At a local restaurant, the amount of time that customers have to wait for their food is normally distributed w

Question

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 28 minutes and a standard deviation of 5 minutes. What is the probability that a randomly selected customer will have to wait less than 33 minutes, to the nearest thousandth?

GPT-5 mini · Answer

We standardize: Z = (X − 28)/5. For X = 33, z = (33 − 28)/5 = 1.0.  
P(X < 33) = P(Z < 1.0) ≈ 0.8413447 ≈ 0.841 (to the nearest thousandth).

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 28 minutes and a standard deviation of 5 minutes. What is the probability that a randomly selected customer will have to wait less than 33 minutes, to the nearest thousandth?

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