Question

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 95 trains to arrive. Assume wait times are independent

Use the Normal approximation to the Binomial distribution (with continuity correction) to find the probability (to 2 decimal places) that 56 or more of the 95 wait times recorded exceed 5minutes

Answer 1

P(X > 5)

= (12 - 5)/(12 - 0)

= 7/12 = 0.583

n = 95

p = 0.583

$\mu$ = np = 95 * 0.583 = 55.385

$\sigma$ = sqrt(np(1 - p))

    = sqrt(95 * 0.583 * (1 - 0.583))

    = 4.8058

P(X > 56)

= P((X - $\mu$)/$\sigma$> (55.5 - $\mu$)/$\sigma$)

= P(Z > (55.5 - 55.385)/4.8058)

= P(Z > 0.02)

= 1 - P(Z < 0.02)

= 1 - 0.5080

= 0.4920

= 0.49