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 w

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.

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Alexandra 3 weeks 2021-11-20T20:40:46+00:00 1 Answer 0 views 0

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    2021-11-20T20:42:37+00:00

    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.

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