jeannie is an experienced business traveler, often traveling back and forth from

Question

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, 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 becuause her total time for catching her plane exceeds one hour?

Explanation / Answer

total available time to jennie=1hr=60min

so test value z=60-mean/std dev=60-46/5=2.8

probability that Jeannie will miss her flight =p(time>60)=p(z>60-46/5)

p(z>2.8)=1-p(z<2.8)=1-0.9974=0.0026

so probability that Jeannie will miss her flight is 0.0026

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