Question

(1 point) The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,15]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent. Part a) What is the approximate probability (to 2 decimal places) that the sum of the 95 wait times you observed is between 670 and 796

Answer 1

Answer:

The probability  that the sum of the 95 wait times you observed is between 670 and 796 is 0.8191

Step-by-step explanation:

a) Average sum of 95 wait times =95*(0+15)/2 =712.5

Standard deviation  =(15-0)*(95/12)1/2 =42.2049

Thus, probability  

P(670<X<796

= P((670-712.5)/42.2049<Z<(796-712.5)/42.2049)

=P(-1.0070<Z<1.9784)

=0.9761-0.1570=0.8191