Cambridge A Level Mathematics 9709 — 2011 May/June Paper 7 · Variant 3

5 Each drink from a coffee machine contains X cm3 of coffee and Y cm3 of milk, where X and Y are independent variables with X ∼N(184, 152) and Y ∼N(50, 82). If the total volume of the drink is less than 200 cm3 the customer receives the drink without charge. (i) Find the percentage of drinks which customers receive without charge. [4] (ii) Find the probability that, in a randomly chosen drink, the volume of coffee is more than 4 times the volume of milk. [5]

Mark scheme: 5 (i) E(T) = 234, Var(T) = 152 + 82 = 289 B1 200− 234 (= –2.000) M1 " 289" Φ(“–2.000”) = 1 – Φ (“2.000”) M1 1 – 0.9772 2.28% A1 [4] (ii) Require P(D > 0) where D = X – 4Y E(D) (= 184 – 4 × 50) = –16 B1 For –16 or +16 or ± (184 – 4 × 50) Var(D) (= 152 + 42 × 82) = 1249 B1 For 1249 or 152 + 42 × 82 0 − ( −16 ) (= 0.453) M1 "1249" 1 – Φ (“0.453”) M1 (= 1 – 0.6747) = 0.325 A1 [5] [Total: 9] 3 3 k ∫ ( 2 ) d ∫ ( 2 ) d

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6 The distance travelled, in kilometres, by a Grippo brake pad before it needs to be replaced is modelled by 10 000X, where X is a random variable having the probability density function −k(x2 −5x + 6) 2 ≤x ≤3, f(x) = 0 otherwise. The graph of y = f(x) is shown in the diagram. y x 0 1 2 3 (i) Show that k = 6. [2] (ii) State the value of E(X) and find Var(X). [4] (iii) Sami fits four new Grippo brake pads on his car. Find the probability that at least one of these brake pads will need to be replaced after travelling less than 22 000 km. [3]

Mark scheme: 6 (i) − k ∫ 2 ( x 2 − 5 x + 6) d x = 1 M1 Integ = 1; ignore − 6 ∫ 2 ( x 2 − 5 x + 6) d x 3 2 3 2 limits (–k( 33 − 5 × 32 + 6 × 3 − [ 23 − 5 × 22 + 6 × 2 ] ) = 1) ignore limits –k × (– 1 ) = 1 or k × 1 = 1 A1 [2] Correctly obtain Correctly obtain 1 6 6 (k = 6 AG) – 16 or 16 CWO No rounded decimals (ii) E(X) = 2.5 B1 Condone 25000 3 − 6 ∫ 2 ( x 4 − 5 x 3 + 6 x 2 ) d x (= –6 × (–1.05)) M1* Integ x2f(x); ignore limits – “2.5”2 Dep Subtr µ2, M1* = 0.05 A1 [4] ISW 2.2 (iii) − 6 ∫ 2 ( x 2 − 5 x + 6 ) d x (= 0.104) M1 Integ with limits 2, 2.2 or 2.2, 3 1 – (1 – “0.104”)4 M1 Or equivalent = 0.355/0.356 A1 [3] [Total: 9] GCE AS/A LEVEL – May/June 2011 9709 73 121 11

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7 Previous records have shown that the number of cars entering Bampor on any day has mean 352 and variance 121. (i) Find the probability that the mean number of cars entering Bampor during a random sample of 200 days is more than 354. [4] (ii) State, with a reason, whether it was necessary to assume that the number of cars entering Bampor on any day has a normal distribution in order to find the probability in part (i). [2] (iii) It is thought that the population mean may recently have changed. The number of cars entering Bampor during the day was recorded for each of a random sample of 50 days and the sample mean was found to be 356. Assuming that the variance is unchanged, test at the 5% significance level whether the population mean is still 352. [4]

Mark scheme: 7 (i) Var( X ) = 121200 or SD of X = 11200 354 − 352 (±) (= ± 2.571) M1 Or with cc attempted. Allow no √ Must 11 200 include 200 or √ 200 A1 2.57(1) or correct expression 1 – Φ(“2.571”) M1 (= 1 – 0.9949) = 0.0051 A1 [4] (ii) (No) “No” must be seen or implied, but gains no marks by itself n is large, B1 n ≥ 30 X (appr) norm distr or CLT applies B1 [2] (SR Both statements correct, but wrong or no conclusion scores B1) (iii) H0: Pop mean = 352 H1: Pop mean ≠ 352 B1 Allow ‘µ’ but not just ‘mean’ 356 − 352 ± ± (= 2.57(1)) M1 Must have √ 50 11 50 A1 Correct statement or 2.57(1) Comp with z = ±1.96 (signs consistent) B1√ Correct comparison, and correct conclusion, Evidence that pop mean has changed [4] follow through one tail test [Total: 10]

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