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Mathematics 9709/53 October/November 2023: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_w23_qp_539709_w23_ms_539709/53/O/N/23

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/53/O/N/23
9709_w23_ms_53.pdf

9709/53 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 16 marks

Question

Becky sometimes works in an office and sometimes works at home. The random variable X denotes the number of days that she works at home in any given week. It is given that P X = x = kx x + 1, where k is a constant and x = 1, 2, 3 or 4 only. (a) Draw up the probability distribution table for X , giving the probabilities as numerical fractions. [3] (b) Find E X and Var X . [3]

Mark scheme

1(a) 1 M1 Using sum of probabilities = 1 to form an equation in k. 2k + 6k + 12k + 20k = 1, k = Accept 1×2×k + 2×3×k + 3×4×k + 4×5×k = 1. 40 X 1 2 3 4 M1 X 1 2 3 4 2 6 12 20 P(X) 2k 6k 12k 20k P(X) 40 40 40 40 Two correctly linked, accurate probabilities. May be in terms of k. 0.05 0.15 0.3 0.5 May not be in a table. A1 Table with correct X values and correct probabilities. 3 1(b) 1× 2 + 2 × 6 + 3 ×12 + 4 × 20 2 + 12 + 36 + 80 [E(X) = 1× 2k + 2 × 6k + 3 ×12k + 4 × 20k = 130k ] [E(X ) =] [E(X ) = ] 40 40 M1 Accept unsimplified expression. May be calculated in variance. FT their table with 3 or more probabilities summing to 1 (0 < p < 1). If there are outcomes in the table without probabilities, condone and treat as p = 0. 12 × 2 + 22 × 6 + 32 ×12 + 42 × 20 [Var(X) = 12 × 2k + 22 × 6k + 32 ×12k + 42 × 20k − (130k ) ] 2 Var(X ) = − (their E(X ))2 = M1 40 Appropriate variance formula using their (E(X))2 value. FT their 1× 2 + 4 × 6 + 9 ×12 + 16 × 20 2 13 table with 3 or more probabilities (0 < p < 1) which need not sum − their 40 4 to 1 and the highlighted calculation (or less simplified) seen, Note: if table is correct, 2 + 24 + 108 + 320 13 2 − their 454 227 40 4 or or any calculation − (their E ( X ))2 implies M1. 40 20 13 1 63 A1 Answers for E(X) and Var(X) must be identified. E(X) may be E(X ) = ,3 , 3.05 Var(X) = , 0.7875 4 4 80 identified by correct use in variance. Condone E, V, μ, σ etc. If A0 earned, SC B1 for identified correct final solutions. 3

Question 26 marks

Question

The weights of large bags of pasta produced by a company are normally distributed with mean 1.5 kg and standard deviation 0.05 kg. (a) Find the probability that a randomly chosen large bag of pasta weighs between 1.42 kg and 1.52 kg. [3] The weights of small bags of pasta produced by the company are normally distributed with mean 0.75 kg and standard deviation 3 kg. It is found that 68% of these small bags have weight less than 0.9 kg. (b) Find the value of 3. [3]

Mark scheme

2(a) 1.42 − 1.5 1.52 − 1.5 M1 Use of ± standardisation formula once with 1.5, 0.05 and either [P(1.42 < X <1.52) =] P( <Z < ) 0.05 0.05 1.42 or 1.52, allow σ 2 or σ , no continuity correction. [= P( −1.6 < Z < 0.4) = Φ ( 0.4 ) + Φ (1.6 ) − 1] M1 Calculating the appropriate probability area (leading to their final answer, expect > 0.5). = 0.6554 + 0.9452 – 1 or 0.6554 – 0.0548 = 0.601 A1 0.6005 < p ⩽ 0.601 SC B1 for 0.601 with no standardisation seen. 3 2(b) 0.9 − 0.75 B1 0.467 < z 0.468 or − 0.468 z < −0.467 seen P ( X < 0.9) = P Z < σ = 0.68 M1 ±standardisation formula with 0.9, 0.75, σ equating to a z-value 0.9 − 0.75 = 0.468 (not 0.32, 0.68, 0.532, 0.7517, 0.2483, 0.6255,) . σ Condone continuity correct ±0.05, not σ 2 , σ . 0.15 Condone ± = 0.468 . σ 25 A1 0.3205 ⩽ σ < 0.3215 σ = 0.321 , SC B1 if M0 www. 78 3

Question 37 marks

Question

Tim has two bags of marbles, A and B. Bag A contains 8 white, 4 red and 3 yellow marbles. Bag B contains 6 white, 7 red and 2 yellow marbles. Tim also has an ordinary fair 6-sided dice. He rolls the dice. If he obtains a 1 or 2, he chooses two marbles at random from bag A, without replacement. If he obtains a 3, 4, 5 or 6, he chooses two marbles at random from bag B, without replacement. (a) Find the probability that both marbles are white. [3] (b) Find the probability that the two marbles come from bag B given that one is white and one is red. [4]

Mark scheme

3(a) [P(WW) = P(AWW) + P(BWW) =] M1 2 8 7 4 6 5 Either × × or × × seen, accept unsimplified. 2 8 7 4 6 5 6 15 14 6 15 14 × × + × × 6 15 14 6 15 14 M1 q r r −1 6 − q s s −1 × × + × × seen, no additional terms, accept 6 15 14 6 15 14 unsimplified. q r r 6−q s s Condone × × + × × , 6 15 15 6 15 15 1 ⩽ q ⩽ 5, 1 < r, s <9. 56 60 4 2 58 A1 SC B1 for 58/315 if either M mark withheld. = 630 + 630 = 45 + 21 = 315 or 0.184 3 3(b) P ( W & R from bag B ) B1 P(W & R from bag B) P(B |WR or RW) = = P ( W and R ) 2 6 7 2 7 6 2 6 7 4 = × × + × × or 2 × × × [= or 0.267] 3 15 14 3 15 14 3 15 14 15 4 6 7 4 7 6 × × + × × Seen alone or as numerator/denominator of conditional 6 15 14 6 15 14 probability. 2 8 4 2 4 8 4 6 7 4 7 6 × × + × × + × × + × × 6 15 14 6 15 14 6 15 14 6 15 14 M1 P(WR or RW) = P(W & R from bag A) + P(W & R from bag B) 2 8 4 4 6 7 = a× × × + a × × × or 4 6 7 6 15 14 6 15 14 2× × × or 6 15 14 2 8 4 2 8 4 4 6 7 = a × × × + their P(W & R from bag B ). 2× × × + 2× × × 6 15 14 6 15 14 6 15 14 a = 1 or 2. 116 [expect or 0.368 ] 315 Seen alone or as numerator/denominator of conditional probability. 168 4 M1 their identified P (W & R frombag B ) 630 = 15 their identified P ( WR or RW ) 232 116 630 315 Accept unsimplified. 168 21 A1 0.7241379 to at least 3SF. = , or 0.724 232 29 4

Question 410 marks

Question

The weights, x kg, of 120 students in a sports college are recorded. The results are summarised in the following table. Weight (x kg) x ≤ 40 x ≤ 60 x ≤ 65 x ≤ 70 x ≤ 85 x ≤ 100 Cumulative frequency 0 14 38 60 106 120 (a) Draw a cumulative frequency graph to represent this information. [2] (b) It is found that 35% of the students weigh more than W kg. Use your graph to estimate the value of W . [2] (c) Calculate estimates for the mean and standard deviation of the weights of the 120 students. [6]

Mark scheme

4(a) M1 At least 3 points plotted accurately at class upper end points (40,0) (60,14) (65,38) (70,60) (85,106) (100,120). Linear cumulative frequency scale 0 ⩽ cf ⩽ 120 and linear weight scale 40 ⩽ weight(kg) ⩽ 100 with at least 3 values identified on each axis. Condone scale reversed. A1 All points plotted correctly, curve drawn (within tolerance) and joined to (40,0). Axes labelled cumulative frequency (cf), weight (w) and kg (kilograms) – or a suitable title. 2 4(b) [120 × 0.65 = ] 78 seen M1 May be implied by use on graph. 76 [kg] A1 75 < hours < 79. Indication of use of graph required. 2 4(c) Frequencies: [0] 14 24 22 46 14 B1 At least 5 correct frequencies seen (condone omission of 0). Midpoints: 20 50 62.5 67.5 77.5 92.5 B1 At least 5 correct midpoints seen (condone omission of 20). Mean = M1 Correct formula for mean using their midpoints and their 0 × 20 + 14 × 50 + 24 × 62.5 + 22 × 67.5 + 46 × 77.5 + 14 × 92.5 8545 frequencies, implied by if correct midpoints & frequencies 120 120 [0] + 700 + 1500 + 1485 + 3565 + 12950 8545 seen. = = 120 120 May be gained in variance calculation. If midpoints not clearly identified, condone midpoints ± 0.5. = 71.2 A1 1709 5 Accept , 71 or 71.208333 to at least 3SF. 24 24 1709 5 If M0 scored, SC B1 for , 71 or 71.208333 to at least 3SF 24 24 www. Variance = M1 Correct formula for variance using their midpoints, their 0 × 20 + 14 × 50 + 24 × 62.5 + 22 × 67.5 + 46 × 77.5 + 14 × 92.5 2 2 2 2 2 2 frequencies and their mean. − 71.22 2 120 625062.5 8545 Implied by − if correct midpoints & [0] + 35000 + 93750 + 100237.5 + 276287.5 + 119787.5 8545 2 − 120 120 120 120 frequencies seen. [= 138.23] Standard deviation = 11.8 A1 11.757016 to at least 3SF. 6

Question 511 marks

Question

The probability that a driver passes an advanced driving test is 0.3 on any given attempt. (a) Dipak keeps taking the test until he passes. The random variable X denotes the number of attempts required for Dipak to pass the test. (i) Find P 2 ≤ X ≤ 6. [2] (ii) Find E X . [1] Five friends will each take their advanced driving test tomorrow. (b) Find the probability that at least three of them will pass tomorrow. [3] 75 people will take their advanced driving test next week. (c) Use an approximation to find the probability that more than 20 of them will pass next week. [5]

Mark scheme

5(a)(i) Method 1 6) = P ( X 6) − P ( X 1) =] 1 − ( 0.7 ) − (1 − 0.7 ) M1 1 – 0.7n seen, n = 5, 6. 6 [P(2 X = 0.582 A1 www 0.582351 to at least 3SF. Method 2 P(X= 2, 3, 4, 5, 6) M1 Sum of first 4 or 5 correct terms – no incorrect terms. = 0.7 × 0.3 + 0.72 × 0.3 + 0.73 × 0.3 + 0.7 4 × 0.3 + 0.75 × 0.3 = 0.21 + 0.147 + 0.1029 + 0.07203 + 0.050421 = 0.582 A1 www 0.582351 to at least 3SF. 2 5(a)(ii) 1 B1 10 1 3 Condone 3.33, 3. 3 or – NOT . 3 3 0.3 1 5(b) Method 1 [P(3, 4, 5) =] 5C3 ( 0.3) ( 0.7 ) + 5C4 ( 0.3) ( 0.7 ) + 5C5 ( 0.3) ( 0.7 ) M1 One term seen 5Cx ( p ) x (1 − p )5− x , 0 < p < 1, x ≠ 0,5. 3 2 4 1 5 0 = 0.1323 + 0.02835 + 0.00243 A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 4077 B1 0.16308 to at least 3SF. = 0.163, 25000 Method 2 [1 – P(0, 1, 2) = M1 One term 5Cx ( p ) x (1 − p )5− x , 0 < p < 1, x ≠ 0,5. 1 − ( 5C0 ( 0.3) ( 0.7 ) + 5C1 ( 0.3) ( 0.7 ) + 5C2 ( 0.3) ( 0.7 ) ) 0 5 1 4 2 3 = 1 – (0.16807 + 0.36015 + 0.3087) A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 4077 B1 0.16308 to at least 3SF. = 0.163, 25000 3 5(c) [Mean = 75 ×0.3 =] 22.5 B1 3 22.5, 22½ and 15.75, 15 seen, allow unsimplified. [Var = 75 ×0.3 × 0.7 =] 15.75 4 3 7 (σ = or 3.9686269… to at least 3SF implies correct variance) 2 20.5 − 22.5 M1 Substituting their µ and σ into ±standardisation formula (any [P( X > 20) =] P Z > number for 20.5), not σ2 not √σ. 15.75 M1 Using continuity correction 19.5 or 20.5 in their standardisation formula. [P( Z > −0.504) = Φ ( 0.504 ) ] M1 Appropriate area Φ, from final process, must be a probability. Expect final answer > 0.5. = 0.693 Note: correct final answer implies this M1. A1 0.6925 < p ⩽ 0.693 5

Question 610 marks

Question

Jai and his wife Kaz are having a party. Jai has invited five friends and each friend will bring his wife. (a) At the beginning of the party, the 12 people will stand in a line for a photograph. (i) How many different arrangements are there of the 12 people if Jai stands next to Kaz and each friend stands next to his own wife? [3] (ii) How many different arrangements are there of the 12 people if Jai and Kaz occupy the two middle positions in the line, with Jai’s five friends on one side and the five wives of the friends on the other side? [2] (b) For a competition during the party, the 12 people are divided at random into a group of 5, a group of 4 and a group of 3. Find the probability that Jai and Kaz are in the same group as each other. [5] 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)(i) Method 1 6! × 26 M1 6! × a, a integer > 1. M1 b × 26 , b integer ⩾ 1. = 46080 A1 Accurate answer required. SC B1 for 46080 if M0 M0 www. Alternative method for question 6(a)(i) 12 × 10 × 8 × 6 × 4 × 2 M1 c×d ×e× f × g ×h 2 ⩽ c,d,e,f,g,h (different integers) ⩽ 12 M1 Correct unsimplified. = 46080 A1 Accurate answer required. SC B1 for 46080 if M0 M0 www. 3 6(a)(ii) 5! × 5! × 2 ×2 M1 5! × 5! × k, k positive integer, 1 may be implied (no adding/subtracting). = 57600 A1 2 6(b) Method 1 probabilities of J & K being placed: 5 4 × 20 5 B1 Correct probability for one identified scenario. In the group of 5 = 132 , 33 12 11 M1 Denominator 12 ×11 for all probabilities, (1, 2 or 3 scenarios). 4 3 × 12 1 In the group of 4 = 132 , 11 12 11 A1 3 correct probabilities, accept unsimplified. 3 2 × 6 1 In the group of 3 = 132 , 22 12 11 5 4 4 3 3 2 M1 Adding probabilities for 3 correct scenarios. × + × + × 12 11 12 11 12 11 19 A1 0.2878787 to at least 3SF. , 0.288 66 6(b) Method 2 number of arrangements of J & K being placed: B1 Correct value of one identified scenario seen, accept unsimplified. In the group of 5 10 C3 × 7C4 [= 120 × 35 = 4200] M1 12 Ca × 12-aCb, a = 3, 4, 5; b = 3, 4, 5 (a ≠ b) In the group of 4 10 C2 × 8C5 [= 45 × 56 = 2520] In the group of 3 10 C1 × 9C5 [= 10 × 126 = 1260] [Total number of ways of arranging the 3 groups =] A1 27720 Seen alone or as denominator of probability –accept 12 C5 ×7C4 = 792 ×35 = 27720 unsimplified. or 12C3 × 9C4 or 12C4 × 8C5 SC B1 if M0. 4200 + 2520 + 1260 = 7980 M1 Values of 3 correct scenarios added, accept unsimplified – or correct. 7980 19 A1 0.2878787 to at least 3SF. [Probability =] , , 0.288 27720 66 5 Note, alternative arrangement calculations possible e.g. In the group of 5 10 C3 × 7C4 [= 120 × 35 = 4200] In the group of 4 10 C5× 5C2 [= 252× 10 = 2520] In the group of 3 10 C5 × 5C4 [= 252 × 5 = 1260]

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On this page, free: the mark scheme for AS & A Level Mathematics 9709/53, October/November 2023 (9709_w23_ms_53), and the question paper (9709_w23_qp_53), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/53 for October/November 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.

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