Cambridge A Level Mathematics 9709 — 2022 May/June Paper 5 · Variant 2

3 The back-to-back stem-and-leaf diagram shows the diameters, in cm, of 19 cylindrical pipes produced by each of two companies, A and B. Company A Company B 4 33 1 2 8 9 8 3 2 0 34 1 6 8 9 9 8 7 5 4 1 1 35 1 2 2 3 9 6 5 2 36 5 6 4 3 1 37 0 3 4 38 2 8 Key: 1 35 3 means the pipe diameter from company A is 0.351cm and from company B is 0.353cm. (a) Find the median and interquartile range of the pipes produced by company A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that for the pipes produced by company B the lower quartile, median and upper quartile are 0.346cm, 0.352cm and 0.370cm respectively. (b) Draw box-and-whisker plots for companies A and B on the grid below. [3] (c) Make one comparison between the diameters of the pipes produced by companies A and B. [1] ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Median = 0.355 B1 Identified condone Q2. [IQR =] 0.366 – 0.348 M1 0.365 ⩽ UQ ⩽ 0.369 – 0.343 ⩽ LQ ⩽ 0.349. Subtraction may be implied by answer. 0.018 A1 If 0/3 scored SC B1 for figs Median = 355 IQR = 18. 3 3(b) Box-and-whisker plot on provided grid B1 All 5 key values for B plotted accurately in standard format using their scale. Labelled B. Check accuracy in the middle of vertical line. B1 FT All 5 key values for A, FT from part 3(a), plotted in standard format accurately using their scale. Labelled A. Check accuracy in the middle of vertical line. B1 Whiskers not through box for both, not drawn at corners of boxes, single linear scale with at least 3 values stated, covering at least 0.34 to 0.38 and labelled diameter (d etc) and cm. Accept as a title. 3 If both plots attempted and plot(s) not labelled, SC B1 for at least 1 fully correct set of values plotted. 3(c) A comparison in context B1 Single comment comparing spread or central tendency in context. Must reference either diameter or pipes. Not a simple numerical comparison of statistical values such as median, range, IQR or min/max. 1

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4 The weights, in kg, of bags of rice produced by Anders have the distribution N 2.02, 0.032 . (a) Find the probability that a randomly chosen bag of rice produced by Anders weighs between 1.98 and 2.03kg. 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The weights of bags of rice produced by Binders are normally distributed with mean 2.55kg and standard deviation 3 kg. In a random sample of 5000 of these bags, 134 weighed more than 2.6kg. (b) Find the value of 3. 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Mark scheme: 4(a) [P(1.98 < X < 2.03) = ]P(1.98 2.02 2.03 2.02) 0.03 0.03 z   [= P( 1.333 0.333) z    ] 1.98 or 2.03 substituted appropriately. Condone 0.032 and continuity correction ±0.005, not √0.03. [=       Φ 0.333 1 Φ 1.333   ] = 0.6304 + 0.9087 – 1 M1 Calculating the appropriate probability area from their z-values. (or 0.6304 – 0.09121 or (0.9087 – 0.5) + (0.6304 – 0.5) etc) 0.539 A1 0.539 ⩽ z < 0.5395 Only dependent upon 2nd M mark. If M0 scored SC B1 for 0.539 ⩽ z < 0.5395. 3 4(b) [P(X > 2.6) = 134 0.0268 5000  ] [P(X<2.6) = 1 – 0.0268 =] 0.9732 B1 0.9732 or 4866 2433 or 5000 2500 seen. 2.6 2.55 1.93    M1 Use of ±standardisation formula with 2.6 and 2.55 substituted, no 2,   or continuity correction. M1 Their standardisation formula with values substituted equated to z-value which rounds to ±1.93. 0.0259  A1 AWRT 0.0259 or 5 193 . If M0 earned, SC B1 for correct final answer. 4

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5 In a large college, 28% of the students do not play any musical instrument, 52% play exactly one musical instrument and the remainder play two or more musical instruments. A random sample of 12 students from the college is chosen. (a) Find the probability that more than 9 of these students play at least one musical instrument. 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A random sample of 90 students from the college is now chosen. (b) Use an approximation to find the probability that fewer than 40 of these students play exactly one musical instrument. 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Mark scheme: 5(a) [P(10, 11, 12) =] 12C10 10 2 0.72 0.28 + 12C11 11 1 0.72 0.28 + 12C12 12 0 0.72 0.28 M1 One term 12Cx 12 1 x x p p   , for 0 < x < 12, 0 < p < 1. = 0.193725 + 0.0905726 + 0.0194084 A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.304 B1 Final answer 0.3036 < p ⩽ 0.304. Alternative method for question 5(a) [1 – P(0,1,2,3,4,5,6,7,8,9) =] 1 – (12C0 0 12 0.72 0.28 + 12C1 1 11 0.72 0.28 + 12C2 2 10 0.72 0.28 + 12C3 3 9 0.72 0.28 + 12C4 4 8 0.72 0.28 + 12C5 5 7 0.72 0.28 + 12C6 6 6 0.72 0.28 + 12C7 7 5 0.72 0.28 + 12C8 8 4 0.72 0.28 + 12C9 9 3 0.72 0.28 ) M1 One term 12Cx   12 1 x x p p   , for 0 < x < 12, 0 < p < 1. A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.304 B1 Final answer 0.3036 < p ⩽ 0.304. 3 5(b) Mean = [   0.52 90 46.8, var 0.52 0.48 90] 22.464       B1 46.8 and 22.464 or 22.46 seen, allow unsimplified, (4.739 < σ ⩽ 4.740 imply correct variance). [P(X < 40) =] P 39.5 46.8 22.464 z         M1 Substituting their mean and their variance into ±standardisation formula (any number for 39.5), not σ2, √ σ. M1 Using continuity correction 39.5 or 40.5 in their standardisation formula. = [P( 1.540)] 1 0. Z   9382 M1 Appropriate area Φ, from final process, must be probability. 0.0618 A1 0.06175 ⩽ p ⩽ 0.0618 5

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6 (a) Find the number of different arrangements of the 9 letters in the word CROCODILE. 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(b) Find the number of different arrangements of the 9 letters in the word CROCODILE in which there is a C at each end and the two Os are not together. 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(c) Four letters are selected from the 9 letters in the word CROCODILE. Find the number of selections in which the number of Cs is not the same as the number of Os. 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(d) Find the number of ways in which the 9 letters in the word CROCODILE can be divided into three groups, each containing three letters, if the two Cs must be in different groups. 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Mark scheme: 6(a) 9! 2!2!        90 720 1 6(b) Method 1 Arrangements Cs at ends – Arrangements Cs at ends and Os together [Os not together =] 7! 6! 2!  [= 2520 – 720] M1 ! 2! w – y, w = 6, 7 y an integer. Condone ! 2 2! w y         . M1 a – 6! or a – 720, a an integer resulting in a positive answer. 1800 A1 Method 2 identified scenarios R ^ ^ ^ R [Os not together =] 6 5 5! 2!   = M1 5! × b, b integer >1. M1 6 6 2 2 6 5 or C or or1 5 2! 2! P c        , c integer > 1. 1800 A1 3 Question Answer Marks Guidance 6(c) CCO _ 5C1 = 5 CC _ _ 5C2 = 10 OOC _ 5C1 = 5 OO _ _ 5C2 = 10 C _ _ _ 5C3 = 10 O _ _ _ 5C3 = 10 B1 Correct outcome/value for 1 identified scenario. Accept unsimplified. WWW M1 Add 5 or 6 values of appropriate scenarios only, no additional incorrect scenarios, no repeated scenarios. Accept unsimplified. Condone use of permutations. [Total =] 50 A1 3 6(d) Both Os in group with a C 5C2 = 10 Both Os in group without a C 5C2 × 3C2 = 30 One O in a C group, one not 5C1 × 4C2 = 30 One O with each C (5C1 × 4C1) ÷2! = 10 B1 A correct scenario calculated accurately. Accept unsimplified. M1 Add 3 or 4 correct scenario values, no incorrect scenarios, accept repeated scenarios. Accept unsimplified. [Total =] 80 A1 Alternative method for question 6(d) CCO O^^ ^^^ = 5C2 = 10 CC^ O^^ O^^ = 5C1 ×4C2 = 30 CC^ OO^ ^^^ = 5C1 × 4C1 = 20 B1 A correct scenario calculated accurately. Accept unsimplified. Total ways of making three groups 9 6 6 3 C C 2 2 3   = 140 140 – (their 10+ their 30+ their 20) M1 Total subtract 2 or 3 correct scenario values, no incorrect scenarios. Accept unsimplified. 80 A1 3

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7 Hanna buys 12 hollow chocolate eggs that each contain a sweet. The eggs look identical but Hanna knows that 3 contain a red sweet, 4 contain an orange sweet and 5 contain a yellow sweet. Each of Hanna’s three children in turn randomly chooses and eats one of the eggs, keeping the sweet it contained. (a) Find the probability that all 3 eggs chosen contain the same colour sweet. 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(b) Find the probability that all 3 eggs chosen contain a yellow sweet, given that all three children have the same colour sweet. 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(c) Find the probability that at least one of Hanna’s three children chooses an egg that contains an orange sweet. 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Mark scheme: 7(a) YYY: 5 4 3 60 1 , 12 11 10 1320 22    OOO: 4 3 2 24 1 , 12 11 10 1320 55    RRR: 3 2 1 6 1 , 12 11 10 1320 220    or a  (a –1)  ( a – 2), a = 5, 4, 3 in numerator seen in at least one expression. A1 One expression 1 2 12 11 10 a a a     , a = 5, 4, 3 (consistent in expression). Correct order of values in the numerator is essential. M1 5 4 3 4 3 2 3 2 1 12 12 12 d e d e d e         , either d = 11, e = 10 or d = 12, e = 12. Condone 1 1 1 OE 22 55 220   [Total =] 90 3 , , 0.0682 1320 44 A1 0.06818. Dependent only upon the second M mark. Question Answer Marks Guidance 7(a) Alternative method for question 7(a) YYY: 5 3 12 3 C 10 1 , 220 22 C  OOO: 4 3 12 3 C 4 1 , 220 55 C  RRR: 3 3 12 3 C 1 220 C  M1 Either 12 3 C in denominator or 3 C a in numerator seen in at least one expression. A1 One expression 3 12 3 C C a a = 5, 4, 3 M1 5 4 3 3 3 3 12 12 12 3 3 3 C C C C C C   Condone 1 1 1 OE 22 55 220   [Total =] 90 3 , , 0.0682 1320 44 A1 0.06818. Dependent only upon the second M mark. 4 7(b) [P(YYY | all same colour) =] 60 90 1320 1320  M1    60 1 P YYY or or 1320 22 90 3 7 a or or 1320 44 their their 2 3 , 0.667 A1 OE 2 Question Answer Marks Guidance 7(c) In each method, the M mark requires the scenarios to be identifiable. This may be implied by a list of scenarios and then the calculations which will be assumed to be in the same order. A correct value/expression will be condoned as identifying the connected scenario. Method 1 [1 – no orange = ]1−8 7 6 12 11 10   or 8 3 12 3 C 1 C  = 14 1 55  B1 8 7 6 12 11 10   or 8 3 12 3 C C seen, condone 336 56 or 1320 220 only, not OE. M1 1 12 f g h d e    Either d = 11, e = 10 or d = 12, e = 12 or 1 - 8 3 12 3 C C . Condone 14 1 55  OE (not 41) 55 . 41 55 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Question Answer Marks Guidance 7(c) Method 2 P(1 O)= 4 3 2 4 5 4 672 12 11 10 12 11 10 3 4 5 3 1320 2 12 11 10                        P(2O) = 4 3 8 288 3 12 11 10 1320    P(3O) = 24 1320 B1 P(1 O)or P(2 O) correct, accept unsimplified. M1 3 correct scenarios added, with at least one 3-term product of form 12 f g h d e   seen, either d = 11, e = 10 or d = 12, e = 12. [Total =] 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Method 3 O Y R = 4 5 3 1 1 1 C C C   = 60 O R R = 4 3 1 2 C C  = 12 O Y Y = 4 5 1 2 C C  = 40 O O Y = 4 5 2 1 C C  = 30 O O R = 4 3 2 1 C C  = 18 O O O = 4 3 C = 4 Total = 164 Prob = 12 3 164 C B1 Number of ways either 1 or 2 orange sweets obtained correctly (112 or 48). Accept unsimplified Note 4C1  8C2 = 112 or 4C2  8C1 = 48 are correct alternatives. M1 3 correct scenarios (1, 2 or 3 orange sweets) added on numerator, denominator 12 3 C 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Question Answer Marks Guidance 7(c) Method 4 P(R R O) = 3 2 4 1 12 11 10 55    P(R O ) = 3 4 1 12 11 11   P(R Y O) = 3 5 4 1 12 11 10 22    P(O ) = 4 1 12 3  P(Y R O) = 5 3 4 1 12 11 10 22    P(Y O ) = 5 4 5 12 11 33   P(Y Y O) = 5 4 4 2 12 11 10 33    B1 P(R ^ ^) = 17 110 or P(Y ^ ^) = 17 66 . Accept unsimplified. M1 3 correct scenarios added, with at least one 3-term product of form 12 f g h d e   seen, either d = 11, e = 10 or d = 12, e = 12. 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Question Answer Marks Guidance 7(c) Method 5 P(O ) = 4 1 12 3  P(^ O ) = 8 4 8 12 11 33   P(^ ^ O) = 8 7 4 28 12 11 10 165    B1 P(^ O ) = 8 33 or P(^ ^ O) = 28 165 . Accept unsimplified. M1 3 correct scenarios added, with at least one 3-term product of form 12 f g h d e   seen, either d = 11, e = 10 or d = 12, e = 12 with correct numerator. 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 3

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