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Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, 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_629709_s20_ms_629709/62/M/J/20
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
The masses, in grams, of plums of a certain type have the distribution N 40.4, 5.22 . The plums are packed in bags, with each bag containing 6 randomly chosen plums. If the total weight of the plums in a bag is less than 220 g the bag is rejected. Find the percentage of bags that are rejected. [4]
Mark scheme
1 N(242.4, 162.24) B1 220 − '242.4' M1 (= –1.759) '162.24' φ ('–1.759) = 1 – φ ('1.759') = 0.0393 M1 3.93% A1 4
Question
A shop obtains apples from a certain farm. It has been found that 5% of apples from this farm are Grade A. Following a change in growing conditions at the farm, the shop management plan to carry out a hypothesis test to find out whether the proportion of Grade A apples has increased. They select 25 apples at random. If the number of Grade A apples is more than 3 they will conclude that the proportion has increased. (a) State suitable null and alternative hypotheses for the test. [1] (b) Find the probability of a Type I error. [3] In fact 2 of the 25 apples were Grade A. (c) Which of the errors, Type I or Type II, is possible? Justify your answer. [2]
Mark scheme
2(a) H0: Proportion = 0.05 B1 H1: Proportion > 0.05 1 2(b) 1 – (0.9525 + 25 × 0.9524 × 0.05 + 25C2 × 0.9523 × 0.052 + 25C3 × 0.9522 × 0.053) M1 Completely correct expression A1 0.0341 A1 3 2(c) Type II B1 Will conclude proportion not increased B1 2
Question
In the data-entry department of a certain firm, it is known that 0.12% of data items are entered incorrectly, and that these errors occur randomly and independently. (a) A random sample of 3600 data items is chosen. The number of these data items that are incorrectly entered is denoted by X . (i) State the distribution of X , including the values of any parameters. [1] (ii) State an appropriate approximating distribution for X , including the values of any parameters. Justify your choice of approximating distribution. [3] (iii) Use your approximating distribution to find P X > 2. [2] (b) Another large random sample of n data items is chosen. The probability that the sample contains no data items that are entered incorrectly is more than 0.1. Use an approximating distribution to find the largest possible value of n. [3]
Mark scheme
3(a)(i) B(3600, 0.0012) B1 1 3(a)(ii) Po(4.32) B2 (B1 for Po. B1 for λ = 4.32) n = 3600 which is large, p = 0.12 which is small and np = 4.32 which is < 5 B1 3 3(a)(iii) 4.322 M1 1 – e–4.32 1 + 4.32 + 2 0.805 (3 sf) A1 2 3(b) e–λ > 0.1 M1 (–λ > ln0.1) (λ < ln10) 0.0012n < ln10 A1 (n < 1918.8) A1 largest n is 1918 3
Question
The score on one spin of a 5-sided spinner is denoted by the random variable X with probability distribution as shown in the table. x 0 1 2 3 4 P X = x 0.1 0.2 0.4 0.2 0.1 (a) Show that Var X = 1.2. [2] The spinner is spun 200 times. The score on each spin is noted and the mean, X , of the 200 scores is found. (b) Given that P X > a = 0.1, find the value of a. [4] (c) Explain whether it was necessary to use the Central Limit theorem in your answer to part (b). [1] (d) Johann has another, similar, spinner. He suspects that it is biased so that the mean score is less than 2. He spins his spinner 200 times and finds that the mean of the 200 scores is 1.86. Given that the variance of the score on one spin of this spinner is also 1.2, test Johann’s suspicion at the 5% significance level. [5]
Mark scheme
4(a) E(X) = 2 B1 0.2 × 1 + 0.4 × 22 + 0.2 × 32 + 0.1 × 42 – 22 (= 1.2) AG B1 2 4(b) a−2 M1 = φ−1 ( 0.9 ) 1.2 ÷ 200 M1 (M1 for LHS, M1 for RHS) a = 2 + 1.2 ÷ 200 × 1.282 M1 2.10 (3 sf) A1 4 4(c) Yes, because X is not normally distributed. B1 1 4(d) H0: pop mean = 2 B1 H1: pop mean < 2 1.86 − 2 M1 1.2 ÷ 200 1.807 A1 comp z = 1.645 M1 There is evidence that the spinner is biased so that mean is less than 2 A1 5
Question
(a) The random variable X has the distribution Po ,. (i) State the values that X can take. [1] It is given that P X = 1 = 3 × P X = 0. (ii) Find ,. [1] (iii) Find P 4 ≤ X ≤ 6. [2] (b) The random variable Y has the distribution Po - where - is large. Using a suitable approximating distribution, it is found that P Y < 46 = 0.0668, correct to 4 decimal places. Find -. [5]
Mark scheme
5(a)(i) 0, 1, 2, 3, . . . . B1 1 5(a)(ii) 3 B1 1 5(a)(iii) 34 35 36 M1 e−3 + + 4! 5! 6! 0.319 (3 sf) A1 2 5(b) Φ–1(0.0668) (= –1.500) M1 N(μ, μ) M1 45.5 − μ M1 = “–1.500” μ μ – “1.500” μ − 45.5 = 0 M1 "1.5"± ("− 1.5")2 + 4 × 45.5 μ= ( = 7.5369 ) 2 μ = 56.8 (3 sf) A1 5
Question
A random variable X has probability density function given by t k 1 ≤ x ≤ a, f x = x2 0 otherwise, where k and a are positive constants. a (a) Show that k = . [3] a−1 (b) Find E X in terms of a. [3] (c) Find the 60th percentile of X in terms of a. [4] 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
6(a) a k M1 1 x2 dx = 1 1a A1 k − = 1 x1 1 k 1 − = 1 a a − 1 A1 k =1 a 1 k = AG a −1 3 6(b) a a 1 M1 a −1 1 x dx a a A1 [ln x ] a −1 1 a ln a A1 a −1 3 6(c) a m 1 3 M1 a −1 1 x 2 dx = 5 a 1m 3 A1 − = a − 1 x 1 5 a 1 3 1− = a − 1 m 5 1 3 ( a − 1) 1 2a + 3 A1 =1− or = m 5a m 5a 5a A1 m= 2a + 3 4
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Mathematics 9709/62 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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