Cambridge A Level Mathematics 9709 — 2014 May/June Paper 7 · Variant 3

1 On average 1 in 25 000 people have a rare blood condition. Use a suitable approximating distribution to find the probability that fewer than 2 people in a random sample of 100 000 have the condition. [3]

Mark scheme: 1 e–4(1 + 4) M1 M1 for P (0 or 1) using Poisson, any λ M1 Expression of correct form correct λ (allow 1 end error) = 0.0916 (3 s.f.) A1 [3] SR Use of Bin(100000, 1/25000) scores M1 for P(0,1) allow one end error. A1 0.0916

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2 x −1 0 1 2 3 4 5 A random variable X takes values between 0 and 4 only and has probability density function as shown in the diagram. Calculate the median of X. [3]

Mark scheme: 2 1 1 ht = seen B1 or y = x 2 8 1  m 1  1 1 1 1 m 2 1 × m ×  × " "  = M1 × m ×( " " m ) = or = o.e. 2  4 2  2 2 8 2 16 2 N.B. B1 M1 must be consistent Or Integrating linear function of form y = kx with limits 0 and m or m and 4 and equated to 0.5 m = √8 or 2√2 or 2.83 (3 s.f.) A1 [3]

More questions on Probability

3 A die is thrown 100 times and shows an odd number on 56 throws. Calculate an approximate 97% confidence interval for the probability that the die shows an odd number on one throw. [4]

Mark scheme: 3 p = 0.56 B1 Used .0 56×.0 44 ‘0.56’ ± z × M1 Equation of correct form condone just +ve or –ve 100 Must be z z = 2.17, or 2.169 or 2.171 B1 0.452 to 0.668 (3 s.f.) A1 [4] Seen Must be an interval

More questions on Sampling and estimation

4 The weights, X kilograms, of rabbits in a certain area have population mean - kg. A random sample of 100 rabbits from this area was taken and the weights are summarised by Σx = 165, Σx2 = 276.25. Test at the 5% significance level the null hypothesis H0 : - = 1.6 against the alternative hypothesis H1 : - ≠1.6. [6]

Mark scheme: 4 x = 1.65 B1 100  276.25 2  est (σ2) =  − .165  B1 99  100  = 0.040404… = 4/99 .165 − 6.1 (±) 100 .165 − 6.1 ".0040404" M1 Without : B1 B0 M1 100 99 ".004" 100 = (±) 2.487/2.488 accept 2.49 Or 0.0065/0.0064 if area comparison = 2.50 A1 done A1 CV Method M1 must use 1.96 A1 for 1.639 or 1.6106 comp with 1.96 For valid comparison (z/z Signs consistent or M1 area/area cv) There is evidence that µ is not 1.6 A1 [6] Accept Reject H0 No contradictions GCE A LEVEL – May/June 2014 9709 73

More questions on Hypothesis tests

5 The lifetime, X years, of a certain type of battery has probability density function given by t k 1 ≤x ≤a, f x = x2 0 otherwise, where k and a are positive constants. (i) State what the value of a represents in this context. [1] a (ii) Show that k = [3] a −1. (iii) Experience has shown that the longest that any battery of this type lasts is 2.5 years. Find the mean lifetime of batteries of this type. [3]

Mark scheme: 5 (i) Longest lifetime B1 [1] Must be in context (ii) a k dx = 1 ∫ M1 Int f(x) and equate to 1. Ignore limits 2 1 x  a k   = 1 A1 Correct integral and limits −x1  1       k  − + 1  = 1  1a      −1 + a  k   = 1 or k (–1 + a) = a  a    a AG A1 [3] Must be convinced (AG) k = a − 1 (iii) 5.2 5.2 1 1 5 d x d x or k ∫ ∫ M1 Int xf(x). Ignore limits x x 3 1 1 5 5.2 5.2 = [lnx] or k[lnx] A1 Correct integral and limits 3 1 1 (Accept “k” or “their k”) 5 = ln2.5 or 1.53 (3 s.f.) A1 [3] 3

More questions on Integration

6 A machine is designed to generate random digits between 1 and 5 inclusive. Each digit is supposed to appear with the same probability as the others, but Max claims that the digit 5 is appearing less often than it should. In order to test this claim the manufacturer uses the machine to generate 25 digits and finds that exactly 1 of these digits is a 5. (i) Carry out a test of Max’s claim at the 2.5% significance level. [5] (ii) Max carried out a similar hypothesis test by generating 1000 digits between 1 and 5 inclusive. The digit 5 appeared 180 times. Without carrying out the test, state the distribution that Max should use, including the values of any parameters. [2] (iii) State what is meant by a Type II error in this context. [1]

Mark scheme: 6 (i) H0: p = 0.2 H1: p < 0.2 B1 (Allow π) P(0 or 1 5s in 25 | H0) M1 0.825 + 25 × 0.824 × 0.2 Use of B(25,1/5) and P(0) or P(1) or both – may be implied by “0.0274” = 0.0274 (3 s.f.) A1 Comp with 0.025 M1 Valid comparison No evidence (at 2.5% level) to A1 [5] No contradictions support claim SR Use of Normal N(5,4) leading to z = 1.75 or 0.0401 B1* H0 µ = 5 H1 µ < 5 B1. Comparison 1.75 < 1.96 or 0.0401 > 0.025 B1* dep (ii) Normal B1 µ = 200, σ2 = 160 or σ =√160 B1 [2] (iii) Concluding that the machine Not concluding that the machine produces too produces the right proportion of 5s, few 5s although it does. Must be in context although it doesn’t. B1 [1] o.e. No contradictions GCE A LEVEL – May/June 2014 9709 73

More questions on Probability

8 In an examination, the marks in the theory paper and the marks in the practical paper are denoted by the random variables X and Y respectively, where X ∼N 57, 13 and Y ∼N 28, 5 . You may assume that each candidate’s marks in the two papers are independent. The final score of each candidate is found by calculating X + 2.5Y. A candidate is chosen at random. Without using a continuity correction, find the probability that this candidate (i) has a final score that is greater than 140, [5] (ii) obtains at least 20 more marks in the theory paper than in the practical paper. [5]

Mark scheme: 8 (i) X + 2.5Y ~ N(127, 44.25) B1 B1 for 127 Allow at early stage (57 + 2.5 × 28) B1 B1 for 44.25 or 6.65 Allow at early stage (13 + 2.52 × 5) May be implied by next line 140 − "127 " (±) M1 For standardising " 44 . 25" = ±(1.954) M1 For area consistent with their working 1 – Φ(“1.954”) = 0.0254/0.0253 (3 s.f.) A1 [5] (ii) X – Y ~ N(29, 18) B1 B1 for 29 Give at early stage (57 – 28) B1 B1 for 18 Give at early stage (13 + 5) May be implied by next line 20 − " 29 " (= –2.121) M1 For Standardising "18" 1 – Φ(“–2.121”) = Φ(“2.121”) M1 For area consistent with their working = 0.983 (3 s.f.) A1 [5]

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