Cambridge A Level Mathematics 9709 — 2021 May/June Paper 5 · Variant 3

1 The heights in cm of 160 sunflower plants were measured. The results are summarised on the following cumulative frequency curve. 160 140 120 100 frequency 80 Cumulative 60 40 20 0 0 40 80 120 160 200 240 Height (cm) (a) Use the graph to estimate the number of plants with heights less than 100 cm. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use the graph to estimate the 65th percentile of the distribution. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use the graph to estimate the interquartile range of the heights of these plants. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 60 B1 Accept 60 or 61. No decimals 1 1(b) 65% of 160 = 104 M1 0.65 × 160 (=104) seen unsimplified or implied by use on graph 136 (cm) A1 Use of graph must be seen. SCB1 correct value (136 only) if neither 104 nor use of graph are evident 2 1(c) UQ: 150 LQ: 76 IQR = 150 – 76 = 74 [cm] M1 UQ – LQ ; 148 ⩽ UQ ⩽ 152; 74 ⩽ LQ ⩽ 78. A1 Must be from 150 - 76 2

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3 A sports club has a volleyball team and a hockey team. The heights of the 6 members of the volleyball team are summarised by Σx = 1050 and Σx2 = 193 700, where x is the height of a member in cm. The heights of the 11 members of the hockey team are summarised by Σy = 1991 and Σy2 = 366 400, where y is the height of a member in cm. (a) Find the mean height of all 17 members of the club. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the standard deviation of the heights of all 17 members of the club. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Mean height 1050 1991 3041 6 11 6 11 17 Σ + Σ + = = = + + x y accept unsimplified. 178.9 A1 Allow 178.88, 15 17817 , 179 2 Question Answer Marks Guidance 3(b) 2 2 193700 366400 6 11 6 11 Σ + Σ + = + + x y M1 Use of appropriate formula with values substituted, accept unsimplified. [ ] 2 2 560100 Sd 1 78.88 948.289 17 their = − = M1 Appropriate variance formula using their mean2, accept unsimplified expression. Standard deviation = 30.8 A1 Accept 30.7 3

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4 Three fair six-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same time, repeatedly. For a single throw of the three dice, the score is the sum of the numbers on the top faces. (a) Find the probability that the score is 4 on a single throw of the three dice. 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(b) Find the probability that a score of 18 is obtained for the first time on the 5th throw of the three dice. 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Mark scheme: 4(a) [Possible cases: 1 1 2, 1 2 1, 2 1 1] Probability = 3 1 3 6  ×     3 1 6  ×     k , where k is an integer. M1 Multiply a probability by 3, not +, – or ÷ 1 72 A1 Accept 3 216 or 0.0138 or 0.0139 3 4(b) P(18) 3 1 1 6 216    = =       B1 P(18 on 5th throw) 4 215 1 216 216   = ×     M1 (1 – p)4p, 0 < their p < 1 0.00454 A1 3

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6 (a) How many different arrangements are there of the 11 letters in the word REQUIREMENT? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) How many different arrangements are there of the 11 letters in the word REQUIREMENT in which the two Rs are together and the three Es are together? [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) How many different arrangements are there of the 11 letters in the word REQUIREMENT in which there are exactly three letters between the two Rs? [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Five of the 11 letters in the word REQUIREMENT are selected. (d) How many possible selections contain at least two Es and at least one R? 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Mark scheme: 6(a) 11! 2!3! M1 11! alone on numerator – must be a fraction. k! × m! on denominator, k = 1, 2, m = 1, 3, 1 can be implied but cannot both = 1. No additional terms 3326400 A1 Exact value only 2 6(b) 8! = 40320 B1 Evaluate, exact value only 1 6(c) 9! 7 3!× M1 9! 3!×k seen, k an integer > 0, no +, – or ÷ M1 7 × an integer seen in final answer, no +, – or ÷ 423360 A1 Exact value only Alternative method for Question 6(c) 9C3 ×7! (× 3! 3! ) M1 9C3×k seen, k an integer > 0, no + or – M1 7! × k seen, , k an integer > 0, no + or – 423360 A1 Exact value only but there must be evidence of 3! 3! × Question Answer Marks Guidance 6(c) cont’d Alternative method for Question 6(c) 8! 3 7 2! × × M1 8! 3 2! × ×k seen, k an integer > 0, no + or – M1 7 × an integer seen in final answer, no +, – or ÷ 423360 A1 Exact value only Alternative method for Question 6(c) 2 9 8 7 1 7 total no. of arrangements 11 10 9 8 7 × × × × × × M1 Product of correct five fractions × k seen, k an integer > 0, no + or – M1 7×’total no of arrangements’ ×k seen, k an integer > 0, no + or – 423360 A1 Exact value only Alternative method for Question 6(c) No E between the Rs – 6 3 3! 7! 100800 3! C × × = 1E between the Rs – 6 2 3! 7! 226800 2! C × × = 2Es between the Rs – 6 1 3! 7! 90720 C × × = 3Es between the Rs – 7! = 5040 M1 Finding the correct number of ways for no, 1 or 2 Es between the Rs, accept unsimplified. M1 Adding the number of ways for 3 or 4 correct scenarios ( ) Total 7! 20 45 18 1 7! 84 423360  = × + + + = × =   A1 CAO 3 Question Answer Marks Guidance 6(d) E E R _ _ 6C2 = 15 E E R R _ 6C1 = 6 E E E R _ 6C1 = 6 E E E R R 6C0 = 1 M1 Identifying four correct scenarios only. B1 Correct number of selections unsimplified for 2 or more scenario. M1 Adding the number of selections for 3 or 4 identified correct scenarios only, accept unsimplified. 3Cx ×2Cy × 6Cz, x+y+z=5 correctly identifies x Es and y Rs [Total =] 28 A1 WWW, only dependent upon 2nd M mark. Alternative method for Question 6(d) – Fixing EER first. No other scenarios can be present anywhere in solution. E E R ^ ^ = 8C2 M1 8Cx seen alone or 8Cx × k, , k = 1 or 2, 0<x<8 Condone 8Px or 8Px × k, k = 1 or 2, 0<x<8 B1 8C2 × k, k = 1 or 2 OE M1 8C2 × k, k = 1 OE and no other terms [Total =] 28 A1 Value stated 4

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7 In the region of Arka, the total number of households in the three villages Reeta, Shan and Teber is 800. Each of the households was asked about the quality of their broadband service. Their responses are summarised in the following table. Quality of broadband service Excellent Good Poor Reeta 75 118 32 Village Shan 223 177 40 Teber 12 60 63 (a) (i) Find the probability that a randomly chosen household is in Shan and has poor broadband service. [1] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Find the probability that a randomly chosen household has good broadband service given that the household is in Shan. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ In the whole of Arka there are a large number of households. A survey showed that 35% of households in Arka have no broadband service. (b) (i) 10 households in Arka are chosen at random. Find the probability that fewer than 3 of these households have no broadband service. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) 120 households in Arka are chosen at random. Use an approximation to find the probability that more than 32 of these households have no broadband service. 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Mark scheme: 7(a)(i) 40 1 or or 0.05 800 20 B1 1 7(a)(ii) 177 223 177 40 + + M1 Their 223 + 177 + 40 seen as denominator of fraction in the final answer, accept unsimplified 177 440 or 0.402 A1 CAO Alternative method for Question 7(a)(ii) ( ) ( ) ( ) 177 177 800 800 G | S 223 177 40 440 800 800 ∩ = = = + + P G S P P S = 177 800 11 or 0.55 20 M1 Their P(S) seen as denominator of fraction in the final answer, accept unsimplified 177 440 or 0.402 A1 CAO 2 7(b)(i) P(0, 1, 2) = 10C0 ( ) 0 0.35 ( ) 10 0.65 + 10C1 ( ) 1 0.35 ( ) 9 0.65 + 10C2 ( ) 2 0.35 ( ) 8 0.65 M1 One term:10Cx px (1 – p)10–x for 0 < x < 10, any 0<p<1 0.013463 + 0.072492 + 0.17565 A1 Correct unsimplified expression, or better 0.262 A1 3 Question Answer Marks Guidance 7(b)(ii) Mean = [ ] 120 0.35 42 × = Variance = [ ] 120 0.35 0.65 27.3 × × = B1 Correct mean and variance seen, allow unsimplified P(X >32) = P( Z > 32.5 42) 27.3 − = P(Z > - 1.818) M1 Substituting their mean and variance into ±standardisation formula (any number), condone σ2 or √σ M1 Using continuity correction 31.5 or 32.5 ( ) Φ 1.818 M1 Appropriate area Φ, from final process, must be probability 0.966 A1 0.965 ⩽ p ⩽ 0.966 5

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