In: Statistics and Probability
Suppose that the waiting time for a bus is normally distributed with a mean of 5 minutes and a standard deviation of 2.5 minutes.
(a) Find the probability that the waiting time for the bus is between 3 minutes and 7 minutes.
(b) If you randomly choose 4 passengers, find the sampling distribution of their average waiting time. Please indicate shape, mean and standard deviation.
(c) For the randomly selected 4 passengers, find the probability that their average waiting time is above 6 minutes. Let X¯ be the average waiting time of randomly selected 4 passengers.
Solution :
Given that ,
mean = = 5
standard deviation = = 2.5
P(3< x < 7) = P[(3-5) /2.5 < (x - ) / < (7-5) /2.5 )]
= P( -0.8< Z <0.8 )
= P(Z < 0.8) - P(Z < -0.8)
Using z table
= 0.7881-0.2119
probability= 0.5762
(B)
n = 4
mean = = 5
standard deviation = = / n = 2.5/ 4 = 1.25
P( >6 ) = 1 - P( <6 )
= 1 - P[( - ) / < (6 -5) /1.25 ]
= 1 - P(z <0.8 )
Using z table
= 1 - 0.7881
= 0.2119
probability=0.2119