Question

7. Each morning Engineer drives from his suburban home to his midtown office and each evening the Engineer returns home. The mean travel for one-way trip is 40 minutes; with a standard deviation of 5 minutes. Assume the distribution of Trip times to be normally distributed What is the probability that a single trip (from home to work) will take longer than 30 minutes? a. b. Each week the Engineer makes this trip 10 times (5 times to work and 5 times back home again). Assume that come mute time is independent trip-to-trip What is the probability that in the Engineer is in her car more than 8 hours (480 minutes) in a given week? Hint: Let T = X1 +X2+. . . . +X10 Where Xi ~ iid Normal ( μ = 40, σ = 5)

Answer 1

Let x be a r.v denotes a single trip of an engineer given x sim N (40, 25) p(x > 30) therefore p(x - 40/5 > 30 - 40/5) = p (z > -2) where z sim N (0, 1) = 1 - p (z less than or equal to -2) = 1 - 1(-2) = 1 - (1 - (2)) = (2) = 0.977 b) let x_1, x_2, ----, x_10 denote the 10 trips therefore let t denotes total time no. of trips t = x_1 + x_2 + ---+x_10 \r\n \ now x sim N (40, 25) therefore t = sigma_i = 1^10 x_i sim N (400, 250) since x_i s are independent now p (t > 480) = p (t - 400/ squareroot 250 > 480 - 400/ squareroot 250) = p (z > 80/ squareroot 250) (z sim N(0, 1)) = 1 - $ (80/ squareroot 250) = 0.0000002.