Question

suppose that about 85% of graduating students attend their graduation. a group of 24 graduating students is randomly chosen. part (a) in words, define the random variable x the number of graduating students that are not attending graduation the number of graduating students that attend graduation the number of graduates the number of people attending graduation part (b) list the values that x may take on x = 1, 2, 3, 83, 84, 85 x = 0, 1, 2, 22, 23, 24 x = 1, 2, 3, 22, 23, 24 x = 0, 1, 2, 83, 84, 85 part (c) give the distribution of x x~()

Explanation:

Part (a)

A random variable \(X\) represents a numerical outcome of an experiment. Here, the experiment is selecting a group of graduating students and observing the number of those who attend graduation. Since the question is about the number of graduating students (from the chosen group of 24) that attend graduation, we can eliminate the other options.

• The first option ("the number of graduating students that are not attending graduation") is incorrect because the problem is focused on those who attend.
• The third option ("the number of graduates") is too general as we have a specific group of 24.
• The fourth option ("the number of people attending graduation") is also incorrect as we are specifically talking about graduating students.

Part (b)

The number of students who can attend graduation from a group of \(n = 24\) students can range from \(0\) (if none of the 24 attend) to \(24\) (if all 24 attend). So \(X\) can take on the values \(X=0,1,2,\cdots,22,23,24\).

• The first option (\(X = 1,2,3,\cdots,83,84,85\)) is wrong because we have only 24 students in the group.
• The third option (\(X=1,2,3,\cdots,22,23,24\)) is incorrect because \(0\) (the case when no student attends) is a valid value.
• The fourth option (\(X = 0,1,2,\cdots,83,84,85\)) is wrong due to the group size of 24.

Part (c)

We have a binomial distribution \(X\sim B(n,p)\) where \(n\) is the number of trials and \(p\) is the probability of success in a single trial.

• Here, \(n = 24\) (the number of graduating students in the group, which is the number of trials).
• \(p=0.85\) (the probability that a graduating student attends their graduation, which is the probability of success).

Answer:

• Part (a): the number of graduating students that attend graduation
• Part (b): \(X = 0,1,2,\cdots,22,23,24\)
• Part (c): \(X\sim B(24,0.85)\)