Cambridge A Level Mathematics 9709 — 2017 May/June Paper 7 · Variant 2

1 In a survey of 2000 randomly chosen adults, 1602 said that they owned a smartphone. Calculate an approximate 95% confidence interval for the proportion of adults in the whole population who own a smartphone. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 0.801 1 0.801 2000 × − (= 0.0000797) M1 0.801± z × "0.0000797" M1 Allow any z-value z = 1.96 B1 0.784 to 0.818 (3 sf) A1 As final answer. Must be an interval Allow 0.783 to 0.819 Total: 4

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3 Household incomes, in thousands of dollars, in a certain country are represented by the random variable X with mean - and standard deviation 3. The incomes of a random sample of 400 households are found and the results are summarised below. n = 400 Σ x = 923 Σx2 = 3170 (i) Calculate unbiased estimates of - and 32. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) A random sample of 50 households in one particular region of the country is taken and the sample mean income, in thousands of dollars, is found to be 2.6. Using your values from part (i), test at the 5% significance level whether household incomes in this region are greater, on average, than in the country as a whole. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) B1 Est(σ2) = 2 400 3170 "2.3075" 399 400   −     OE M1 = 2.60696 or 2.61 (3 sf) A1 (Note: Biased Var= 2.600 scores M0) Total: 3 3(ii) H0: Pop mean (or µ) = "2.31" or "2310" H1: Pop mean (or µ) > "2.31" or "2310" B1 FT ± 2.6 "2.310" 2.60696 50 − ÷ = 1.27 M1 A1 Standardising using their values, Accept 1.28 Comp 1.645 (OE) M1 Valid comparison z values or areas No evidence that incomes in the region greater A1 FT OE FT their z. No contradictions (No FT for 2 tail test – max score B0 M1 A1 M1 for comp 1.96 A0) Note: Accept alternative CV method Total: 5

More questions on Hypothesis tests

4 It is claimed that 1 in every 4 packets of certain biscuits contains a free gift. Marisa and Andr´e both suspect that the true proportion is less than 1 in 4. (i) Marisa chooses 20 packets at random. She decides that if fewer than 3 contain free gifts, she will conclude that the claim is not justified. Use a binomial distribution to find the probability of a Type I error. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Andr´e chooses 25 packets at random. He decides to carry out a significance test at the 1% level, using a binomial distribution. Given that only 1 of the 25 packets contains a free gift, carry out the test. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) M1 = 0.0913 A1 As final answer Total: 2 4(ii) H0: Pop proportion=0.25 H1: Pop proportion<0.25 B1 Allow p or π, not "proportion" (Accept anywhere in the question) 0.7525 + 25 × 0.7524 × 0.25 M1 Must be B(25,0,25) No end errors = 0.00702 A1 comp 0.01 M1 Valid comparison There is evidence that the claim is not justified A1 FT OE. No contradictions Total: 5

More questions on Probability

6 Old televisions arrive randomly and independently at a recycling centre at an average rate of 1.2 per day. (i) Find the probability that exactly 2 televisions arrive in a 2-day period. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use an appropriate approximating distribution to find the probability that at least 55 televisions arrive in a 50-day period. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Independently of televisions, old computers arrive randomly and independently at the same recycling centre at an average rate of 4 per 7-day week. (iii) Find the probability that the total number of televisions and computers that arrive at the recycling centre in a 3-day period is less than 4. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) 2 2.4 2.4 2! e− × M1 = 0.261 (3 sfs) A1 Total: 2 6(ii) N(60, 60) B1 seen or implied 54.5 60 60 − (= ̶ 0.710) M1 allow with wrong or missing cc 1 ̶ φ(" ̶ 0.710") = φ("0.710") M1 For area consistent with their working = 0.761 (3 sf) A1 Total: 4 6(iii) λ = 3.6 + 12 ÷ 7 (= 186/35) (= 5.314) M1 ( ) 2 3 5.314 5.314 5.314 2 3! 1 5.314 e− + + + M1 Allow incorrect λ. Allow one end error. = 0.224 (3 sfs) A1 Total: 3

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