Final answer:
The random variable X represents the number of seniors in a group of 60 who participated in after-school sports all four years and follows a binomial distribution. The expected value is 4.8 students. It would be surprising to have none, and a numerical comparison of binomial probabilities for four or five students would clarify which is more likely.
Step-by-step explanation:
Approximately 8 percent of students at a local high school participate in after-school sports all four years of high school. The random variable X is defined as the number of seniors out of a randomly chosen group of 60 who participated in after-school sports all four years of high school. The values that X may take on are 0, 1, 2, ..., 60, as there can be anywhere from none to all students who participated in sports all four years.
The distribution of X follows a binomial distribution with parameters n = 60 and p = 0.08, which is notated as X ~ B(60, 0.08). The expected number (expected value) of seniors who have participated in after-school sports all four years is calculated as the product of the total number of trials (n) and the probability of success (p), which is E(X) = np = 60 * 0.08 = 4.8. This means, on average, we expect about 5 seniors to have participated in sports all four years.
In a numerical context, having none of the seniors participating in after-school sports all four years would be surprising, given the expected number is 4.8. To justify this numerically, we would calculate the probability for X=0 using the binomial probability formula, which would likely be a small number, indicating a low probability event.
When comparing whether it is more likely that four or five of the seniors participated in after-school sports all four years, we would calculate the individual binomial probabilities for X=4 and X=5. The larger of the two probabilities will show which event is more likely numerically. Without calculating these explicitly, we cannot determine which is more likely, but since the expected value is near five, both probabilities could be reasonably close.