Question

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 half
1/2 hour.

Answer 1

0.209,0.0885,0.7054,22

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)
`P(X\geq 0.30) =P(Z\geq 0.81)\\=0.2090`

b)
`P(X\geq 25) =P(Z\geq 1.35)\\=0.0885`

c)
`P(X\geq 25) = P(Z\geq -0.54)\\=0.7054`

d) 10th percentile =
`-1.28(3.7)+27\\=22.264`

e) Since each trip is independent we have binomial with p

=
`P(X\geq 90) = 0.0000`

Reqd prob =
`3C2 (0) (1) = 0`