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Mathematics 9709/52 October/November 2020: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_w20_qp_529709_w20_ms_529709/52/O/N/20

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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/20
9709_w20_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

A fair six-sided die, with faces marked 1, 2, 3, 4, 5, 6, is thrown repeatedly until a 4 is obtained. (a) Find the probability that obtaining a 4 requires fewer than 6 throws. [2] On another occasion, the die is thrown 10 times. (b) Find the probability that a 4 is obtained at least 3 times. [3]

Mark scheme

1(a) 5 5 M1 1 – pn n = 5,6 1− or p + pq + pq2+pq3+ pq4 (+ pq5) 6 0 < p < 1, p + q = 1, 2 3 4 1 5 1 5 1 5 1 5 1 or + × + × + × + × 6 6 6 6 6 6 6 6 6 4651 A1 0·598, 7776 2 1(b) (1 – P(0, 1, 2)) M1 10 10 − x Cx p x (1 − p ) , 0 < p < 1, any p, x ≠ 0,10 5 10 1 5 9 1 5 2 8 1 – +10C1 + 10C2 6 6 6 6 6 1 – (0·1615056 + 0·3230111 + 0·290710) A1 Correct expression, accept unsimplified, condone omission of final bracket 0·225 A1 0·2247 < p ≤ 0·225, WWW 3

Question 27 marks

Question

A bag contains 5 red balls and 3 blue balls. Sadie takes 3 balls at random from the bag, without replacement. The random variable X represents the number of red balls that she takes. (a) Show that the probability that Sadie takes exactly 1 red ball is 15 56 . [2] (b) Draw up the probability distribution table for X . [3] (c) Given that E X = 15 8 , find Var X . [2]

Mark scheme

2(a) 5 3 2 M1 a b c 5 3 2 P(1 red) = × × ×3 × × × k or × × × 3 , 1⩽ a,b,c ⩽ 5, d ,e ,f ⩽8, a, b, c, 8 7 6 8 7 6 d e f d, e, f, k all integers.1 < k ⩽ 3, 15 A1 AG, WWW 56 Alternative method for question 2(a) 5 C1 × 3 C2 M1 a C1 × b C2 5 C d × 3 Ce 8 8 or 8 or C3 C3 C3 5 ( Cd × 3 Ce or a C1 × b C 2 ) , 5 C3 × 3 C0 + 5 C2 × 3 C1 + 5 C1 × 3 C 2 + 5 C0 × 3 C3 a + b = 8, d + e = 3 15 A1 15 AG, WWW, must be seen 56 56 2 2(b) B1 Probability distribution table with correct outcomes with at least x 0 1 2 3 one probability less than 1, allow extra outcome values if Prob. probability of zero stated. 1 15 30 15 10 5 = = 56 56 56 28 56 28 B1 2 of P(0), P(2) and P(3) correct 0·0179 0·268 0·536 0·179 B1 FT 4th probability correct or FT sum of 3 or more probabilities = 1, with P(1) correct 3 2(c) (02 × 1) + 12 × 15 + 22 × 30 + 32 × 10 15 2 M1 Substitute their attempts at scores in correct variance formula, Var(X) = − must have ‘ – mean2’ (FT if mean calculated) (condone 56 8 probabilities not summing to 1 for this mark) 2 15 120 90 15 = + + − 56 56 56 8 225 A1 , 0·502 448 2

Question 39 marks

Question

Pia runs 2 km every day and her times in minutes are normally distributed with mean 10.1 and standard deviation 1.3. (a) Find the probability that on a randomly chosen day Pia takes longer than 11.3 minutes to run 2 km. [3] (b) On 75% of days, Pia takes longer than t minutes to run 2 km. Find the value of t. [3] (c) On how many days in a period of 90 days would you expect Pia to take between 8.9 and 11.3 minutes to run 2 km? [3]

Mark scheme

3(a) 11.3 − 10.1 M1 Using ± standardisation formula, P(X > 11.3) = P( z > ) = P( z > 0 ⋅ 9231) 1.3 no σ or σ 2 , continuity correction 1 – 0.822 M1 Appropriate area Φ , from standardisation formula P(z>…) in final solution 0·178 A1 0.1779… 3 3(b) z = −0 ⋅ 674 B1 ±0.674 seen (critical value) t − 10.1 M1 An equation using ±standardisation formula with a z-value, = −0 ⋅ 674 1.3 condone σ or σ 2 , continuity correction. t = 9·22 A1 AWRT. Only dependent on M1 3 3(c) P(8.9 < X < 11.3) = 1 – 2 × their 3(a) B1 FT FT from their 3(a) < 0·5 or correct, accept unevaluated ≡ 2(1 – their 3(a)) – 1 probability ≡ 2(0·5 – their 3(a)) OE =0.644 Number of days = 90 × 0·644 M1 90 × their p seen, 0 < p < 1 = 57·96 So 57 (days) A1 FT Accept 57 or 58, not 57·0 or 58·0, no approximation/rounding stated FT must be an integer value Alternative method for question 3(c) 8 ⋅ 9 − 10 ⋅ 1 11.3 − 10.1 B1 Accept unevaluated probability P <z< 1⋅ 3 1⋅ 3 = Φ ( 0 ⋅ 9231) − (1 − Φ ( 0 ⋅ 9231) ) oe = 0·822 – (1 – 0·822) = 0·644 Number of days = 90 × 0·644 M1 90 × their p seen, 0 < p < 1 = 57·96 So 57 (days) A1 FT Accept 57 or 58, not 57·0 or 58·0, no approximation/rounding stated FT must be an integer value 3

Question 49 marks

Question

In a certain country, the weather each day is classified as fine or rainy. The probability that a fine day is followed by a fine day is 0.75 and the probability that a rainy day is followed by a fine day is 0.4. The probability that it is fine on 1 April is 0.8. The tree diagram below shows the possibilities for the weather on 1 April and 2 April. (a) Complete the tree diagram to show the probabilities. [1] 1 April 2 April Fine Fine Rainy Fine Rainy Rainy (b) Find the probability that 2 April is fine. [2] Let X be the event that 1 April is fine and Y be the event that 3 April is rainy. (c) Find the value of P X › Y . [3] (d) Find the probability that 1 April is fine given that 3 April is rainy. [3]

Mark scheme

4(a) B1 All probabilities correct, may be on branch or next to ‘Fine/Rainy’ Ignore additional branches. Fine 0.75 0.8 Fine 0.25 Rainy 0.4 Fine 0.2 Rainy 0.6 Rainy 1 4(b) 0·8 × 0·75 + 0·2 × 0·4 (= 0·6 + 0·08) M1 Correct or FT from their diagram unsimplified, all probabilities 0 < p < 1. Partial evaluation only sufficient when correct. Accept working in 4(b) or by the tree diagram. 17 A1 From supporting working 0.68, 25 2 4(c) 0·8 × 0·75 × 0·25 + 0·8 × 0·25 × 0·6 M1 a × b × c + a × 1-b × d, 0 < c, d ⩽ 1, a, b consistent with their tree diagram or correct, no additional terms 0·15 + 0·12 A1 At least one term correct, accept unsimplified 0.27 A1 Final answer 3 4(d) P(Y) = their (c) + 0·2 × 0·4 × 0·25 + 0·2 × 0·6 × 0·6 B1 FT their (c) + e × f × g + e × (1-f) × h, 0 < g, h ⩽ 1, e, f consistent (= 0·362) with their tree diagram, or correct their ( c ) 0.27 M1 their 4(c) (or correct)/their previously calculated and identified P(X|Y) = = P(Y) or a denominator involving 3 or 4 3-factor probability their P ( Y ) 0.362 terms consistent with their tree diagram & third factor 0 < p < 1 373 135 A1 (0.7458…) 0·746, or 500 181 3

Question 59 marks

Question

The following table gives the weekly snowfall, in centimetres, for 11 weeks in 2018 at two ski resorts, Dados and Linva. Dados 6 8 12 15 10 36 42 28 10 22 16 Linva 2 11 15 16 0 32 36 40 10 12 9 (a) Represent the information in a back-to-back stem-and-leaf diagram. [4] (b) Find the median and the interquartile range for the weekly snowfall in Dados. [3] (c) The median, lower quartile and upper quartile of the weekly snowfall for Linva are 12, 9 and 32 cm respectively. Use this information and your answers to part (b) to compare the central tendency and the spread of the weekly snowfall in Dados and Linva. [2]

Mark scheme

5(a) B1 Correct stem can be upside down, ignore extra values Dados Linva B1 Correct Dados labelled, leaves in order and lined up vertically 8 6 0 0 2 9 (less than midway to next column), no commas etc, no extra 6 5 2 0 0 1 0 1 2 5 6 terms 8 2 2 B1 Correct Linva on opposite side of stem labelled, leaves in order and lined up vertically (less than midway to next column), no 6 3 2 6 commas etc, no extra terms 2 4 0 B1 Correct single key for their diagram, need both resorts identified and ‘cm’ stated at least once here or in leaf headings or title. KEY 6| 3| 2 means 36 cm (snow) in Dados SC If 2 separate diagrams drawn, SCB1 if both keys meet these and 32 cm (snow) in Linva criteria B0B1B0SCB1 max. 4 5(b) Median or Q2 = 15 (cm) B1 Correct UQ or Q3 = 28 cm, LQ or Q1 = 10 cm M1 22 ⩽ UQ ⩽ 36 – 8 ⩽ LQ ⩽ 10 IQR = 28 – 10 18 (cm) A1 WWW 3 5(c) On average the snowfall in Davos is higher B1 FT FT from their 5(b) values for Dados. Statement comparing central tendency in context The amount of snowfall in Linva varies more than in Davos B1 FT Statement comparing spread in context Note: simply stating and comparing the values is not sufficient. 2 9

Question 611 marks

Question

Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children, are visiting an activity centre together. They will divide into groups for some of the activities. (a) In how many ways can the 9 people be divided into a group of 6 and a group of 3? [2] 5 of the 9 people are selected at random for a particular activity. (b) Find the probability that this group of 5 people contains all 3 of the Baker children. [3] All 9 people stand in a line. (c) Find the number of different arrangements in which Mr Ahmed is not standing next to Mr Baker. [3] (d) Find the number of different arrangements in which there is exactly one person between Mr Ahmed and Mr Baker. [3] 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

6(a) C6 (× 3C3) M1 9 Ck × n, k = 6, 3, n = 1,2 oe Condone 9C6 + 3C3, 9P6 × 3P3 84 A1 Accept unevaluated. 2 6(b) Number with 3 Baker children = 6C2 or 15 B1 Correct seen anywhere, not multiplied or added Total no of selections = 9C5 or 126 M1 Seen as denominator of fraction number of selections with 3 Baker children Probability = total number of selections 15 A1 5 , 0·119 OE, e.g. 126 42 Alternative method for question 6(b) 5 3 2 1 6 5 5 B1 C3 (OE) or 10 seen anywhere, multiplied by fractions only, not × × × × × C3 added 9 8 7 6 5 M1 3 2 1 6 5 × × × × × k , 1 ⩽ k, k integer 9 8 7 6 5 15 A1 5 , 0·119 OE, e.g. 126 42 3 6(c) [Total no of arrangements = 9!] M1 9! – k or 362880 – k, k an integer<362 880 [Arrangements with men together = 8! × 2] Not together: 9! – 8! × 2 B1 8! × 2(!) or 80 640 seen anywhere 282 240 A1 Exact value Alternative method for question 6(c) 7! × 8 × 7 B1 7! × k, k positive integer > 1 M1 m × 8 × 7, m × 8P2, m × 8C2 m positive integer > 1 282 240 A1 Exact value 3 6(d) 7! × 2 × 7 M1 7! × k, k positive integer > 1 If 7! not seen, condone 7 × 6 × 5 × 4 × 3 × 2 × (1) × k or 7 × 6! × k only M1 m × 2 × 7, m positive integer > 1 70 560 A1 3

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Where can I find the 9709/52 October/November 2020 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, October/November 2020 (9709_w20_ms_52), and the question paper (9709_w20_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 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.

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Yes. Upload your answers to 9709/52/O/N/20 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.