Question

A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 23 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. Click here to view page 1 of the standard normal distribution table. view pag standard normal distribution table (a) What is the probability that a trip will take at least shour? (Round to four decimal places as needed.)

Answer 1

Define the standard random variable Z as $Z={X-23\over 3.7}$

Then using normal table we have

$P(X\geq 30) =P\left (Z\geq {30-23\over 3.7}\right) =P(Z\geq 1.89)=0.0294$

So required probability is 0.0294.