Ask Question
24 October, 06:22

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.

+1
Answers (1)
  1. 24 October, 07:34
    0
    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 0.42) = 0.0630 + (1 - 0.6628) = 0.4002

    The probability is approximately 0.40.
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers