The distribution of the number of items washed in a standard load of laundry is skewed left with a mean of 41 items and a standard deviation of 7 point 7 items. What is the probability that 50 randomly selected loads of laundry have a mean of more than 39 point 5 items?

For 50 loads, the sampling distribution of the mean is approximately normal, with mean 41 and standard error 7.7/√50 ≈ 1.09. The z-score for 39.5 items is (39.5 − 41)/1.09 ≈ −1.38. Thus, the probability of a sample mean greater than 39.5 is about 0.9162, so the correct choice is 0.9162.

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