1. The time taken for a paperboy to deliver his papers is normally distributed with mean 52 minutes and standard deviation 6.5 minutes. Find the probability on any given day the paperboy takes: a) longer than 1 hour to deliver (3 marks) b) less than 45 minutes (3 marks) c) between 48 and 56 minutes (3 marks) (Marks available: 9) Answer Answer outline and marking scheme for question: 1 1. X ~ N(52, 6.52) a) P(X > 60) = P(Z > 60 - 52) add bell on ans sheet 3 A1 6.5 = P(Z > 1.23) = 1 - F(1.23) A1 = 1 - 0.8907 = 0.1093 A1 (3 marks) b) P(X < 45) = P(Z < 45 - 52) add bell on ans sheet 3 A1 6.5 = P(Z < -1.08) A1= 1 - F(1.08) = 1 - 0.8599 = 0.1401 A1 (3 marks) c) P(48 < X < 56) = P(48 - 52 < Z < 56 - 52) A1 6.5 6.5 = P(-0.62 < Z < 0.62) bell on ans sheet 3 = F(0.62) - [1-F(0.62)] A1 = 2F(0.62) - 1 = 2 ' 0.7324 - 1 = 0.4648 A1 (3 marks) (Marks available: 9) 2. An Olympic high jumper jumps distances which are normally distributed with mean, μ = 2.45m and variance 0.49m. a) find the probability the jumper manages a height over 2.35m(2 marks) b) What distance is exceeded by 5% of his jumps? (4 marks) (Marks available: 6) Answer Answer outline and marking scheme for question: 2 X ~ N(2.45,0.49) we are given the variance so s = v0.49= 0.7 a) P(X > 2.35) = P(Z > 2.35 - 2.45) add bell on sheet 3 0.7 = P(Z > -0.14) = F(0.14) = 0.5557 (2 marks) b) we need the distance d where P(X > d) = 0.05 P(Z > d - 2.45) = 0.05 sheet 3 for bell 0.7 1 - F(d - 2.45) = 0.05 hence F(d - 2.45) = 0.95 0.7 0.7 so d - 2.45 = 1.645 hence d = 3.6m ( a very high jump!) 0.7 (4 marks) (Marks available: 6)