Consider the pdf for total waiting time Y? ? or two buses introduced in Exercise 8. a.? ?Compute and sketch the cdf of ?Y?. [?Hint?: Consider separately 0?y<5 and 5?y?10 in computing ?F?(?y?). A graph of the pdf should be helpful.] b.? ?Obtain an expression for the (100?p?)th percentile. [?Hint?: Consider separately 0 c.? ?Compute ?E?(?Y ?) and ?V?(?Y?). How do these compare with the expected waiting time and variance for a single bus when the time is uniformly distributed on [0, 5]? Reference exercise 8 In commuting to work, a professor must first get on a bus near her house and then transfer to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with A = 0 and B = 5 , then it can be shown that the total waiting time Y? ?has the pdf a.?? ketch a graph of the pdf of ?Y?. b.?? erify that . c.?? hat is the probability that total waiting time is at most 3 min? d.?? hat is the probability that total waiting time is at most 8 min? e.?? hat is the probability that total waiting time is between 3 and 8 min? f.? ?What is the probability that total waiting time is either less than 2 min or more than 6 min?

Answer Step 1 of 3 a)For 0y<5 y 1 1 2 y f(y)= 25y dy= 50 [ ] 0 0 1 2 = 50y Step 2 of 6 For 5y<10 5 y f(y)= 1 y dy+ 2 - 1 y dy 25 5 25 0 5 2 y 1 y 2 5 2y y = 50 [ ] 0+ [ 5 50] 5 2y y2 = 5 - 50 -1 Step 3 of 6 The cdf of y is f(y)=0, y<0 1 2 = 50y , 0y<5 2y y2 = 5 - 50 -1, 5y<10 =1, y>0 Step 4 of 6 Step 5 of 6 b)The 100% percentile is 2y y 2 1= - -1 5 50 Rearrange the equation then we get y -20y+100=0 Solve for y We get y=10 Step 6 of 6 5 10 1 2 2 1 2 c)E(y)= yf(y)dy = 25y dy+ ( 5 y- 25 y )dy 0 5 1 3 5 2y 2 y3 10 = 75 [ ] 0+ [ 10 75] 5 =5 2 5 10 E(y )= y f(y)dy = 25 y dy+ ( 5 y -25y )dy 0 5 5 3 4 10 = 1 [ ]4 0 + [ 2y y ] 75 10 75 5 =29.167 V(y)=E(y )-E(y) 2 =29.167-25 =4.167 Problem 2 Reference exercise 8 Step 1 of 7 a) The probability distribution is x 0 1 2 3 4 5 6 7 8 9 10 f(x) 0 0.04 0.08 0.12 0.16 0.2 0.16 0.12 0.08 0.04 0 Step 2 of 7 Step 3 of 7 5 10 b) f(y)dy = 1 y dy+ ( 2- 1 y )dy 0 25 5 5 25 5 2 3 10 = 1 [ ]2 +[ y y ] 50 0 5 50 5 1 1 = 2((4-2)-(2- )2½+½=1 Step 4 of 7 c) The probability that total waiting time is at most 3 min =p(x3)=0+0.04+0.08+0.12=0.24