Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 7 · Variant 3

1 The lengths of logs are normally distributed with mean 3.5 m and standard deviation 0.12 m. Describe fully the distribution of the total length of 8 randomly chosen logs. [3]

Mark scheme: 1 Normal with mean 28 B1 Both Var = 0.122 × 8 M1 square & × by 8 or sd = 0.12 × √8 = 0.115 (3 sfs) A1 [3] or sd = 0.339 (3 sfs) clearly stated var / sd Total [3]

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3 Joshi suspects that a certain die is biased so that the probability of showing a six is less than 6.1 He plans to throw the die 25 times and if it shows a six on fewer than 2 throws, he will conclude that the die is biased in this way. (i) Find the probability of a Type I error and state the significance level of the test. [3] Joshi now decides to throw the die 100 times. It shows a six on 9 of these throws. (ii) Calculate an approximate 95% confidence interval for the probability of showing a six on one throw of this die. [4]

Mark scheme: 3 (i) 25 24 M1 Allow end errors, but just P(2) implies M0  5   5   1    + 25    Accept p/q mix  6   6   6  = 0.0629 final answer A1 Sig level = 6.29% B1ft [3] ft their P(X < 1) with Binomial used. Allow 6.3% or 6% (ii) .009 × .091 Var (p) ≈ 100 M1 For pq /100 seen ( any p/q ) ( must be probs ) (= 0.000819) B1 z = 1.96 .009 × .091 0.09 ± z M1 For correct form of C.I. ( any p/q ) ( must be probs ) 100 = 0.034 to 0.146 (3 dps) A1 [4] Total [7] GCE A LEVEL – October/November 2012 9709 73

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4 The masses of a certain variety of potato are normally distributed with mean 180 g and variance 1550 g2. Two potatoes of this variety are chosen at random. Find the probability that the mass of one of these potatoes is at least twice the mass of the other. [7]

Mark scheme: 4 Use of X1 – 2X2 or similar Or use of ½ X1 – X2 E(X1–2X2) = 180 – 360 ( = –180 ) B1 E(2X1–X2) = 360 – 180 ( = 180 ) Or E(½ X1 – X2 ) = 90 – 180 = ( –90 ) Var(X1–2X2) = 5×1550 or 7750 M1 for 1550 + 4 × 1550 or ¼ × 1550 + 1550 A1 7750 or 1937.5 0 − ( −180) 0−180 M1 Allow incorrect var (dep > 0 & ≠ 1550), no or '7750' '7750' Standardising – no mixed methods (= ±2.045) Or ± (0 – –90)/ √1937.5 1 – c(‘2.045’) M1 For finding correct area (consistent with working) = 0.0205 or 0.0204 A1 Ans 0.041 (2 sf) B1ft [7] Allow double their prob Total [7]

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5 It is claimed that, on average, people following the Losefast diet will lose more than 2 kg per month. The weight losses, x kilograms per month, of a random sample of 200 people following the Losefast diet were recorded and summarised as follows. n = 200 Σ x = 460 Σ x2 = 1636 (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) Test the claim at the 1% significance level. [5]

Mark scheme: 5 (i) Est(µ) = 2.3 B1 2  2  200  1636  460  200  1636  460      − or 1.7043 for M1   − Est(ë2) = M1 Allow   199  199  200  200   200  200       Or 1/199 ( 1636 – 4602/200 ) = 2.90 (3 sf) or 2.91 or 578/199 A1 [3] (ii) H0: Pop mean wt loss = 2 kg H1: Pop mean wt loss > 2 kg B1 Allow ‘µ’ but not just ‘mean’ 3.2 − 2 3.2 − 2 Stand’ise with √200. Accept sd/var M1 .2'9045' .1'7043' 200 200 mixes Or xcrit = 2 + 2.326√( 2.9045/200 ) = 2.489 or ± 2.49 A1 or 0.0064 / 0.9936 for area comparison or xcrit = 2.28(03) comp z = 2.326 M1 For valid comparison ( z or area or xcrit ) Evidence that mean wt loss > 2 kg A1ft No contradictions Reject H0 / accept H1 only if H0 / H1 correctly defined 200 If not used in (i): var = 2.89, sd = 1.7, [5] 199 cr z = 2.496 can score all marks Total [8] GCE A LEVEL – October/November 2012 9709 73

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6 Darts are thrown at random at a circular board. The darts hit the board at distances X centimetres from the centre, where X is a random variable with probability density function given by 2 x 0 ≤x ≤a, a2 f(x) =   0 otherwise, where a is a positive constant. (i) Verify that f is a probability density function whatever the value of a. [3] It is now given that E(X) = 8. (ii) Find the value of a. [3] (iii) Find the probability that a dart lands more than 6 cm from the centre of the board. [3]

Mark scheme: 6 (i) f(x) [ 0 for all x defined B1 a 2 x ) dx with limits 0, a . Must be a. ∫ 0 xdx M1 Attempt ∫ f( a 2   2 x 2  a  =   A1 Or equivalent methods ( e.g. by areas )  0   2 a 2    = 1 [3] (ii) a 2 2 x dx (= 8) , ignore limits ∫ 0 M1 Attempt ∫ xf( x ) d x a 2 2  x 3  a 2   (= 8) A1 Correct integrand and limits a  3  0  2a   = 8  A1  3  a = 12 [3] (iii) 2 2 1 – 6∫0 144 xdx or 12∫6 144 xd x M1 Correct expr’n incl limits; ft their ‘a’ 1  x 2  6 =1 – 1 −   or A1ft Correct integrand and limits; ft their ‘a’ 72 2 0   1 x 2  12   72 2 6   3 = 4 A1ft [3] ft their ‘a’, dep 0 < ans < 1 Total [9] GCE A LEVEL – October/November 2012 9709 73

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7 The number of workers, X, absent from a factory on a particular day has the distribution B(80, 0.01). (i) Explain why it is appropriate to use a Poisson distribution as an approximating distribution for X. [2] (ii) Use the Poisson distribution to find the probability that the number of workers absent during 12 randomly chosen days is more than 2 and less than 6. [3] Following a change in working conditions, the management wishes to test whether the mean number of workers absent per day has decreased. (iii) During 10 randomly chosen days, there were a total of 2 workers absent. Use the Poisson distribution to carry out the test at the 2% significance level. [5]

Mark scheme: 7 (i) n > 50 B1 Accept n large np = 0.8, which is < 5 B1 [2] Accept p small (ii) λ = 9.6 B1 6.9 3 6.9 4 6.9 5 e–9.6( + + ) M1 Any λ Accept end errors. !3 !4 !5 = 0.0800 (3 sfs) A1 [3] Allow 0.08 (iii) H0: Pop mean for 10 days = 8 or Pop mean for 1 day = 0.8 H1: Pop mean for 10 days < 8 B1 Pop mean for 1 day < 0.8 Allow λ or µ but not just ‘mean’ 82 e–8(1 + 8 + ) M1 Any λ. Accept end errors. !2 NB P(2) only used scores M0M0 Accept CR method = 0.0138 or 0.0137 A1 CR = 0, 1, 2 all working must be shown Compare 0.02 M1 Valid comparison with 0.02 or CR Evidence that mean number of A1ft No contradictions absentees has decreased Reject H0 / accept H1 only if H0 / H1 correctly [5] defined Total [10] Total for paper [50]

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