Question

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 22 minutes and a standard deviation of 2 minutes. What is the probability that a randomly selected customer will have to wait between 26 minutes and 27 minutes, to the nearest thousandth?

Answer 1

Let X ~ N(22, 2^2). Convert to z-scores:

z1 = (26−22)/2 = 2.0, z2 = (27−22)/2 = 2.5.

P(26 ≤ X ≤ 27) = Φ(2.5) − Φ(2.0) ≈ 0.99379 − 0.97725 = 0.01654 ≈ 0.017.

So the probability ≈ 0.017 (about 1.7%).