Cambridge A Level Mathematics 9709 — 2011 May/June Paper 7 · Variant 1

1 On average, 2 people in every 10 000 in the UK have a particular gene. A random sample of 6000 people in the UK is chosen. The random variable X denotes the number of people in the sample who have the gene. Use an approximating distribution to calculate the probability that there will be more than 2 people in the sample who have the gene. [4]

Mark scheme: 1 Poisson B1 λ = 1.2 B1 1.2 seen 2 1 – e–1.2(1 + 1.2 + 2.1 ) M1 1 – Poisson P(0, 1, 2, 3) attempted, any λ, allow 2 A1 1 end error = 0.121 [4] SC: using Bin, ans 0.120: B1 32 6

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2 (a) The time taken by a worker to complete a task was recorded for a random sample of 50 workers. The sample mean was 41.2 minutes and an unbiased estimate of the population variance was 32.6 minutes2. Find a 95% confidence interval for the mean time taken to complete the task. [3] (b) The probability that an α% confidence interval includes only values that are lower than the population mean is 16.1 Find the value of α. [2]

Mark scheme: 32 6. 2 (a) 41.2 ± z × 50 M1 z = 1.96 B1 [39.6, 42.8] (3 sfs) A1 Allow any brackets or none, or < or “to” etc [3] (b) 2 × 1 or 1 or 0.125 or 12.5% M1 or 0.875 16 8 α = 87.5% A1 [2] 857. −85

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3 Past experience has shown that the heights of a certain variety of rose bush have been normally distributed with mean 85.0 cm. A new fertiliser is used and it is hoped that this will increase the heights. In order to test whether this is the case, a botanist records the heights, x cm, of a large random sample of n rose bushes and calculates that x = 85.7 and s = 4.8, where x is the sample mean and s2 is an unbiased estimate of the population variance. The botanist then carries out an appropriate hypothesis test. (i) The test statistic, ß, has a value of 1.786 correct to 3 decimal places. Calculate the value of n. [3] (ii) Using this value of the test statistic, carry out the test at the 5% significance level. [3]

Mark scheme: 3 (i) 8.4 (= 1.786) M1 n n = ( .1786 × 8.4 ) 2 A1 Correct equation in n 7.0 = 150 A1 [3] (ii) H0: µ = 85.0 H1: µ > 85.0 B1 z = 1.645 M1 Comparison 1.786 and 1.645 Allow 1.96 if H1: µ ≠ 85.0 Evidence that µ increased A1f Correct conc. No contradictions. ft H1 [3]

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4 (a) g( )x h( )x 1 1 x x 0 1 2 0 1 2 –1 –1 The diagrams show the graphs of two functions, g and h. For each of the functions g and h, give a reason why it cannot be a probability density function. [2] (b) The distance, in kilometres, travelled in a given time by a cyclist is represented by the continuous random variable X with probability density function given by 30 10 ≤x ≤15, x2 f(x) =   0 otherwise. (i) Show that E(X) = 30 ln 1.5. [3] (ii) Find the median of X. Find also the probability that X lies between the median and the mean. [5]

Mark scheme: 4 (a) g: Area ≠ 1 or > 1 B1 h: pdf cannot be neg B1 [2] 15 30 (b) (i) dx M1 Attempt integ xf(x), ignore limits ∫ x 10 = [30 ln x ] 1510 A1 Correct integrand and limits = 30(ln15 – ln10) A1 or 30ln(15/10) (= 30ln1.5 AG) [3] m 30 (ii) 2 dx = 0.5 M1 Integ f(x) = 0.5, limits 10 to unknown ∫ x 10 [− 30 x −1 ] 10m = 0.5 A1 Correct integrand, limits and = 0.5 −m30 − ( − 1030 ) = 0.5 m = 12 A1 30 ln1.5 30 2 dx M1 ∫ x '12 ' = 0.0337 (3 sfs) A1 [5] GCE AS/A LEVEL – May/June 2011 9709 71

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5 Cans of drink are packed in boxes, each containing 4 cans. The weights of these cans are normally distributed with mean 510 g and standard deviation 14 g. The weights of the boxes, when empty, are independently normally distributed with mean 200 g and standard deviation 8 g. (i) Find the probability that the total weight of a full box of cans is between 2200 g and 2300 g. [6] (ii) Two cans of drink are chosen at random. Find the probability that they differ in weight by more than 20 g. [5]

Mark scheme: 5 (i) W ~N(2240, 848) B2 B1 each parameter 2200 − 2240 (= –1.374) 848 Φ(“–1.374”) = 1– Φ(“1.374”) (= 0.0847) 2300 − 2240 (= 2.060) 848 Φ(“2.060”) (= 0.9803) M1A1 Standardise either value and evaluate correctly Φ(“2.060”) – (1 – Φ(“1.374”)) M1 Correct combination of Φ’s = 0.896 (3 sfs) A1 [6] (ii) X1 – X2 ~N(0, 392) B1 May be implied 20 − 0 (= 1.010) M1 392 (Φ(“1.010” = 0.8438) P(X > 20) = 1 – Φ(“1.010”) (= 0.1562) A1 2 × P(X > 20) M1 = 0.312 (3 sfs) A1 [5]

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6 The number of injuries per month at a certain factory has a Poisson distribution. In the past the mean was 2.1 injuries per month. New safety procedures are put in place and the management wishes to use the next 3 months to test, at the 2% significance level, whether there are now fewer injuries than before, on average. (i) Find the critical region for the test. [5] (ii) Find the probability of a Type I error. [1] (iii) During the next 3 months there are a total of 3 injuries. Carry out the test. [3] (iv) Assuming that the mean remains 2.1, calculate an estimate of the probability that there will be fewer than 20 injuries during the next 12 months. [5]

Mark scheme: 6 (i) mean = 6.3 B1 B1 for 6.3 P(X < 1) = e–6.3(1 + 6.3) = 0.0134 M1 Allow incorrect λ in both probs 3.6 2 P(X < 2) = e–6.3(1 + 6.3 + ) = 0.0498 M1A1 2 CR is X < 1 A1 A1 for both values [5] (ii) P(Type I error) = P(X < 1) = 0.0134 B1 [1] (iii) H0: λ = 6.3 H1: λ < 6.3 B1 Can be scored in (i). Accept λ = 2.1(per month) 3 not in CR M1 or P(X < 3) = 0.126 > 0.02 No evidence mean no. of injuries has decreased A1 Correct conclusion [3] (iv) N(25.2, 25.2) B2 B1 for N & µ = 25.2. B1 for σ2 = 25.2 May be implied 195. − 252. (= –1.135) M1 Allow with wrong or no cc or no √ 252. Φ(“–1.135”) = 1 – Φ(“1.135”) M1 Correct area = 0.128 (3 sfs) A1 [5]

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