The wait time (after a scheduled arrival time) in minutes for a train to arrive

Question

The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 100 trains to arrive. Assume wait times are independent.

Part c) Find the probability (to 2 decimal places) that 97 or more of the 100 wait times exceed 11 minute. Please carry answers to at least 6 decimal places in intermediate steps.

Part d) Use the Normal approximation to the Binomial distribution (with continuity correction) to find the probability (to 2 decimal places) that 56 or more of the 100 wait times recorded exceed 55 minutes.

Explanation / Answer

c.

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    12      
          
          
Note that P(x>c) = P(c<x<b) = (b-c)/(b-a). Thus, as          
          
c = critical value =    1      
          
Then          
          
P(x>c) =    0.916666667  

We first get the z score for the critical value:          
          
x = critical value =    96.5      
u = mean = np =    91.6666667      
          
s = standard deviation = sqrt(np(1-p)) =    2.763853987      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    1.74876579      
          
Thus, the right tailed area is          
          
P(z >   1.74876579   ) =    0.040165756 [ANSWER]
  

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D.

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    12      
          
          
Note that P(x>c) = P(c<x<b) = (b-c)/(b-a). Thus, as          
          
c = critical value =    5      
          
Then          
          
P(x>c) =    0.583333333      

We first get the z score for the critical value:          
          
x = critical value =    55.5      
u = mean = np =    58.3333333      
          
s = standard deviation = sqrt(np(1-p)) =    4.930066486      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -0.574704886      
          
Thus, the right tailed area is          
          
P(z >   -0.574704886   ) =    0.717254549 [ANSWER]

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