My bus is scheduled to depart at noon However in reality the

My bus is scheduled to depart at noon. However, in reality the departure time varies randomly, with average departure time 12 o\'clock noon and a standard deviation of 6 minutes. Assume the departure time is normally distributed. If I get to the bus stop 5 minutes past noon, what is the probability that the bus has not yet departed?

Solution

Let noon be u = 0.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    5      
u = mean =    0      
          
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) / s =    0.833333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.833333333   ) =    0.202328381 [ANSWER]