Question 13 marks
Question
For n values of the variable x, it is given that Σ x − 50 = 144 and Σ x = 944. Find the value of n. [3]
Mark scheme
1 Σ x − 50 n = 144 B1 50n + 144 = 944 M1 n = 16 A1 3
Mathematics 9709/52 May/June 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s20_qp_529709_s20_ms_529709/52/M/J/20
- Paper
- Paper 5 Probability & Statistics 1
- Total marks
- 50
- Time
- 1 hour 15 minutes
- Code on the paper
- 9709/52/M/J/20
9709/52 mark scheme, question by question
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question 25 marks
Question
A total of 500 students were asked which one of four colleges they attended and whether they preferred soccer or hockey. The numbers of students in each category are shown in the following table. Soccer Hockey Total Amos 54 32 86 Benn 84 72 156 Canton 22 56 78 Devar 120 60 180 Total 280 220 500 (a) Find the probability that a randomly chosen student is at Canton college and prefers hockey. [1] (b) Find the probability that a randomly chosen student is at Devar college given that he prefers soccer. [2] (c) One of the students is chosen at random. Determine whether the events ‘the student prefers hockey’ and ‘the student is at Amos college or Benn college’ are independent, justifying your answer. [2]
Mark scheme
2(a) 56 14 B1 or or 0.112 500 125 1 2(b) P ( D ∩ S) 120 M1 P ( D | S) = = P ( S) 280 120 3 A1 or 280 7 2 2(c) 220 M1 P(hockey) = = 0.44 500 242 P(Amos or Benn) = = 0.484 500 104 P(hockey ∩ A or B) = = 0.208 500 P(H) × P(A U B) = P(H ∩ (A U B)) if independent 220 242 1331 A1 × = so not independent 500 500 6250 2
Question 38 marks
Question
Two machines, A and B, produce metal rods of a certain type. The lengths, in metres, of 19 rods produced by machine A and 19 rods produced by machine B are shown in the following back-to-back stem-and-leaf diagram. A B 21 1 2 4 7 6 3 0 22 2 4 5 5 6 8 7 4 3 1 1 23 0 2 6 8 9 9 5 5 5 3 2 24 3 3 4 6 4 3 1 0 25 6 Key: 7 22 4 means 0.227 m for machine A and 0.224 m for machine B. (a) Find the median and the interquartile range for machine A. [3] It is given that for machine B the median is 0.232 m, the lower quartile is 0.224 m and the upper quartile is 0.243 m. (b) Draw box-and-whisker plots for A and B. [3] (c) Hence make two comparisons between the lengths of the rods produced by machine A and those produced by machine B. [2]
Mark scheme
3(a) Median = 0.238 B1 UQ = 0.245, LQ = 0.231, M1 So IQR = 0.245 – 0.231 0.014 A1 3 3(b) LQ M UQ A 0.231 0.238 0.245 0.220 0.254 FT FT FT B 0.211 0.224 0.232 0.243 0.256 Medians and quartiles correctly plotted for A or B B1 End points correct for A or B B1 Completely correct, including scale B1 3 3(c) Lengths of rods produced by machine A are longer. B1 (B1 for comparison of central tendency) Lengths of rods produced by machine A are less spread out B1 (B1 for comparison of spread) 2
Question 48 marks
Question
Trees in the Redian forest are classified as tall, medium or short, according to their height. The heights can be modelled by a normal distribution with mean 40 m and standard deviation 12 m. Trees with a height of less than 25 m are classified as short. (a) Find the probability that a randomly chosen tree is classified as short. [3] Of the trees that are classified as tall or medium, one third are tall and two thirds are medium. (b) Show that the probability that a randomly chosen tree is classified as tall is 0.298, correct to 3 decimal places. [2] (c) Find the height above which trees are classified as tall. [3]
Mark scheme
4(a) 25 − 40 M1 P ( X < 25 ) = P z < = P ( z < − 1.25 ) P ( X < 25 ) = P ( z < ) 12 1 – 0.8944 M1 0.106 A1 3 4(b) 0.8944 divided by 3 M1 (M1 for 1 - their (a) divided by 3) 0.298 AG A1 2 4(c) 0.2981 gives z = 0.53 B1 h − 40 M1 = 0.53 12 h = 46.4 A1 3
Question 58 marks
Question
A fair three-sided spinner has sides numbered 1, 2, 3. A fair five-sided spinner has sides numbered 1, 1, 2, 2, 3. Both spinners are spun once. For each spinner, the number on the side on which it lands is noted. The random variable X is the larger of the two numbers if they are different, and their common value if they are the same. 7. (a) Show that P X = 3 = 15 [2] (b) Draw up the probability distribution table for X . [3] (c) Find E X and Var X . [3]
Mark scheme
5(a) M1 1 1 2 2 3 1 1 1 2 2 3 2 2 2 2 2 3 3 3 3 3 3 3 7 A1 AG 15 2 5(b) B1 x 1 2 3 Probability 2 6 7 15 15 15 P(1) or P(2) correct B1 3rd probability correct, FT sum to 1 B1 3 5(c) 2 + 12 + 21 35 7 B1 E(X) = = = 15 15 3 12 × 2 + 22 × 6 + 32 × 7 7 2 M1 Var(X) = − 15 3 22 A1 ( 0.489 ) 45 3
Question 69 marks
Question
(a) Find the number of different ways in which the 10 letters of the word SUMMERTIME can be arranged so that there is an E at the beginning and an E at the end. [2] (b) Find the number of different ways in which the 10 letters of the word SUMMERTIME can be arranged so that the Es are not together. [4] (c) Four letters are selected from the 10 letters of the word SUMMERTIME. Find the number of different selections if the four letters include at least one M and exactly one E. [3]
Mark scheme
6(a) 8! M1 3! 6720 A1 2 6(b) 10! B1 Total number = ( 302400 ) (A) 2!3! 9! B1 With Es together = ( 60480 ) ( B) 3! Es not together = their (A) – their (B) M1 241920 A1 Alternative method for question 6(b) _^_^_^_^_^_^_^_^_ 8! 9 × 8 × 3! 2 8! × k in numerator, k integer ≥ 1, denominator ≥ 1 B1 3! × m in denominator, m integer ≥ 1 B1 8! M1 Their Multiplied by 9C2 (OE) only (no additional terms) 3! 241920 A1 4 6(c) Scenarios: M1 5 EMMM C0 = 1 5 EMM_ C1 = 5 5 EM__ C2 = 10 Summing the number of ways for 2 or 3 correct scenarios M1 Total = 16 A1 3
Question 79 marks
Question
On any given day, the probability that Moena messages her friend Pasha is 0.72. (a) Find the probability that for a random sample of 12 days Moena messages Pasha on no more than 9 days. [3] (b) Moena messages Pasha on 1 January. Find the probability that the next day on which she messages Pasha is 5 January. [1] (c) Use an approximation to find the probability that in any period of 100 days Moena messages Pasha on fewer than 64 days. [5] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) 1 – P(10, 11, 12) M1 = 1 – [12C10 0.72100.282 + 12C11 0.72110.281+0.7212] 1 – (0.19372 + 0.09057 + 0.01941) A1 0.696 A1 3 7(b) 0.283 × 0.72 = 0.0158 B1 1 7(c) Mean = 100 × 0.72 = 72 M1 Var = 100 × 0.72 × 0.28 = 20.16 63.5 − 72 M1 P(less than 64) = P z < 20.16 (M1 for substituting their µ and σ into ±standardisation formula with a numerical value for ‘63.5’) Using either 63.5 or 64.5 within a ±standardisation formula M1 Appropriate area Φ, from standardisation formula P(z<…) in final solution M1 = P(z < –1.893) 0.0292 A1 5
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Mathematics 9709/52 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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