An engineer commutes daily from his suburban home to his midtown office and need
ID: 3060367 • Letter: A
Question
An engineer commutes daily from his suburban home to his midtown office and needs to arrive at 9 a.m. A normal, one-way trip (including walking to and from the car) is 25 minutes, with a standard deviation of 6 minutes. A shortcut can be taken, and the time savings (reduced from the total trip) has an average of 7 minutes and a standard deviation of 5 minutes. Both travel time and time savings are normally distributed and can be assumed uncorrelated.
What is the probability the trip takes more than 30 minutes if the engineer takes the shortcut?
What is the probability the trip takes more than 30 minutes if the engineer does not take the shortcut?
If the engineer takes the shortcut 30% of the time, how often does the trip take more than 30 minutes?
At what time should the engineer leave in order to have a 97% probability of arriving on time, if he doesn't plan to take the shortcut?
Explanation / Answer
1)
probability the trip takes more than 30 minutes if the engineer takes the shortcut:
2)
probability the trip takes more than 30 minutes if the engineer does not take the shortcut:
3)
If the engineer takes the shortcut 30% of the time, how often does the trip take more than 30 minutes
=P(normal trip and longer than 30 minutes+shortcut and longer than 30 minutes)
=0.7*0.2023+0.3*0=0.1416
4)
at 97th percentile ; z =1.88
threfore corresponding time =25+6*1.88 =~36.3 minutes
therefore he must leave at 8:23 am
for normal distribution z score =(X-)/ here mean= = 7 std deviation == 5.0000