A snack company makes packages of grapes and honeydew slices. The weight of an individual package of grapes, G, is approximately Normally distributed with a mean of 3.25 ounces and a standard deviation of 0.91 ounces. The weight of an individual package of honeydew slices, H, is approximately Normally distributed with a mean of 4.51 ounces and a standard deviation of 2.02 ounces. Assume G and H are independent random variables. Let D = G – H. What is the probability that a randomly selected package of grapes weighs more than a randomly selected package of honeydew slices?
Because G and H are independent, the difference D = G − H is Normal with mean 3.25 − 4.51 = −1.26 ounces and standard deviation √(0.91² + 2.02²) ≈ 2.22 ounces. The probability that grapes weigh more is P(D > 0), which corresponds to a z-score of (0 − (−1.26))/2.22 ≈ 0.57. The upper-tail probability is about 0.284, so the correct choice is 0.284.
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