Cambridge A Level Mathematics 9709 — 2025 May/June Paper 5 · Variant 1

1 The masses of the bags of rice made by a company are normally distributed with mean n kg and standard deviation 0.14 kg. The probability that the mass of a randomly chosen bag of this rice is less than 1.48 kg is 0.22. Find the value of n. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 1.48 −  B1 0.771 < z < 0.773 or −0.773 < z < −0.771 seen. P ( Z  ) = 0.22 0.14 1.48 −  M1 Use of the ±standardisation formula with µ, 1.48 and 0.14 and = − 0.772 equating to a z-value (not 0.78, 0.22, 0.5871, 0.7823, 0.228). 0.14 Condone σ2,  and continuity correction ±0.005. = 1.59 A1 1.585 < µ ⩽ 1.59. If M0 scored SC B1 for correct answer WWW. 3 2(a) Method 1 total arrangements with As together – arrangements with As together and Os together 7! M1 7! − 6! – k , where k is an integer ⩾ 1. 2! 2 n – 6! M1 n – 6!, where n is an integer > 720. 1800 A1 3 Method 2 arrangements of other 6 letters with the As together and then the Os placed 6  5 M1 5! × p, where p is an integer > 1. 5! 2 M1 6  5 q  , q × 6C2, q × 6P2, q × 6 × 5, q an integer > 1. 2 1800 A1 3

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2 (a) Find the number of different arrangements of the 8 letters in the word KANGAROO in which the two As are together and the two Os are not together. [3] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... A fair 8-sided dice has faces labelled K, A, N, G, A, R, O, O. The dice is rolled repeatedly. (b) Find the probability that fewer than 6 rolls of this dice are required to obtain an A. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... (c) Find the probability that the second A is obtained on the 6th roll of the dice. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 2(c) 2 4 M1 2 4  1  3 405 p (1 − p )  5 , 0 < p < 1, p ≠ 1 – p. =    5  4  4  4096 = 0.0989 A1 AWRT 2

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3 Last Sunday, teams of runners took part in a charity event. The time taken, in seconds, to run 50 m was recorded, correct to 1 decimal place, for each runner. The times recorded for 11 runners from each of the Gulls and the Herons are shown in the table. Gulls 7.9 8.2 8.3 8.6 8.6 8.8 9.2 9.7 9.8 10.0 10.4 Herons 9.5 9.9 8.5 8.1 9.2 10.8 8.3 9.7 9.3 9.9 8.7 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with Gulls on the left-hand side. [4] (b) Find the median and the interquartile range of the times of the runners from the Gulls. 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Two other teams of runners, the Eagles and the Swifts, also took part in the event. The recorded times in seconds for 20 runners from the Eagles and 30 runners from the Swifts are denoted by x and y respectively. It is given that / x = 175.0 and that the mean of y is 8.4. (c) Find the mean of the times taken by all 50 runners. 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It is given that / x 2 = 1823.0 . It is also known that the standard deviation of the times taken by all 50 runners is 1.38 seconds. (d) Find the value of / y2 , correct to 1 decimal place. 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Mark scheme: 3(a) B1 Correct stem, ignore extra values (not in reverse, not split). Gulls Herons If a split stem-and-leaf plot is used (i.e. stem values are repeated), the remaining B marks are available. 9 7 B1 Correct Gulls labelled on left, leaves in order from right to left and 8 6 6 3 2 8 1 3 5 7 lined up vertically (less than halfway to next column), no commas or other punctuation. 8 7 2 9 2 3 5 7 9 9 B1 Correct Herons labelled on same diagram, leaves in order and 4 0 10 8 lined up vertically (less than halfway to next column), no commas or other punctuation. Key: 7|9|5 means 9.7 seconds for Gulls and 9.5 seconds for Herons Penalise each error only once in question. E.g. commas in both sets of data If the correct data for Gulls & Herons is transposed, treat as a single error in Gulls and condone in Herons. B1 Correct key for their diagram, need both clubs labelled and ‘sec’ or ‘s’ stated at least once here, or in leaf headings or title. SC: If 2 separate diagrams drawn, max marks: B1 if both stems correct, B1 if Gulls is correct to the left of the stem B0 B1 if both keys correct including ‘sec’ or ‘s’ 4 3(b) Median = 8.8 (seconds) B1 Clearly identified, e.g. Q2, med, m.. [LQ 8.3, UQ 9.8] M1 9.7 ⩽ UQ ⩽ 10.0 – 8.2 ⩽ LQ ⩽ 8.6. 9.8 – 8.3 Implied if both quartile values are stated and the appropriate IQR is calculated accurately. 1.5 A1 WWW If M0 scored SCB1 for 1.5 WWW. 3 3(c) 175.0 + 30  8.4 175.0 + 252.0 M1 Mean = or 50 50 = 8.54 A1 If M0 scored, SCB1 for 8.54 WWW. 2 3(d) 2 2 M1 Substitute values into correct variance formula, any form. 1823.0 +  y 2 ( )  427.0  1.38 = − their   50  50  Solve to find  y 2 DM1 Rearrange equation to obtain  y 2 .  y 2 2  427.0  2 1823.0 = 1.38 + their   − . 50  50  50  y 2 = 1918.8 A1 Mark final answer. If one or both M marks not awarded, SCB1 for 1918.8 seen as final answer WWW mark final answer. 3

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4 Every Saturday, a particular community holds a ‘Puzzle’ event to raise money for a new Leisure Centre. Competitors attempt to solve a puzzle as quickly as possible. Last Saturday, 600 competitors took part. The times taken to complete the puzzle were normally distributed with mean 32.4 minutes and standard deviation 2.5 minutes. (a) How many competitors would you expect to have times within 1.2 minutes of the mean time? 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In this Saturday’s event, 60% of the competitors had times less than 36.0 minutes. (b) 9 competitors who took part in this Saturday’s event are selected at random. Find the probability that at least 2 and fewer than 8 of these competitors had times less than 36.0 minutes. 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(c) 80 competitors who took part in this Saturday’s event are selected at random. Use a suitable approximation to find the probability that more than 50 of these competitors had times less than 36.0 minutes. 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Mark scheme: 4(a) −1.2 1.2 M1 OE P(  Z  ) 2 2.5 2.5 Using ± standardisation formula, not  , not σ, no continuity correction Use of ±standardisation formula once with 32.4, 2.5 and either 31.2 or 33.6. No continuity correction, not ,2 not  1.2 1.2 Implied by either − or seen 2.5 2.5 [ = Φ ( 0.48 ) + Φ ( −0.48 ) = 2Φ ( 0.48 ) − 1 ] M1 Calculating the correct probability area (leading to their final probability). 2 × 0.6844 – 1 or 2 × (0.6844 – 0.5) or 0.6844 – 0.3156 This may be implied by the correct or appropriate probability area. 0.3688 A1 Expected number = 0.3688  600 = 221.28 so 221 or 222 B1 FT FT their 4-figure probability to obtain a single integer answer. No approximation indicated, condone use of 3sf probability here if more accurate answer seen earlier. 4 4(b) Method 1 [P(2 ⩽ X < 8) = 1 – P(0, 1, 8, 9) = ] M1 One term of the form 9Cx ( p ) x (1 − p )9 − x , 0  p  1, x  0 or 9 1 – (0.49 + 9C1 0.48 0.6 + 9C8 0.410.68 + 0.69) = A1 Correct un-simplified expression. Condone omission of last bracket only. If both brackets omitted in [1 – 0.000262144 – 0.00353894 – 0.060466176 – 0.010077696 =] un-simplified expression, allow recovery for final stated calculation of 1 – 0.07434… or better. 0.926 B1 0.925 < p ⩽ 0.926 from correct working. Method 2 [P(2 ⩽ X < 8) = P(2, 3, 4, 5, 6, 7) = ] M1 One term of the form 9Cx ( p ) x (1 − p )9 − x , 0  p  1, x  0 or 9. 9C2 0.47 0.62 +9C3 0.46 0.63 +9C4 0.45 0.64 +9C5 0.44 0.65 +9C6 0.43 A1 Correct un-simplified expression. 0.66 +9C7 0.42 0.67 0.926 B1 0.925 < p ⩽ 0.926 from correct working. 3 4(c) Mean = 80 0.6 = 48 B1 48 and 19.2 (CAO) seen, allow un-simplified. Variance = 80  0.6  0.4 = 19.2 May be in standardisation formula. (4.38178… to at least 4SF identified as σ implies correct variance). Do not condone clear incorrect identification. 50.5 − M1 Substituting their 48 and their 19.2 into the ±standardising [P(X  50 ) =] P( Z  48) formula (any number for 50.5), allow σ2 or √σ. 19.2 M1 Use continuity correction 49.5 or 50.5 in their standardisation formula.  2.5   2.5  Note: If no working    or   seen gains B2.  19.2   4.382  [P( Z  0.571 ) = 1 − Φ ( 0.571) = ] M1 Appropriate probability area, from final process, must be a probability. 1 – 0.7160 = May be implied by a sketch of the required probability area. Expect final answer < 0.5. 0.284 A1 AWRT 5

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5 In a group of 20 musicians, there are 9 guitarists, 6 pianists and 5 drummers. 6 musicians are selected from these 20 to perform at a concert. (a) Find the number of different ways in which the 6 musicians can be selected if there must be at least 3 guitarists, at most 2 pianists and exactly 1 drummer. 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Three bands will be selected from the original group of 20 musicians. Each band will consist of 3 guitarists, 1 pianist and 1 drummer. No musician can be in more than one band. The first band selected will play at a concert in France, the second band selected will play in Italy and the third band selected will play in Spain. (b) Find the number of different ways in which these three bands can be selected. 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Mark scheme: 5(a) 3G 2P 1D: 9C3 6C2  5C1 = 6300 B1 Correct outcome/value for either the 1st or 2nd scenario clearly identified, accept un-simplified, WWW. 4G 1P 1D: 9C4 6C1  5C1 = 3780 M1 2 correct outcomes/values obtained, accept un-simplified 5G 0P 1D: 9C5 [6C0]  5C1 = 630 M1 Sum of 3 correct scenarios, may be identified by un-simplified expression. Condone 5C1 =5 and 6C1 = 6 Total: 10710 A1 CAO If one or both M marks not awarded, SCB1 for 10710 WWW. 4 5(b) Ways of selecting 1st band: 9C3 6C1  5C1 = 2520 M1 9C3 6C1  5C1 or 9C3 × 6 × 5 seen, Ways of selecting 2nd band: 6C3 5C1  4C1 = 400 condone 3 or 3! . Ways of selecting 3rd band: 1 4C1  3C1 = 12 M1 their 2520 × their 400 × their 12 seen, accept un-simplified, [Total number of ways =] 2520 400  12 = condone 3 or 3! . 12096000 A1 Condone 12100000. If one or both M marks not awarded, SCB1 for 12096000 (CAO) WWW. 3

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6 A bag contains 10 marbles, of which 4 are red and 6 are blue. Four marbles are selected from the bag at random, without replacement. The random variable X denotes the number of blue marbles selected. (a) Show that P ( X = 2) = 3 . 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(b) Draw up the probability distribution table for X. 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(c) Find the probability that at least 2 of the marbles chosen are blue, given that at least 1 red marble and at least 1 blue marble are chosen. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. ..................................................................................................................................................................

Mark scheme: 6(a) Method 1 6 5 4 3 4! 3 M1 AG. P(X = 2) =     = 10 9 8 7 2!2! 7 6 5 4 3 4 3 6 5     k or     k for k an integer, k > 1. 10 9 8 7 10 9 8 7 A1 4! 4C2 may be seen for 2!2!. Method 2 6 C 2  4 C 2  15  6  3 M1 AG. = = 6 4   10C4  210  7 C 2  C 2 seen as the numerator of a fraction. Condone use of permutations if used consistently. A1 2 6(b) B1 Table with correct values of x and at least one further non-zero probability correct. x 0 1 2 3 4 Condone extra x values if probability stated as 0. P(X = x) 1 4 3 8 1 , , , , B1 Third probability correct. 210 35 7 21 14 Accept probabilities not in table if clearly identified. 0.00476 0.114 0.381 0.0714 B1 Fourth probability correct. Accept probabilities not in table if clearly identified. B1 Fifth probability correct. Accept probabilities not in table if clearly identified. 4 SCB1 for 4 further non-zero probabilities adding to , 0.5712 if 7 B2 max scored. 4 6(c) 3 8 M1 3 8 + + their seen as the numerator of a fraction. 7 21 7 21 [P(2B, 3B | 3B1R or 2B2R or 1B 3R) =] 4 3 8 + + 35 7 21 B1 FT 4 3 8 their + + their seen as the denominator of a fraction. 35 7 21  17  A1 Accept 0.87628…to at least 3SF.  21 =  = 170 , 85 , 0.876 97 194 97    105  3

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