6 × 1400 140 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. 1600 + 7500 + 16 800 + 23 800 + 12 600 + 8400 70 700 Accept or . 140 140 70 770 Condone for M1. 140 505 A1 WWW Variance = M1 Values substituted into variance formula using (their mean)2 and their 16 × 1002 + 30 × 2502 + 42 × 4002 + 34 × 7002 + 12 × 10502 + 6 × 14002 midpoints and their frequencies (including for denominator). Accept unsimplified. Condone 1 data error. 140 2 Accept: −505 160 000 + 1875 000 + 6 720 000 + 16 660 000 + 13 230 000 + 11 760 000 [ 140 50 405 000 or or 360 035.7143] − 5052 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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Where can I find the 9709/52 October/November 2021 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, October/November 2021 (9709_w21_ms_52), and the question paper (9709_w21_qp_52), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/52 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.

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