Question

1. A lawyer travels daily from his home (in the suburbs) to his office in the city center. On average, the journey takes 24 minutes with a standard deviation of 3.8 minutes. Assume that the distribution of transfer times are normally distributed.

1. if the office opens at 9:00 a.m. and he leaves at 8:45 a.m. What percentage of the time are you late to work daily?
2. If you leave home at 8:35 a.m. and coffee is served in the office between 8:50 and 9 a.m., what is the probability that you will be served coffee?
3. Find the period above which is the 15% of the slowest transfers.
4. Suppose a sample of 5 trips taken by the lawyer is taken. If the attorney always leaves your home at 8:30 a.m., find the probability that on at least 2 trips he arrived between 9:05 a.m. and 9:20 a.m.

Answer 1

Mean, = 24 minutes

Standard deviation, = 3.8 minutes

Let X(in minutes) denote the transfer time

(a) You are late to work is X > 15

Thus, the required percentage of time = P(X > 15)

= P{Z > (15 - 24)/3.8}

= P(Z > -2.368)

= 0.9911 = 99.11%

(b) You will be served coffee if X lies in the range 15 and 25

The probability that you will be served coffee

= P(15 ≤ X ≤ 25)

= P{(15 - 24)/3.8 ≤ Z ≤ (25 - 24)/3.8}

= P(-2.368 ≤ Z ≤ 0.263)

= 0.5948

(c) Corresponding to the 15% lowest transfers, the z value = 1.0345

= 27.93

The period above 27.93 minutes has the 15% of the slowest transfers

(d) Probability of arriving between 9:05 and 9:20 when you leave at 8:30 i.e P(35 ≤ X ≤ 50)

.= P{(35 - 24)/3.8 ≤ Z ≤ (50 - 24)/3.8}

= P(2.895 ≤ Z ≤ 6.842)

= 0.0019

The probability that on at least 2 trips out of 5, he arrived between 9:05 am and 9:20 am

=

= 0.00004