Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 7 · Variant 3

2 An airline has found that, on average, 1 in 100 passengers do not arrive for each flight, and that this occurs randomly. For one particular flight the airline always sells 403 seats. The plane only has room for 400 passengers, so the flight is overbooked if the number of passengers who do not arrive is less than 3. Use a suitable approximation to find the probability that the flight is overbooked. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Poisson B1 seen or implied λ= 4.03 B1 seen or implied e–4.03(1 + 4.03 + 4.032! 2 ) M1 any λ; e.g. allow λ = 4 no extra or missing terms = 0.234 (3 sf) A1 4

More questions on Probability

3 After an election 153 adults, from a random sample of 200 adults, said that they had voted. Using this information, an !% confidence interval for the proportion of all adults who voted in the election was found to be 0.695 to 0.835, both correct to 3 significant figures. Find the value of !, correct to the nearest integer. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1533 200 × 200200−153 M1 153 200 + z × = 0 . 8 35 2 00 (Var(Ps) = 0.000898875) (s.d. 0.02998) z = 2.335 A1 allow 2.33 or 2.34 2Φ ( z ) − 1 M1 or equivalent method indep α = 98 A1 allow 98.0 but not e.g. 98.04 4

More questions on Sampling and estimation

4 The lengths, in millimetres, of rods produced by a machine are normally distributed with mean - and standard deviation 0.9. A random sample of 75 rods produced by the machine has mean length 300.1 mm. (i) Find a 99% confidence interval for -, giving your answer correct to 2 decimal places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The manufacturer claims that the machine produces rods with mean length 300 mm. (ii) Use the confidence interval found in part (i) to comment on this claim. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) 300.1 ± z × 0.9 M1 allow any value of z 75 z = 2.576 B1 allow 2.574 to 2.579 299.83 to 300.37 (2 dps) A1 answer must be seen to 2 dps need an interval 3 4(ii) CI includes 300 so claim supported or B1 FT or equivalent justified or probably true FT from CI in (i) 1

More questions on Sampling and estimation

6 The numbers of barrels of oil, in millions, extracted per day in two oil fields A and B are modelled by the independent random variables X and Y respectively, where X ∼N 3.2, 0.42 and Y ∼N 4.3, 0.62 . The income generated by the oil from the two fields is $90 per barrel for A and $95 per barrel for B. (i) Find the mean and variance of the daily income, in millions of dollars, generated by field A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that the total income produced by the two fields in a day is at least $670 million. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) Mean = 3.2 × 90 = 288 B1 Variance = 0.42 × 902 M1 = 1296 A1 3 6(ii) Mean = ‘288’ + 4.3 × 95 = 696.5 B1 FT Variance = ‘1296’ + 0.62 × 952 = 4545 B1 FT FT their (i) 670 − 696.5 (= -0.393) M1 FT Var provided both given Vars used 4545 standardising (ignore cc) no sd / Var mix 1 – φ(‘–0.393’) = φ(‘0.393) M1 correct area consistent with their working ( i.e. their mean ) = 0.653 (3 sf) A1 5

More questions on Discrete random variables

8 In order to test the effect of a drug, a researcher monitors the concentration, X, of a certain protein in the blood stream of patients. For patients who are not taking the drug the mean value of X is 0.185. A random sample of 150 patients taking the drug was selected and the values of X were found. The results are summarised below. n = 150 Σ x = 27.0 Σ x2 = 5.01 The researcher wishes to test at the 1% significance level whether the mean concentration of the protein in the blood stream of patients taking the drug is less than 0.185. (i) Carry out the test. [7] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that, in fact, the mean concentration for patients taking the drug is 0.175, find the probability of a Type II error occurring in the test. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(i) x = 27/150 (= 0.18) B1 150 5.01 2 M1 or var = 1/149(5.01 – 27.02/150) s = × − 0.18 or variance 149 150 (= 0.031729) (var = 3/2980 = 0.0010067) H0: Pop mean = 0.185 B1 allow just ‘µ’ H1: Pop mean < 0.185 0.18 − 0.185 M1 standardising, need 150 '0.031729' 150 = ( – ) 1.930 (3 sfs) or 1.93 A1 Comp with z = ( – ) 2.326 M1 consistent signs or using probs 0.0268 > 0.01 or 0.9732 < 0.99 or using xcrit 0.18 > 0.17897 There is no evidence (at 1% level) that A1 FT conclusion FT concentration with drug is less than no contradictions without drug 7 8(ii) cv − 0.185 M1 must use 0.185 and 150 ( = – 2.326 ) '0.031729' 150 = 0.17897 or 0.179 A1 acceptance region ( for H0 ) is > 0.179 "0.17897"− 0.175 M1 must use 0.175 and 150 (=1.534) '0.031729' 150 1 – φ(“1.534”) M1 indep mark = 0.0625 (3 sf) A1 Accept 0.0610 to 0.0628 5

More questions on Hypothesis tests