2. A lawyer commutes daily from her suburban home to her midtown office. The ave

Question

2. A lawyer commutes daily from her suburban home to her midtown office. The average time for a one-way trip is 23 minutes, with a standard deviation of 5 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? (1 pt) (b) If the office opens at 9am and the lawyer leaves her house at 8:45am daily, what percentage of the time is she late for work? (1 pt) (c) If she leaves the house at 8:35am and coffee is served at the office from 8:50am until 9am, what is the probability that she arrives while coffee is being served? (1 pt) (d) Find the length of time above which we find the slowest 15% of the trips. (1 pt) (e) Find the probability that exactly 2 of the next 3 trips will take at least ½ hour. Assume the trip times are independent. (1 pt)

Explanation / Answer

From information given, Xi=30 minutes, Xbar=23 minutes, s=5 minutes. Substitute the values in following equation to compute the z score. The area corresponding to z score gives the required probbaility.

z=(Xi-Xbar)/s

a) z=(30-23)/5=1.4, the probbaility is 0.4192.

b) Xi=15, Xbar=23, s=5

z=(15-23)/5=-1.6 . The percentage is area corresponding to z=-1.6 multiplied with 100. She is arond 5.48% of time late for work.

c) Xi=25, z=(25-23)/5=0.4 The probability that she reaches office while coffee is served is 0.1554.

d) z=0.15, Xbar=23, s=5, find Xi

0.15=(Xi-23)/5; Xi=23.75 minutes [ans]

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