A late delivery

Question

Delivery times for a courier are normally distributed with a mean of 30 minutes and a standard deviation of 6 minutes. What is the probability that a delivery takes longer than 33 minutes?

Accepted Answer

Convert the time to a z-score, the number of standard deviations above the mean: z = \frac{x - \mu}{\sigma}. For x = 33, mean \mu = 30, spread \sigma = 6, z = \frac{33 - 30}{6} = 0.5. So 33 minutes is only half a standard deviation above average, a fairly ordinary value. The standard normal table gives the area to the left of z = 0.5 as about 0.6915, so 69.15% of deliveries beat 33 minutes. The chance of running longer is the remaining tail: P(X 33) = 1 - 0.6915 = 0.3085. About 31% of deliveries take more than 33 minutes. Because the cutoff is close to the mean, the tail is large, nowhere nea…

A late delivery

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