The time needed to broil cauliflower is approximately Normally distributed with a mean of 6 minutes and a standard deviation of 1.9 minutes. If a random sample of 100 batches of cauliflower is selected, what is the probability the mean broiling time is more than 6.5 minutes?

For a sample of 100 batches, the standard deviation of the sample mean is 1.9/√100 = 0.19 minutes. A mean of 6.5 minutes is (6.5 − 6)/0.19 ≈ 2.63 standard deviations above the population mean. The probability of being above that z-score is about 0.0042, so the correct choice is 0.0042.

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