Question

A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of trip times to be normally distributed.
(a) What is the probability that a trip will take at least ½ hour?
(b) If the office opens at 9:00 A.M. and he leaves his house at 8:45 A.M. daily, what percentage of the time is he late for work?
(c) If he leaves the house at 8:35 A.M. and coffee is served at the office from 8:50 A.M. until 9:00 A.M., what is the probability that he misses coffee?
​(d) Find the length of time above which we find the slowest 10​% of trips.
(e) Find the probability that 2 of the next 3 trips will take at least one half
1/2 hour.

Answer 1

Answer:

Step-by-step explanation:

a) Probability-Above 30 min = 5.72% = .0572

b) Probability-Above 15 min =  99.11% = .9911

c) *Probability-Between  1 - 59.49% = .4051

d) 19.136 minutes  z = -1.28