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Mathematics 9709/52 February/March 2022: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_m22_qp_529709_m22_ms_529709/52/F/M/22

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/52/F/M/22
9709_m22_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 red spinner has edges numbered 1, 2, 2, 3. A fair blue spinner has edges numbered −3, −2, −1, −1. Each spinner is spun once and the number on the edge on which each spinner lands is noted. The random variable X denotes the sum of the resulting two numbers. (a) Draw up the probability distribution table for X . [3] (b) Given that E X = 0.25, find the value of Var X . [2]

Mark scheme

1(a) B1 Table with correct X values and at least one probability X −2 −1 0 1 2 0 < p <1. Condone any additional X values if probability stated as 0. No repeated X values. P(X) 1 3 5 5 2 16 16 16 16 16 B1 3 correct probabilities linked with correct outcomes, may not be in table. 0.0625 0.1875 0.3125 0.3125 0.125 B1 2 further correct probabilities linked with correct outcomes, may not be in table No repeated X values. SC if less than 3 correct probabilities seen, award SCB1 Sum of their probabilities, 0 < p <1, of 4,5 or 6 X values = 1 (condone summing to 1±0∙01 or better). 3 1(b) 1 M1 Appropriate variance formula using (E(X))2 value, accept 1 2 2 3 2 5 2 5 2 2 2 × −2 + × −1 + × 0 + × 1 + × 2 − unsimplified. FT their table with at least 3 different X values 16 16 16 16 16 4 even if probabilities not summing to 1, 0 < p < 1. 1× 4 + 3 × 1 + 5 × 0 + 5 × 1 + 2 × 4 Condone 1 error providing all probabilities <1 and 0.252 used − 0.252 16 5 1 19 A1 Condone 1.188 or 1.19 WWW = 4 − 16 = 16 , 1.1875 2

Question 26 marks

Question

In a certain country, the probability of more than 10 cm of rain on any particular day is 0.18, independently of the weather on any other day. (a) Find the probability that in any randomly chosen 7-day period, more than 2 days have more than 10 cm of rain. [3] (b) For 3 randomly chosen 7-day periods, find the probability that exactly two of these periods have at least one day with more than 10 cm of rain. [3]

Mark scheme

2(a) [P(>2) = 1 – P(0,1,2) =] M1 One term 7Cx p x (1 − p )7 − x , 0 < p < 1, 0 < x < 7 1 – (7C0 0.180 0.827 + 7C1 0.181 0.826 + 7C2 0.182 0.825 ) = 1 – (0.249285 + 0.383048 + 0.252251) A1 Correct unsimplified expression or better = 1 – 0.88458 Condone omission of brackets if recovered 0.115 B1 WWW. 0∙115 ⩽ p < 0∙1155 not from wrong working 3 2(b) [P(at least 1 day of rain) = 1 – P(0) = 1 − ( 0.82 ) =] 0.7507 7 B1 AWRT 0.751 seen [P(exactly 2 periods) =] 0.7507 2 × (1 − 0.7507 ) × 3 M1 FT their 1− p 7 or their 0.7507 if identified, not 0.18, 0.82 Accept ×3Cr, r=1,2 or ×3P1 for ×3 Condone ×2 0.421 A1 Accept 0.421 ⩽ p ⩽ 0.4215 SC B1 if 0/3 scored for final answer only 0.421 ⩽ p ⩽ 0.4215 3

Question 36 marks

Question

At a summer camp an arithmetic test is taken by 250 children. The times taken, to the nearest minute, to complete the test were recorded. The results are summarised in the table. Time taken, in minutes 1 − 30 31 − 45 46 − 65 66 − 75 76 − 100 Frequency 21 30 68 86 45 (a) Draw a histogram to represent this information. [4] (b) State which class interval contains the median. [1] (c) Given that an estimate of the mean time is 61.05 minutes, state what feature of the distribution accounts for the median and the mean being different. [1]

Mark scheme

3(a) M1 At least 4 frequency densities calculated Class 30 15 20 10 25 A1 All heights correct on graph Width Frequency 0.7 2 3.4 8.6 1.8 B1 Bar ends at 0∙5, 30∙5, 45∙5, 65.5, 75.5, 100.5 Density (at axis), 5 bars drawn, condone 0 in first bar 0.5 ⩽ time axis ⩽ 100.5, linear scale with at least 3 values indicated. B1 Axes labelled: Frequency density (fd), time (t) and mins (or appropriate title). Linear fd scale, with at least 3 values indicated 0 ⩽ fd axis ⩽ 8∙6 4 3(b) 66 – 75 B1 Condone 65.5 – 75.5 1 3(c) Distribution is not symmetrical B1 Or skewed, ignore nature of skew 1

Question 411 marks

Question

The weights of male leopards in a particular region are normally distributed with mean 55 kg and standard deviation 6 kg. (a) Find the probability that a randomly chosen male leopard from this region weighs between 46 and 62 kg. [4] The weights of female leopards in this region are normally distributed with mean 42 kg and standard deviation 3 kg. It is known that 25% of female leopards in the region weigh less than 36 kg. (b) Find the value of 3. [3] The distributions of the weights of male and female leopards are independent of each other. A male leopard and a female leopard are each chosen at random. (c) Find the probability that both the weights of these leopards are less than 46 kg. [4]

Mark scheme

4(a) 46 − 55 62 − 55 M1 46 or 62, 55 and 6 substituted into ±standardisation formula P( 46 < X < 62) = P <Z < once. Condone 62 and continuity correction ±0.5 6 6 7 B1 Both standardisation values correct, accept unsimplified = P −1.5 < Z < 6 7 M1 Calculating the appropriate area from stated Φs of z-values, =Φ 6 − (1 − Φ (1.5 ) ) must be probabilities. = 0.8784 + ( 0.9332 − 1) 0.812 A1 0∙8115 < p ⩽ 0∙812 4 4(b) z = ±0.674 B1 CAO, critical z-value 36 − 42 M1 36 and 42 substituted in ±standardisation formula, no = −0.674 continuity correction, not σ2, √ σ, equated to a z-value σ σ = 8.9 [ 0] A1 WWW. Only dependent on M. 3 4(c) P(male < 46) = 1−their 0.9332 = 0.0668 M1 FT value from part (a) or 46 − 55 2 Correct: 1 − Φ ,condone continuity correction, σ , 6 √ σ, and probability found. Condone unsupported correct value stated. 46 − 42 M1 46, 42 and their 4(b) σ (or correct σ) substituted in P(female < 46) = P( Z < ) = Φ ( 0.449 ) their 8.90 ±standardisation formula, condone continuity correction, σ2, = 0.6732 √ σ, and probability found 4 Condone . their 8.90 P(both) = 0.0668 ×0.6732 M1 Product of their 2 probabilities (0 < both < 1) Not 0.25 or their final answer to 4(a) used. 0.0450 or 0.0449 A1 0∙0449 ⩽ p ⩽ 0∙0450 4 5

Question 510 marks

Question

A group of 12 people consists of 3 boys, 4 girls and 5 adults. (a) In how many ways can a team of 5 people be chosen from the group if exactly one adult is included? [2] (b) In how many ways can a team of 5 people be chosen from the group if the team includes at least 2 boys and at least 1 girl? [4] The same group of 12 people stand in a line. (c) How many different arrangements are there in which the 3 boys stand together and an adult is at each end of the line? [4]

Mark scheme

5(a) C1 × 7C4 M1 7 C4 × k, k integer ⩾ 1 Condone 5P1 for M1 only 175 A1 2 3 5(b) 2B 1G 2A C2 × 4C1 × 5C2 = 120 M1 3 Cx × 4Cy × 5Cz , x + y + z = 5, x,y,z integers ⩾1 3 2B 2G 1A C2 × 4C2 × 5C1 = 90 Condone use of permutations for this mark 3 2B 3G C2 × 4C3 = 12 3 4 5 3B 1G 1A C3 × C1 × C1 = 20 B1 2 appropriate identified outcomes correct, allow unsimplified 3 3B 2G C3 × 4C2 =6 M1 Summing their values for 4 or 5 correct identified scenarios only (no repeats or additional scenarios), condone identification by unsimplified expressions [Total =] 248 A1 Note: Only dependent upon M marks 4 5(c) 8! × 3! × 5P2 M1 8! × m, m an integer ⩾ 1 Accept 8 × 7! for 8! M1 3! × n, n an integer > 1 M1 p × 5P2, p × 5C2 × 2, p × 20, p an integer > 1 If extra terms present, maximum 2/3 M marks available 4 838 400 A1 Exact value required 4

Question 612 marks

Question

A factory produces chocolates in three flavours: lemon, orange and strawberry in the ratio 3 : 5 : 7 respectively. Nell checks the chocolates on the production line by choosing chocolates randomly one at a time. (a) Find the probability that the first chocolate with lemon flavour that Nell chooses is the 7th chocolate that she checks. [1] (b) Find the probability that the first chocolate with lemon flavour that Nell chooses is after she has checked at least 6 chocolates. [2] ‘Surprise’ boxes of chocolates each contain 15 chocolates: 3 are lemon, 5 are orange and 7 are strawberry. Petra has a box of Surprise chocolates. She chooses 3 chocolates at random from the box. She eats each chocolate before choosing the next one. (c) Find the probability that none of Petra’s 3 chocolates has orange flavour. [2] (d) Find the probability that each of Petra’s 3 chocolates has a different flavour. [3] (e) Find the probability that at least 2 of Petra’s 3 chocolates have strawberry flavour given that none of them has orange flavour. [4] 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) 3 1 B1 0.0524288 rounded to more than 3SF if final answer [Probability of lemon = = ] 15 5 4 6 1 4096 × = , 0.0524 5 5 78125 1 6(b) 1 6 M1 4 6 1 1 − or . FT their or correct. From final answer 5 5 5 5 5 6 4 1 4 4 Condone or × + 5 5 5 5 4096 A1 0.262144 rounded to more than 3SF , 0.262 15625 Alternative method for question 6(b) [1 – P(1,2,3,4,5,[6]) =] M1 From final answer 5 1 4 1 4 2 1 4 3 1 4 4 1 4 5 1 4 1 1 – + × + × + × + × + × Condone omission of × 5 5 5 5 5 5 5 5 5 5 5 5 5 4096 A1 0.262144 rounded to more than 3SF , 0.262 15625 2 6(c) 10 9 8 M1 a a −1 a − 2 × × × × , no additional terms 15 14 13 15 14 13 24 A1 0.263736 rounded to more than 3SF , 0∙264 91 Alternative method for question 6(c) 3 2 1 3 2 7 3 7 6 7 6 5 M1 [3Ls + 2Ls1S + 1L2Ss + 3Ss] × × + 3× × × +3 × × × + × × Condone one numerator error. 15 14 13 15 14 13 15 14 13 15 14 13 Condone no multiplications seen if tree diagram complete with probabilities on each branch, scenarios listed and attempt at evaluation 24 A1 0.263736 rounded to more than 3SF , 0∙264 91 Alternative method for question 6(c) 5 4 3 5 4 10 5 10 9 M1 1 ‒ P(3,2,1 oranges) 1− × × + 3× × × + 3× × × Condone one numerator error. 15 14 13 15 14 13 15 14 13 24 A1 0.263736 rounded to more than 3SF , 0∙264 91 Alternative method for question 6(c) 10 C3 M1 15 c3 24 A1 0.263736 rounded to more than 3SF , 0∙264 91 2 6(d) 7 5 3 M1 7 5 3 × × × 3! All probabilities of the form: × × , 13 ⩽ a,b,c ⩽ 15 15 14 13 a b c M1 e g i × × ×3! e,f,g,h,i.j positive integers forming f h j probabilities or 6 identical probability calculations or values added, no additional terms 3 A1 0∙230769 rounded (not truncated) to more than 3SF , 0∙231 13 Alternative method for question 6(d) 3 C1 × 5 C1 × 7 C1 M1 3 C1 × 5 C1 × 7 C1 15 , k integer > 1 C3 k Condone use of permutations M1 3 C a × 5 Cb × 7 C c 15 , 0<a<3, 0<b<5, 0<c<7, C3 Condone use of permutations 3 A1 0∙230769 rounded (not truncated) to more than 3SF , 0∙231 13 3 6(e) 7 6 5 3 7 6 B1 3 7 6 × × + × × ×3 × × × 3 seen (SSL, SLS, LSS) 15 14 13 15 14 13 14 24 15 14 13 their ( c ) = 65 ÷ 91 3 126 SC B1 × 3, × 3 seen 65 2730 B1 7 6 5 × × seen in numerator (SSS) 15 14 13 210 1 SCB1 , seen in numerator 2730 13 M1 Fraction with their (c) or correct in denominator 720 24 , , 0.263736 2730 91 49 A1 Accept 0.816 = , 0∙817 60 Alternative method for question 6(e) 7 C 2 × 3 C1 + 7 C3 B1 7 C2 × 3 C1 seen (SSL, SLS, LSS) 10 C3 SCB1 21 × 3 seen or use of permutations B1 7 C3 seen in numerator (SSS) SCB1 35 seen in numerator or use of permutations M1 Fraction with 10 C or consistent with their numerator of 6(c) 3 in denominator 49 A1 Accept 0.816 = , 0∙817 60 4

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Where can I find the 9709/52 February/March 2022 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, February/March 2022 (9709_m22_ms_52), and the question paper (9709_m22_qp_52), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/52 for February/March 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.

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