Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 7 · Variant 2

1 In a survey, 36 out of 120 randomly selected voters in Hungton said that if there were an election next week they would vote for the Alpha party. Calculate an approximate 90% confidence interval for the proportion of voters in Hungton who would vote for the Alpha party. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Var(Ps) = 0.3(1120− 0.3) (= 0.00175) M1 Attempt correct values in correct formula 0.3 ± z "0.00175" M1 must be a z-value, not a prob z = 1.645 B1 CI = 0.231 to 0.369 (3 sf) A1 Total: 4

More questions on Sampling and estimation

2 Karim has noted the lifespans, in weeks, of a large random sample of certain insects. He carries out a test, at the 1% significance level, for the population mean, -. Karim’s null hypothesis is - = 6.4. (i) Given that Karim’s test is two-tail, state the alternative hypothesis. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Karim finds that the value of the test statistic is z = 2.43. (ii) Explain what conclusion he should draw. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Explain briefly when a one-tail test is appropriate, rather than a two-tail test. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) (H1): µ ≠ 6.4 B1 Total: 1 2(ii) comp 2.43 with a z-value M1 oe valid comparison z = 2.576 AND No evidence that µ is not 6.4 A1 Allow “Accept µ = 6.4” or do not reject µ = 6.4 Must mention µ, not just “H0” or “H1” Total: 2 2(iii) Testing for an increase in µ, or for a B1 Any equiv statement decrease in µ, rather than a change Total: 1

More questions on Hypothesis tests

4 At a doctors’ surgery, the number of missed appointments per day has a Poisson distribution. In the past the mean number of missed appointments per day has been 0.9. Following some publicity, the manager carries out a hypothesis test to determine whether this mean has decreased. If there are fewer than 3 missed appointments in a randomly chosen 5-day period, she will conclude that the mean has decreased. (i) Find the probability of a Type I error. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State what is meant by a Type I error in this context. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability of a Type II error if the mean number of missed appointments per day is 0.2. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) (λ =) 4.5 B1 e–4.5(1 + 4.5 + 4.52!2 ) M1 Allow any λ. Allow one end error = 0.174 A1 Total: 3 4(ii) Accept reduction in mean no. of B1 or Mean is 0.9 (or 4.5) but < 3 missed missed appts although untrue appts. In context Total: 1 4(iii) P(X ⩾ 3) M1 Attempted = 1 – e–1(1 + 1 + 212! ) M1 Allow any λ except 4.5 or 0.9, Allow one end error = 0.0803 (3 sfs) A1 Total: 3

More questions on The Poisson distribution

6 The masses, in kilograms, of cartons of sugar and cartons of flour have the distributions N 78.8, 12.62 and N 62.0, 10.02 respectively. (i) The standard load for a certain crane is 8 cartons of sugar and 3 cartons of flour. The maximum load that can be carried safely by the crane is 900 kg. Stating a necessary assumption, find the percentage of standard loads that will exceed the maximum safe load. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that a randomly chosen carton of sugar has a smaller mass than a randomly chosen carton of flour. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) Assume cartons are random sample(s) B1 or masses of cartons are independent of each other oe E(T) = 816.4 B1 Both Var(T) = 1570.08 z = 900 − "816.4" (= 2.110) M1 "1570.08" 1 – Φ(“2.110”) M1 = 0.0174 = 1.74% (3 sfs) A1 % only (accept 1.7% if 0.0174 seen) Total: 5 6(ii) P(F – S > 0) stated or implied M1 P(S – F < 0) 62.0 – 78.8 (= –16.8) B1 78.8 – 62.0 (= 16.8) & 10.02 + 12.62 (= 258.76) & 12.62 + 10.02 (= 258.76) z = 0 −− (" 16.8") (= 1.044) M1 z = 0"258.76"− "16.8" (= –1.044) "258.76" 1 – Φ(“1.044”) M1 Φ(“–1.044”) = 1 – Φ(“1.044”) (= 1 – 0.8517) A1 = 0.148 (3 sfs) Total: 5

More questions on Discrete random variables

7 The number of planes arriving at an airport every hour during daytime is modelled by the random variable X with distribution Po 5.2 . (i) State two assumptions required for the Poisson model to be valid in this context. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) (a) Find the probability that the number of planes arriving in a 15-minute period is greater than 1 and less than 4, [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Find the probability that more than 3 planes will arrive in a 40-minute period. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (iii) The airport has enough staffto deal with a maximum of 60 planes landing during a 10-hour day. Use a suitable approximation to find the probability that, on a randomly chosen 10-hour day, staffwill be able to deal with all the planes that land. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) Planes arrive at constant mean rate B1 Planes arrive at random B1 or Planes arrive independently Must be in context Total: 2 7(ii)(a) (λ =) 5.2 ÷ 4 M1 e–1.3( 1.32 2 + 1.33!3 ) M1 Allow any λ , allow one end error = 0.330 (3 sfs) A1 Accept 0.33 Total: 3 7(ii)(b) 1– e–3.467×(1+3.467 + 3.4672! 2 + 3.4673! 3 ) M1 Allow any λ except 5.2 or 1.3, allow one end error = 0.456 (3 sfs) A1 Total: 2 7(iii) N(52, 52) stated or implied B1 60.5 − 52 (= 1.179) M1 ft their mean and var. 52 Allow wrong or no cc or no √ Φ(“1.179”) M1 = 0.881 (3 sf) A1 Total: 4

More questions on The Poisson distribution