(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

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

answer

point o.

step-by-step explanation: