Question

The actual arrival time of the scheduled 10:20 a.m. bus at the alpine and Broadway stop is a uniformly distributed random variable ranging from 10:18 to 10:23. (a) What is the average arrival time of the 10:20 a.m. bus? (b) The standard deviation? (c) What is the probability that the bus is early? (d) What is the probability that the bus arrives between 10:19 and 10:21 a.m.?

Answer 1

It means unoform distribution is between 18 to 23 minutes

a)

here a=   18
b=   23

mean =    (a+b)/2 =    20.5

Average arrval time = 10.20 minutes and 30 seconds

b)

variance =    (b-a)²/12 =    2.083333333
      
std dev =   √ variance =    1.44

c)

x = 20

P(X ≤ x) =    (x-a)/(b-a) =    0.4

d)

x1 =   19
x2=   21

P (    19   ≤ X ≤    21   ) =(x2-x1)/(b-a) =    0.4

Please revert in case of any doubt.

Please upvote. Thanks in advance