Mathematics 9709/53 May/June 2021: question paper and mark scheme

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 M1 UQ – LQ ; 148 ⩽ UQ ⩽ 152; 74 ⩽ LQ ⩽ 78. IQR = 150 – 76 = 74 [cm] A1 Must be from 150 - 76 2

Mark scheme

2 p + p + 0.1 + q + q = 1 B1 Sum of probabilities = 1 0.1 + 2q = 3(2p) B1 Use given information Attempt to solve two correct equations in p and q M1 Either use of Substitution method to form a single equation in either p or q and finding values for both unknowns. Or use of Elimination method by writing both equations in same form (usually ap + bq = c ) and + or – to find an equation in one unknown and finding values for both unknowns. 1 13 A1 CAO, both WWW p= or 0.125 and q = or 0.325 8 40 4

Mark scheme

3(a) Σx + Σy 1050 + 1991 3041 M1 Use of appropriate formula with values substituted, Mean height = = = accept unsimplified. 6 + 11 6 + 11 17 178.9 A1 15 Allow 178.88, 178 , 179 17 2 3(b) Σx 2 + Σy 2 193700 + 366400 M1 Use of appropriate formula with values substituted, = accept unsimplified. 6 + 11 6 + 11 560100 M1 Appropriate variance formula using their mean2, Sd 2 = − their 178.882 [ = 948.289] accept unsimplified expression. 17 Standard deviation = 30.8 A1 Accept 30.7 3

Mark scheme

4(a) [Possible cases: 1 1 2, 1 2 1, 2 1 1] M1 3 1 1 3 × k , where k is an integer. Probability = × 3 6 6 M1 Multiply a probability by 3, not +, – or ÷ 1 A1 3 Accept or 0.0138 or 0.0139 72 216 3 4(b) 3 B1 1 1 P(18) = = 6 216 4 M1 (1 – p)4p, 0 < their p < 1 215 1 P(18 on 5th throw) = × 216 216 0.00454 A1 3

Mark scheme

5(a) 4−μ B1 1.378 ⩽ z1 ≤ 1.379 or –1.379 ⩽ z1 ⩽ –1.378 z1 = = −1.378 δ 10 − μ B1 0.841 ⩽ z2 ⩽ 0.842 or –0.842 ⩽ z2 ⩽ –0.841 z2 = = 0.842 σ Solve to find at least one unknown: M1 Use of ±standardisation formula once with µ, σ, a 4−μ z-value and 4 or 10, allow continuity correction, not σ2 = −1.378 σ or σ 10 − μ = 0.842 M1 Use either the elimination method or the substitution σ method to solve two equations in µ and σ . σ = 2.70 μ = 7.72 A1 2.70 σ 2.71 7.72 μ 7.73 5 5(b) Φ(2) − Φ (−2) = 2Φ ( 2 ) − 1 M1 Identifying 2 and –2 as the appropriate z-values 2 × their 0.9772 − 1 B1 Calculating the appropriate area from stated phis of z-values which must be ± the same number 0.9544 or 0.9545 A1 Accept AWRT 0.954 0.9544 × 800 = 763.52 B1 FT FT their 4SF (or better) probability, final answer must 763 or 764 be positive integer 4

Mark scheme

6(a) 11! M1 11! alone on numerator – must be a fraction. 2!3! 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! M1 9! ×7 × k seen, k an integer > 0, no +, – or ÷ 3! 3! M1 7 × an integer seen in final answer, no +, – or ÷ 423360 A1 Exact value only Alternative method for Question 6(c) 3! M1 9C3× k seen, k an integer > 0, no + or – 9 C3 × 7! (× ) 3! M1 7! × k seen, , k an integer > 0, no + or – 423360 A1 3! Exact value only but there must be evidence of × 3! 6(c) cont’d Alternative method for Question 6(c) 8! M1 8! 3× 7 × 3× × k seen, k an integer > 0, no + or – 2! 2! 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 M1 Product of correct five fractions × k seen, k an integer 7× × × × × × total no. of arrangements > 0, no + or – 11 10 9 8 7 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) 6 C3 × 3!× 7! M1 Finding the correct number of ways for no, 1 or 2 Es No E between the Rs – = 100800 between the Rs, accept unsimplified. 3! 6 C2 × 3!× 7! M1 Adding the number of ways for 3 or 4 correct scenarios 1E between the Rs – = 226800 2! 2Es between the Rs – 6C1 × 3!× 7! = 90720 3Es between the Rs – 7! = 5040 Total = 7!× ( 20 + 45 + 18 + 1) = 7!× 84 = 423360 A1 CAO 3 6 6(d) EER__ C2 = 15 M1 Identifying four correct scenarios only. 6 EERR_ C1 = 6 6 EEER_ C1 = 6 B1 Correct number of selections unsimplified for 2 or 6 EEERR C0 = 1 more scenario. M1 Adding the number of selections for 3 or 4 identified correct scenarios only, accept unsimplified. 3 Cx ×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 8 Cx seen alone or 8Cx × k, , k = 1 or 2, 0<x<8 Condone 8Px or 8Px × k, k = 1 or 2, 0<x<8 8 B1 C2 × k, k = 1 or 2 OE 8 M1 C2 × k, k = 1 OE and no other terms [Total =] 28 A1 Value stated 4

Mark scheme

7(a)(i) 40 1 B1 or or 0.05 800 20 1 7(a)(ii) 177 M1 Their 223 + 177 + 40 seen as denominator of fraction 223 + 177 + 40 in the final answer, accept unsimplified 177 A1 CAO or 0.402 440 Alternative method for Question 7(a)(ii) 177 177 177 M1 Their P(S) seen as denominator of fraction in the final P (G ∩ S ) 800 answer, accept unsimplified P ( G | S) = = = 800 = 800 P(S ) 223 + 177 + 40 440 11 or 0.55 800 800 20 177 A1 CAO or 0.402 440 2 7(b)(i) P(0, 1, 2) = M1 One term:10Cx px (1 – p)10–x for 0 < x < 10, any 0<p<1 C0 ( 0.35 ) ( 0.65) + 10C1 ( 0.35) ( 0.65) + 10C2 ( 0.35) ( 0.65) 10 0 10 1 9 2 8 0.013463 + 0.072492 + 0.17565 A1 Correct unsimplified expression, or better 0.262 A1 3 7(b)(ii) Mean = 120 × 0.35 [ = 42 ] B1 Correct mean and variance seen, allow unsimplified Variance = 120 × 0.35 × 0.65 [ = 27.3] 32.5 − 42 M1 Substituting their mean and variance into P(X >32) = P( Z > ) = P(Z > - 1.818) ±standardisation formula (any number), condone σ2 or 27.3 √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