4 pts The wait time after a scheduled arrival time in minute

(4 pts) The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent.

Find the probability (to 2 decimal places) that 92 or more of the 95 wait times exceed 1 minute. Please carry answers to at least 6 decimal places in intermediate steps.

Solution

The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent.

Let X: Number of times wait time exceeds 1 minute

So here X will follow binomial with n = 95 and p = 11/12.

Probability mass function of binomial distribution with parameters n and p is given as follows:

P(X = x) = ⁿC_x p^x (1-p)^n-x

The probability that 92 or more of the 95 wait times exceed 1 minute = P( X >= 92)

                                                                                                    = P(X=92) + P(X=93) + P(X=94) + P(X=95)

                                                                                                   = ⁹⁵C₉₂ (11/12)⁹² (1-(11/12))^95-92 +

                                                                                                     ⁹⁵C₉₃ (11/12)⁹³ (1-(11/12))^95-93 +

                                                                                                     ⁹⁵C₉₄ (11/12)⁹⁴ (1-(11/12))^95-94 +

                                                                                                      ⁹⁵C₉₅ (11/12)⁹⁵ (1-(11/12))^95-95

                                                                                               = 0.04 ... (rounded to 2 decimal places)