Jeannie is an experienced business traveler, often traveling back and forth from San Francisco to the East Coast several times per month. To catch her flights from San Francisco she leaves her office one hour before her flight leaves. Her travel time from her office to the departing gate at the San Francisco airport, including the time to park and go through security screening, is normally distributed with a mean of 46 minutes and a standard deviation of 5 minutes. What is the probability that Jeannie will miss her flight because her total time for catching her plane exceeds sixty minutes? Round your final answer to 4 decimal places.

Given : Jeannie's travel time from her office to the departing gate at the San Francisco airport, including the time to park and go through security screening, is normally distributed with

Mean: [tex]\mu=46\text{ minutes }[/tex]

Standard deviation : [tex]\sigma=5\text{ minutes }[/tex]

Let X be a random variable that represents the her travel time from her office to the departing gate at the San Francisco airport, including the time to park and go through security screening.

Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 60

[tex]z=\dfrac{60-46}{5}\approx2.8[/tex]

Now by using standard normal distribution table , the probability that Jeannie will miss her flight because her total time for catching her plane exceeds sixty minutes :-

[tex]P(x>60)=P(z>2.8)=1-P(z\leq2.8)\\\\=1-0.9974449\approx0.0026[/tex]

Hence, the probability that Jeannie will miss her flight because her total time for catching her plane exceeds sixty minutes = 0.0026