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 7.7 items. What is the probability that 50 randomly selected loads of laundry have a mean of more than 39.5 items?

For 50 loads, the sampling distribution of the mean is approximately normal by the Central Limit Theorem, with mean 41 and standard error 7.7/√50 ≈ 1.089. The standardized value for 39.5 is z = (39.5 − 41)/1.089 ≈ −1.38. Thus, P(mean > 39.5) = P(Z > −1.38) ≈ 0.9158, so the answer is 0.9158.

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