Question

If a person takes the bus 30 times a month commuting between his dorm and the Dining Hall. It takes the bus 10 minutes to run one loop. The waiting time, in minutes, for a bus to arrive is uniformly distributed on the interval [0, 10]. Suppose that waiting times on different occasions are independent. What is the standard deviation of the mean waiting time in minutes of a month?

Round your answer to three decimal digits. What is the probability of the average waiting time of a month is greater than 6 minutes?

Answer 1

Let X be the waiting times on different occasions. Then X ~ Unif(0, 10)

E(X) = (0 + 10) / 2 = 5

Var(X) = (10 - 0)² / 12 = 8.33

By Central limit theorem, the sampling distribution of mean waiting time in minutes of a month follows Normal distribution with $\mu$ = 5 minutes and variance, $\sigma^2$ =  8.33 / 30 = 0.2777

Standard deviation of the mean waiting time in minutes of a month = = 0.527 minutes

V0.2777

Probability of the average waiting time of a month is greater than 6 minutes = P($\bar{X}$ > 6)

= P[Z > (6 - 5)/0.527]

= P[Z > 1.90]

= 0.0287