Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 5 · Variant 3

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

Mark scheme: 2(a) − 1.5 1.52 − M1 Use of ± standardisation formula once with 1.5, 0.05 and either [P(1.42 < X <1.52) =] P(1.42  Z  1.5) 2 0.05 0.05 1.42 or 1.52, allow  or , 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 Z  = 0.68 X  0.9 ) = P    P (       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,) .  Condone continuity correct ±0.05, not 2, . 0.15 Condone  = 0.468 .  25 A1 0.3205 ⩽ σ < 0.3215 = 0.321 , SC B1 if M0 www. 78 3

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3 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. 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(b) Find the probability that the two marbles come from bag B given that one is white and one is red. 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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. = + = + = or 0.184    630 630 45 21  315 3 3(b)  P ( W & R from bag B )  B1 P(W & R from bag B)  P(B |WR or RW) = =  2 6 7 2 7 6 2 6 7 4  P ( W and R )  =   +   or 2    [= 0.267] 3 15 14 3 15 14 3 15 14 15or 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    6 15 14 2 8 4 or = a    + their P(W & R from bag B ). 2 8 4 4 6 7 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 Accept unsimplified. 630 315 168 21 A1 0.7241379 to at least 3SF. = , or 0.724 232 29 4

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4 The weights, xkg, of 120 students in a sports college are recorded. The results are summarised in the following table. Weight (xkg) 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 2 + 14  50 2 + 24  62.5 2 + 22  67.5 2 + 46  77.5 2 + 14  92.5 2 2 frequencies and their mean. − 71.2 120 625062.5  8545  2 2 Implied by −  if correct midpoints & [0] + 35000 + 93750 + 100237.5 + 276287.5 + 119787.5  8545  120  120  −  120  120  frequencies seen. [= 138.23] Standard deviation = 11.8 A1 11.757016 to at least 3SF. 6

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6 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? 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(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? 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(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. 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Mark scheme: 6(a)(i) Method 1 6!  2 6 M1 6!  a, a integer > 1. M1 b 2 6 , binteger ⩾ 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,kpositive 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  = ,   12 11  132 33  M1 Denominator 12 11 for all probabilities, (1, 2 or 3 scenarios). 4 3  12 1  In the group of 4  = ,   12 11  132 11  A1 3 correct probabilities, accept unsimplified. 3 2  6 1  In the group of 3  = ,   12 11  132 22  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 10C3  7C4 [= 120  35 = 4200] M1 12Ca  12-aCb, a = 3, 4, 5; b = 3, 4, 5 (a ≠ b) In the group of 4 10C2  8C5 [= 45  56 = 2520] In the group of 3 10C1  9C5 [= 10  126 = 1260] [Total number of ways of arranging the 3 groups =] A1 27720 Seen alone or as denominator of probability –accept 12C5 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 10C3  7C4 [= 120  35 = 4200] In the group of 4 10C5 5C2 [= 252 10 = 2520] In the group of 3 10C5  5C4 [= 252  5 = 1260]

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