Question

A particular variety of watermelon weighs on average 26.8 pounds with a standard deviation of 1.79 pounds. Consider the sample mean weight of 91 watermelons of this variety. Assume the individual watermelon weights are independent. What is the expected value of the sample mean weight?

Answer 1

The expected value of the sample mean weight for the watermelons is 26.8 pounds, which is the same as the population mean.

Step-by-step explanation:

The expected value of the sample mean weight of watermelons, when given a population with a mean weight of 26.8 pounds and a standard deviation of 1.79 pounds, is simply the population mean weight. This is due to the fact that the expected value of the sample mean is equal to the population mean according to the properties of expected values. Therefore, regardless of the sample size, in this case with 91 watermelons, the expected value of the sample mean weight remains 26.8 pounds.