A lawyer commutes daily from his suburban home to his midtown office. The averag

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 1/2 hour?

b.Find the probability that at most of the next 10 trips will take at least 1/2 hour.

c.If the office opens at 9:00 A.M. and the lawyer leaves his house at 8:45 A.M. daily, what percentage of the time is he late for work?

d.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?

e.Find the length of time below which we find the slowest 15% of the trips.

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Explanation / Answer

mean = 24 , s = 3.8
a) P(X > 30)
= P(Z > (30 - 24)/3.8)
= P(Z > 1.58)
= 0.05705343

b)
P(X > 30)
= P(Z > (30 - 24)/(3.8/sqrt(10))
= P(Z > 4.99)
= 0

c) find P(X > 15)
= P( Z > (15 - 24) / 3.8)
= P( Z > -2.37)
= 0.991106

d) find P(X > 25)
= P( Z > (25 - 24) / 3.8)
= P( Z > 0.26)
= 0.3974319

e) z = (x - mean) / s
1.04 = ( x - 24) / 3.8
x = 27.952

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