Question

A) The Transportation Department determines that in a certain metropolitan area the average one-way commute takes 1 hour 10 minutes (70 minutes). The distribution of commute times is normally distributed with a standard deviation of 8 minutes. If you select a person randomly, what is the probability that his or her commute takes less than
55 minutes?

B) Using this same distribution; If you select a person randomly, what is the probability that you find that
his or her commute takes more than 1 hour 25 minutes?

C) Using this same distribution; If you select a person randomly, what is the probability that you find that
his or her commute takes between 1 hour 5 minutes and 1 hour and 25 minutes?

Answer 1

mean = 70.00 = (60+10) mins
sd = 8

a)
P( X<55) = ?
  
I know that,
z = (X-mean)/(sd)  
z = (55-70)/8)
z = -1.8750

Hence,  
P( X<55) = P(Z<-1.875)
P( X<55) = NORMSDIST(-1.875)
P( X<55) = 0.0304

b)
more than 1 hr 25 mins is equivalent to more than 60+25 = 85 mins
P( X>85) = 1 - P(X<85) = ?

I know that, z = (X-mean)/(sd)  
z1 = (85-70)/8) = 1.8750

Hence,
P(X>85) = 1- P(Z<1.875)
P(X>85) = 1 - NORMSDIST(1.875)
P(X>85) = 0.030396

c)
P(65 < X < 85)  
= P(X<85) - P(X<65)  
  
I know that, z = (X-mean)/(sd)  
z1 = (65-70)/8) =   -0.6250
z2 = (85-70)/8) =   1.8750
  
hence,  
P(65 < X < 85) = P(Z<1.875) - P(Z<-0.625)
P(65 < X < 85) = NORMSDIST(1.875) - NORMSDIST(-0.625)
P(65 < X < 85) = 0.7036