Question

On average, commuters in Phoenix, Arizona, area require m= 40.0 minutes to get to work. Assume that for all commuters the times to get to work are normally distributed with the standard deviation of s= 10 minutes. Joe is an average Phoenix resident and goes to work every day.

What is the probability that on any given day it will take Joe over 45 minutes to get to work?

What is the probability that it will take Joe exactly 40.0 minutes to get to work on any given day?

Joe has just left home to attend a meeting with the CEO in 30 minutes. If the CEO routinely fires employees who are tardy for meetings, what is the probability that Joe will still be employed tomorrow? Joe is so punctual, that he leaves home for work at exactly same time. His work starts at 8:00 AM sharp. Due to traffic jams during morning rush hours Joe historically was tardy for work on 38.2% of days. What time does Joe leave home for work every morning? Briefly explain your train of thoughts in arriving to the answer.

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	40
std deviation   =σ=	10

1)

probability that on any given day it will take Joe over 45 minutes to get to work :

probability =P(X>45)=P(Z>(45-40)/10)=P(Z>0.5)=1-P(Z<0.5)=1-0.6915=0.3085

2)

probability that it will take Joe exactly 40.0 minutes to get to work on any given day =0.000

(Since point probability on a continuous distribution is 0)

3)

probability that Joe will still be employed tomorrow :

probability =P(X<30)=(Z<(30-40)/10)=P(Z<-1)=0.1587

4)

for top 38.2% values fall at (100-38.2=61.8 percentile) for which critical z =0.30

therefore since 8:00 am is at z score =0.3 , therefore

time from his ;leaving home to 8:00 AM=40+0.3*10 =43 minutes

from above it appear that he leave Home on 7:17 AM every morning.