> GCE Further Mathematics B (MEI) Y422/01: Statistics major Advanced GCE Mark Scheme for Autumn 2021

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Mark Scheme October 2021 Question Answer Marks AOs Guidance 1 (a) 34.711 B1 1.1 ± 1.96 M1 3.3 1.53 50 × M1 1.1 = 34.711 ± 0.424 or (34.287, 35.135) A1 3.4 Allow 34.29 to 35.13 or 35.14 [4] ... 1 (b) 50 is a sufficiently large sample to apply the CLT which states that for large samples the distribution of the sample mean is approximately Normal B1* *B1 [2] 2.2b 2.4 For mention of central limit theorem For full statement (including CLT) No credit if CLT not mentionedY422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 2 (a) 6 1 1 P( 0) X = = × × 6 6 6 M1 3.1a Allow M1 for 1 1 1 6 6 36 × = 1 = 36 A1 1.1 AG [2] 2 (b) B1 B1 [2] 1.1 1.1 For heights For axes and labels Roughly correct but must have linear scale Do not allow just P on vertical axis 2 (c) The distribution has (slight) negative skew B1 [1] 1.1 Allow ‘roughly symmetrical’ or ‘unimodal’ Not ‘Normal distribution’ 2 (d) DR 1 2 1 2 5 5 E( ) 0 1 2 3 4 5 X = × + × + × + × + × + × 36 36 9 4 9 36 M1 1.1a 105 35 = = = 36 12 2.9166 A1 1.1 Allow fraction or decimal form 2 2 2 2 2 2 2 1 2 1 2 5 5 E( ) 0 1 2 3 4 5 X = × + × + × + × + × + × 36 36 9 4 9 36 371 = = 36 10.3055 M1 1.1 2 Var( ) 10.3055 (2.9166 ) X = −   M1 1.2 259 = = 144 1.80 (1.7986…) A1 1.1 [5] 2 (e) Variance 30 1.7986 1619 = × = 2  (pence2) B1 1.1 [1] 2 (f) Average amount received = 30 × 2.916… = 87.5 B1 3.1a k – 87.5 = 12.5 ⇒ k = 100 B1 1.1 [2] 0.00 0.10 0.20 0.30 0 1 2 3 4 5 P(X) = r rY422/01 Mark Scheme October 2021 Question Answer Marks AOs Guidance 3 (a) Using B(50, 0.04) M1 3.3 P(X = 2) = 0.276 A1 1.1 BC [2] 3 (b) 0.96 0.04 0.0277 9 × = B1 1.1 Allow 0.028 [1] 3 (c) 0.96 0.442 20 = B1 1.1 [1] 3 (d) Expected value for one misunderstood = 1 25 0.04 = B1 2.1 Must quote probabilities to get full marks Because geometric E1 2.4 For 3 misunderstood expected number = 25 + 25 + 25 = 75 E1 1.1 [3] 3 (e) Require P(2 misunderstood in first 59) × 0.04 B1 3.1a For identifying required probability so using B(59, 0.04) gives P(X = 2) = 0.267 M1 2.2a Use of correct binomial 0.267 × 0.04 = 0.0107 A1 1.1 BC [3] 4 (a) Nuclei decay randomly and decays are independent with constant probability 200 000 1 The number of decays out of 1 000 000 is being counted, so a binomial distribution is appropriate Because n = 1 000 000 is large and p = 200 000 1 is small a Poisson distribution is also appropriate E1 E1 E1 [3] 2.4 2.4 2.4 For partial explanation of binomial For full explanation For explanation of Poisson [Show More]

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