Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 5 · Variant 1

2 The probability that a student at a large music college plays in the band is 0.6. For a student who plays in the band, the probability that she also sings in the choir is 0.3. For a student who does not play in the band, the probability that she sings in the choir is x. The probability that a randomly chosen student from the college does not sing in the choir is 0.58. (a) Find the value of x. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Two students from the college are chosen at random. (b) Find the probability that both students play in the band and both sing in the choir. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 0·6 × 0·7 + 0·4(1 – x) = 0·58 ≡ 0·42 + 0·4(1 – x) = 0·58 M1 Equation of form 0·6 × a + 0·4 × b = 0·58; a = 0·3, 0·7, b = x, (1 – x) B1 Single correct product seen, condone 0·42, in an equation of appropriate form x = 0·6 A1 Alternative method for question 2(a) 0·6 × 0·3 + 0·4x = 0·42 ≡ 0·18 + 0·4x = 0·42 M1 Equation of form 0·6 x a + 0·4 x b = 0·42; a = 0·3, 0·7, b = x, (1 – x) B1 Single correct product seen, condone 0·18, in an equation of appropriate form x = 0·6 A1 3 2(b) ( ) 2 0.6 0.3 × M1 (a × b)2, a = 0·6, 0·4 and b = 0·7, 0·3, x, (1–x) or 0·182, alone. 0.0324 A1 2 M1 pn, n = 6, 7 0 < p < 1

More questions on Probability

3 Kayla is competing in a throwing event. A throw is counted as a success if the distance achieved is greater than 30 metres. The probability that Kayla will achieve a success on any throw is 0.25. (a) Find the probability that Kayla takes more than 6 throws to achieve a success. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that, for a random sample of 10 throws, Kayla achieves at least 3 successes. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) P(X > 6) = 6 0.75 0.178, 729 4096 A1 0·17797… 2 Question Answer Marks Guidance 3(b) 1 – P(0, 1, 2) = 1 – ( 10 0.75 + 10C1 1 9 0.25 0.75 + 10C2 2 8 0.25 0.75 ) M1 Binomial term of form 10Cx ( ) 10 1 , x x p p − − 0 < p < 1, any p, x ≠ 0, 10 1 – (0·0563135 + 0·1877117 + 0·2815676) A1 Correct unsimplified expression 0·474 A1 0·474 ⩽ p ⩽ 0·4744 3

More questions on Probability

4 The random variable X takes each of the values 1, 2, 3, 4 with probability 1 Two independent values 4. of X are chosen at random. If the two values of X are the same, the random variable Y takes that value. Otherwise, the value of Y is the larger value of X minus the smaller value of X. (a) Draw up the probability distribution table for Y. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that Y = 2 given that Y is even. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) y 1 2 3 4 prob 7 16 5 16 3 16 1 16 B1 1 2 3 4 1 1 1 2 3 2 1 2 1 2 3 2 1 3 1 4 3 2 1 4 Probability distribution table with correct scores with at least one probability, allow extra score values if probability of zero stated’ B1 One probability (linked with correct score) correct B1 2 more probs (linked with correct scores) correct B1 FT 4th prob correct, FT sum of 3 or 4 terms = 1 4 Question Answer Marks Guidance 4(b) P(2|even) = 5 16 6 16 M1 ( ) ( ) ( ) P 2 P 2 P 4 their their their + seen or correct outcome space. 5 6 or 0·833 A1 2

More questions on Discrete random variables

5 The time in hours that Davin plays on his games machine each day is normally distributed with mean 3.5 and standard deviation 0.9. (a) Find the probability that on a randomly chosen day Davin plays on his games machine for more than 4.2 hours. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) On 90% of days Davin plays on his games machine for more than t hours. Find the value of t. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate an estimate for the number of days in a year (365 days) on which Davin plays on his games machine for between 2.8 and 4.2 hours. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) P(X > 4.2) = P( 4.2 3.5) 0.9 z − > = P( 0.7778) z > M1 Using ± standardisation formula, no 2 or σ σ , continuity correction 1 – 0·7818 M1 Appropriate area Φ , from standardisation formula P(z>…) in final solution 0·218 A1 3 5(b) 1.282 z = − B1 ±1.282 seen (critical value) 3.5 1.282 0.9 t − = − M1 An equation using ±standardisation formula with a z-value, condone 2 , σ σ and continuity correction 2.35 t = A1 AWRT, only dependent on M mark 3 Question Answer Marks Guidance 5(c) P(2.8 < X < 4.2) = 1 – 2 × their 5(a) ≡ 2(1 – their 5(a)) – 1 ≡ 2(0·5 – their 5(a)) = 0·5636 B1 FT FT from their 5(a) < 0.5 or correct Accept unevaluated probability OE Accept 0·564 Number of days = 365 × 0·5636 = 205·7 M1 365 × their p So, 205 (days) A1 FT Accept 205 or 206, not 205·0 or 206·0 no approximation/ rounding stated FT must be an integer value Alternative method for question 5(c) P 2.8 3.5 4.2 3.5 0.9 0.9 z − −   < <     ( ) ( ) Φ 0.7778 1 Φ0.7778 = − − = 0·7818 – (1 – 0·7818) = 0·5636 B1 0·5635 < p ⩽ 0·564 OE Number of days = 365 × 0·5636 = 205·7 M1 365 × their p So, 205 (days) A1 FT Accept 205 or 206, not 205·0 or 206·0 no approximation/ rounding stated FT must be an integer value 3

More questions on The normal distribution

6 The times, t minutes, taken by 150 students to complete a particular challenge are summarised in the following cumulative frequency table. Time taken (t minutes) t ≤20 t ≤30 t ≤40 t ≤60 t ≤100 Cumulative frequency 12 48 106 134 150 (a) Draw a cumulative frequency graph to illustrate the data. [2] (b) 24% of the students take k minutes or longer to complete the challenge. Use your graph to estimate the value of k. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate estimates of the mean and the standard deviation of the time taken to complete the challenge. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) M1 At least 4 points plotted at upper end points, with both scales linear with at least 3 values indicated Correct cumulative frequency curve A1 All plotted correctly with curve drawn joined to (0, 0), axes labelled cumulative frequency, time, minutes 2 6(b) 150 × 0·76 = 114 M1 114 SOI, may be on graph k = 45 (mins) A1 FT Clear indication that their graph has been used, tolerance ±1mm 2 Question Answer Marks Guidance 6(c) Frequencies: 12 36 58 28 16 B1 Correct frequencies seen Mean = 10 12 25 36 35 58 50 28 80 16 150 × + × + × + × + × B1 At least 4 correct midpoints seen and used 120 900 2030 1400 1280 150 + + + + M1 Correct formula with their midpoints (not upper boundary, lower boundary, class width or frequency density). 38.2, 38 1 5 A1 Variance = 2 2 2 2 2 2 12 10 36 25 58 35 28 50 16 80 150 mean × + × + × + × + × − = 2 1200 22500 71050 70000 102400 150 mean + + + + − M1 Substitute their midpoints and frequencies (condone use of cumulative frequency) in correct variance formula, must have ‘– their mean2’ (Standard deviation = 321.76 ) = 17.9 A1 6

More questions on Representation of data

7 (a) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that all 3 Es are together. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that the Ps are not next to each other. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that a randomly chosen arrangement of the 10 letters of the word SHOPKEEPER has an E at the beginning and an E at the end. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Four letters are selected from the 10 letters of the word SHOPKEEPER. (d) Find the number of different selections if the four letters include exactly one P. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) 8! 2! M1 ( ) 8! 7! 8 ! ,where , , where 2 ! a k a k k × ≡ ∈ ∈   20160 A1 2 Question Answer Marks Guidance 7(b) Total number of ways: ( ) 10! 302400 2!3! = (A) B1 Accept unsimplified With Ps together: 9! ( 3! = 60 480) (B) B1 Accept unsimplified With Ps not together: 302 400 – 60 480 M1 10! 9! m n − , m, n integers or (A) – (B) if clearly identified 241 920 A1 Alternative method for question 7(b) 8! 3! B1 k × 8! in numerator, k a positive integer, no ± B1 m × 3! in denominator, m a positive integer, no ± 9 8 2 × × M1 Their 8! 3!multiplied by 9C2 or 9P2 no additional terms 241 920 A1 Exact value, WWW 4 Question Answer Marks Guidance 7(c) Probability = Number of ways Es at beginning and end Total number of ways Probability = 8! 2! 10! 2! 3! × = 20160 302400 M1 8! ! 10! ! ! k k l      1 ⩽ k, l ∈ ℕ ⩽ 3, FT denominator from 7(b) or correct 1 15 , 0·0667 A1 Alternative method for question 7(c) Probability = 3 2 10 9 × M1 1 3,2 10 9 a a a − × = 1 15 , 0·0667 A1 Alternative method for question 7(c) Probability = 1 1 3! 10 9 × × M1 1 1 !, 3,2 10 9 m m × × = 1 15 , 0.0667 A1 2 Question Answer Marks Guidance 7(d) Scenarios: P E E E 5C0 = 1 P E E _ 5C1 = 5 P E _ _ 5C2 = 10 P _ _ _ 5C3 = 10 M1 5Cx seen alone, 1 ⩽ x ⩽ 4 M1 Summing the number of ways for 3 or 4 correct scenarios (can be unsimplified), no incorrect scenarios Total = 26 A1 3

More questions on Permutations and combinations