Asked by b
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 4 minutes. What is the probability that a randomly selected customer will have to wait between 18 minutes and 28 minutes, to the nearest thousandth?
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GPT-5 mini
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Let X ~ N(28, 4^2). Convert to Z:
z1 = (18 - 28)/4 = -2.5, z2 = (28 - 28)/4 = 0.
P(18 ≤ X ≤ 28) = P(-2.5 ≤ Z ≤ 0) = Φ(0) − Φ(−2.5) = 0.5 − (1 − Φ(2.5)) = Φ(2.5) − 0.5 ≈ 0.9938 − 0.5 = 0.4938.
To the nearest thousandth: 0.494.
z1 = (18 - 28)/4 = -2.5, z2 = (28 - 28)/4 = 0.
P(18 ≤ X ≤ 28) = P(-2.5 ≤ Z ≤ 0) = Φ(0) − Φ(−2.5) = 0.5 − (1 − Φ(2.5)) = Φ(2.5) − 0.5 ≈ 0.9938 − 0.5 = 0.4938.
To the nearest thousandth: 0.494.
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