A lawyer travels daily from his home to his oce in town. On average, the one-way trip takes 20 minutes, with a standard deviation of 3 minutes. What is the probability that the lawyer's one way trip time is between 18 and 27 minutes. Assume the distribution of trip times to be normally distributed.
Let "X" be a random variable representing the lawyer's one way trip then, "\\mu=20 minutes" and "\\sigma=3minutes". We need to find the probability that the lawyer's one way trip time is between 18 and 27 minutes given as,
"p(18\\lt X\\lt27)" and we standardize in order to use the standard normal tables to find the required probability.
Now,
"p(18\\lt X\\lt27)=p((18-\\mu)\/\\sigma \\lt Z\\lt(27-\\mu)\/\\sigma)=p((18-20)\/3\\lt Z\\lt (27-20)\/3)=p(0.67\\lt Z\\lt 2.33)"
We can obtain this probability from standard normal tables as,
"p(0.67\\lt Z\\lt 2.33)=\\phi(2.33)-\\phi(0.67)=0.9901-0.7486=0.2415"
Therefore, the probability that the lawyer's one way trip time is between 18 and 27 minutes is 0.2415.
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