Question

11. The finish times for marathon runners during a race are normally distributed with a mean of 195 minutes and a standard deviation of 25 minutes. a) What is the probability that a runner will complete the marathon within 3 hours? b) Calculate to the nearest minute the tine by which the first 8% runners have completed the marathon c) What proportion of the runners will complete the marathon between 3 hours and 4 hours?

Answer 1

Let X be the random variable denoting the running time of the marathon in minutes.

Thus, X ~ N(195, 25) i.e. (X - 195)/25 ~ N(0,1)

a) The probability that the runner will complete the marathon in 3 hours = P(X < 180) = P[(X - 195)/25 < (180 - 195)/25] = P[(X - 195)/25 < - 0.6] = $\Phi$(-0.6) = 0.2743

[$\Phi$(.) is the cdf of N(0,1)]

b) Let the required time be a mins.

Thus, P(X < a) = 0.08 i.e. P[(X - 195)/25 < (a - 195)/25] = 0.08 i.e. $\Phi$[(a - 195)/25] = 0.08 i.e. (a - 195)/25 = (0.08) = - 1.405 i.e. a = 159.875 mins.

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c) We have P(180 < X < 240) = P[(180 - 195)/25 < (X - 195)/25 < (240 - 195)/25] = P[-0.6 < (X - 195)/25 < 1.8] = $\Phi$(1.8) - $\Phi$(-0.6) = 0.9741 - 0.2743 = 0.6998

Hence, 69.98% of runners finish the marathon between 3 and 4 hours.