A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 27 minutes, with a standard deviation of 3.7 minutes. Assume the distribution of trip times to be normally distributed. Complete parts (a) through (e) below.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 half1/2 hour.
Accepted Solution
A:
Answer:0.209,0.0885,0.7054,22Step-by-step explanation:Given that a lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 27 minutes, with a standard deviation of 3.7 minutes.a) [tex]P(X\geq 0.30) =P(Z\geq 0.81)\\=0.2090[/tex]b) [tex]P(X\geq 25) =P(Z\geq 1.35)\\=0.0885[/tex]c) [tex]P(X\geq 25) = P(Z\geq -0.54)\\=0.7054[/tex]d) 10th percentile = [tex]-1.28(3.7)+27\\=22.264[/tex]e) Since each trip is independent we have binomial with p= [tex]P(X\geq 90) = 0.0000[/tex]Reqd prob = [tex]3C2 (0) (1) = 0[/tex]