Every working day, John conies to the bus stop exactly at 7am and takes the firs

Question

Every working day, John conies to the bus stop exactly at 7am and takes the first bus that arrives. The waiting time (T) for the bus is an exponential random variable with mean 20 minutes, i.e. T Exp{A = 1/20). Also, every working day, and independently, Mary comes to the same bus stop at a random time, uniformly distributed between 7 and 7:30am. That is, Mary's arrival time X ~ Uni f(0,30). What is the probability that tomorrow John will wait for more than 30 minutes': What is the probability that Mary arrives before 7:20am?

Explanation / Answer

a)

mean = u =    20      

The right tailed area in an exponential distribution is          
          
P(x>x1) = e^(-x1/u)          
          
As          
          
x1 = critical value =    30      
          
          
Then          
          
P =    0.22313016   [ANSWER]

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b)

Let x = 0 be 7am, and x = 30 be 7:30am.

Hence, we want P(x<20).

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    30      
          
Note that P(x<c) = P(a<x<c) = (c-a)/(b-a). Thus, as          
          
c = critical value =    20      
          
Then          
          
P(x<c) =    0.666666667   [ANSWER]  

  

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