Question

The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with a standard deviation of 3.6 minutes. Assume wait time is normally distributed.
When a patron arrives at the restaurant for dinner, find the probability that the patron will have to wait less than 18 minutes or more than 25 minutes.

Answer 1

Answer:

0.40

Step-by-step explanation:

Find the z-scores.

z = (x − μ) / σ

z₁ = (18 − 23.5) / 3.6

z₁ = -1.53

z₂ = (25 − 23.5) / 3.6

z₂ = 0.42

Use a table or calculator to find the probability.

P(Z < -1.53) + P(Z > 0.42) = 0.0630 + (1 − 0.6628) = 0.4002

The probability is approximately 0.40.