Question

The probability density function of the time a customer arrives at a terminal (in minutes after 8:00 A.M.) is rx) = 0.5 e-x/2 for x > 0, Determine the probability that (a) The customer arrives by 11:00 A.M. (Round your answer to one decimal place (e.g. 98.7) (b) The customer arrives between 8:16 A.M. and 8:31 A.M. (Round your answer to four decimal places (e.g. 98.7654)) (c) Determine the time (in hours A.M. as decimal) at which the probability of an earlier arrival is 0.50. (Round your answer to two decimal places (e.g. 98.76)) (d) Determine the cumulative distribution function and use the cumulative distribution function to determine the probability that the customer arrives between 8:16 A.M. and 8:31 A.M (Round your answer to four decimal places (e.g. 98.7654)) (e) Determine the mean and (f) standard deviation of the number of minutes until the customer arrives. (Round your answers to one decimal place (e.g. 98.7))

Answer 1

a) by 11A.M means 3 hours = 3(60) = 180 min integral^180_0 f(x) dx = integral^180_0 0.5 e^-x/2 = 1 b) B = 16 to 8: 31 there is 15 min integral^31_10 f(x) dx = integral^31_16 0.5 e^-x/2 = [-e^-x/2]^31_16 = 0.00034 c) Here integral^a_0 f(x) dx = 0.5 [-e^-x/2]^a_0 = 0.5 + e^-a/2 = 0.5 e^-a/2 = 0.5 a/2 = ln(0.5) = 0.69 a = 1.38 here a = 1 hr + 38 min = 8: 00 + 1 hr 38m = 9: 38 hours d) The cumulative density function f(x) = integral^x_0 0.5 e^-x/2 dx = -e^-x/2 P(8: 16 lessthanorequalto x lessthanorequalto 8: 31) = p(x lessthanorequalto 8: 31) - P(x lessthanorequalto 8: 16) e) Mean = integral^infinity_0 x f(x) dx = integral^infinity_0 x 0.5 e^-x/2 dx = 2 f) E(x^2) = integral^infinity_0 x^2 f(x)dx = integral^infinity_0 x^2 0.5 e^-x/2 dx variance = E(x^2) - [E(x)]^2 = 8 - (2)^2 = 8 - 4 = 4 S.D = squareroot n = 2