Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 6 · Variant 3

2 The number of enquiries received per day at a customer service desk has a Poisson distribution with mean 45.2. If more than 60 enquiries are received in a day, the customer service desk cannot deal with them all. Use a suitable approximating distribution to find the probability that, on a randomly chosen day, the customer service desk cannot deal with all the enquiries that are received. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 N(45.2, 45.2) B1 SOI 60.5 45.2 45.2 − [= 2.276] M1 Allow with wrong or no continuity correction. 1 –ɸ(‘2.276’) M1 0.0114 A1 4

More questions on The Poisson distribution

3 A random sample of 75 students at a large college was selected for a survey. 15 of these students said that they owned a car. From this result an approximate !% confidence interval for the proportion of all students at the college who own a car was calculated. The width of this interval was found to be 0.162. Calculate the value of ! correct to 2 significant figures. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 est(p) = 0.2 accept 15 75 2 × z × 0.2 0.8 75 × = 0.162 M1 Expression of the correct form. Condone missing 2x. z 75 0.081 0.2 0.8   = ×   ×   = 1.754 A1 Correct z. Condone 3sf accuracy. Φ(‘1.754’) = 0.96[03] '0.96' – (1 – '0.96') M1 OE. Using their z to find alpha. α = 92 A1 Following correct working. 5

More questions on Discrete random variables

7 The masses, in grams, of apples from a certain farm have mean - and standard deviation 5.2. The farmer says that the value of - is 64.6. A quality control inspector claims that the value of - is actually less than 64.6. In order to test his claim he chooses a random sample of 100 apples from the farm. (a) The mean mass of the 100 apples is found to be 63.5 g. Carry out the test at the 2.5% significance level. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Later another test of the same hypotheses at the 2.5% significance level, with another random sample of 100 apples from the same farm, is carried out. Given that the value of - is in fact 62.7, calculate the probability of a Type II error. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) H0: μ = 64.6 H1: μ < 64.6 B1 Allow population mean, not just ‘mean’. [ ]63.5 64.6 5.2 100 − ± ÷ M1 Standardising. Must have 100 . [ ] ± –2.115 A1 Accept -2.12 (3sf) ‘2.115’ > 1.96 or ‘–2.115’ < –1.96 [do not accept H0] M1 Valid comparison (0.0172 < 0.025 for area comparison). There is evidence that μ < 64.6 A1 FT Not definite, e.g. not ‘μ < 64.6’. in context. No contradictions. Accept critical value method leading to 63.5 < 63.58 or 64.6 > 64.52. 5 7(b) 64.6 5.2 100 m − ÷ = –1.96 M1 Finding the critical value using N 5.2 64.6, 100       and a z value. m = 63.5808 A1 63.5808 62.7 5.2 100 − ÷ [= 1.694] M1 Standardising using N 5.2 62.7, 100       and a critical value. 1 – Φ(‘1.694’) M1 For area consistent with their values. 0.0451 A1 Accept answers that round to 0.045. 5

More questions on Hypothesis tests