Question

Suppose the mean and the standard deviation of the waiting times of passengers at the bus station near the Cross Harbour Tunnel are 11.5 minutes and 2.2 minutes, respectively.

(a) Assume the waiting times are normally distributed. How likely a randomly selected passenger waits less than 15 minutes at the bus station?

(b) Assume the waiting times are normally distributed. 82% of the passengers at the bus station are expected to wait more than k minutes. Find k.

(c) For a random sample of 36 passengers, find the probability that their mean waiting time will be less than 11 minutes. Does your calculation require assuming the waiting times to be normally distributed? Explain.

Answer 1

a) Let the waiting time be denoted by X.

So, X ~ N(11.5,2.22)

So, P( X <15) =

P(Z < (15-11.5)/2.2)

= P(Z < 1.59)

= 0.944.

b) The waiting time k for which 82% of the people wait.

So, P(X < k ) = 0.82

= P(z < (k-11.5)/2.2) = 0.82

= (k-11.5)/2.2 = 0.915

= k = 13.513 minutes.

So, we can say that 82% people have to wait for more than 13.513 minutes.

c) Probability that the mean waiting time for 36 people is less than 11 minutes.

Let X be the mean waiting time for the 36 passengers.

P(X < 11)

P(Z < (11-11.5)/2.2)

= P(Z < -0.2273)

= 0.410.

So, the probability that the mean waiting time will be less than 11 is 0.41.

Our calculation require the mean waiting time to be normally distributed with mean of 11.5 minutes and standard deviation of 2.2 minutes. After this we calculate the value of the strandard normal variable which has a mean of 0 and variance of 1. Using the value of the standard normal variable, we calculate the value of the probability.

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