In a randomly selected group of 300 students at a college where 10% of students are left-handed, we would, on average, expect 30 students to be left-handed. The standard deviation, indicating the typical variation or 'spread' around this mean, would be approximately 9.1.
The subject of this problem is statistics, specifically the mean and standard deviation of a proportion. In this case, if 10% of 5000 students are left-handed, that means there are 500 left-handed students. When we select a group of 300 students, we expect (on average) 10% of this group to be left-handed, which is 30 students.
The mean (or expected value) here refers to the average number of left-handed students we would expect to see if many groups of 300 students were randomly selected. Thus, the mean is 30.
The standard deviation describes the typical variation or 'spread' around the mean. In a binomial scenario like this one, the standard deviation formula is √(n × p × (1-p)), where n is the number of trials (the group size, 300), p is the probability of success (proportion of left-handed students, 0.1), and (1 - p) is the probability of failure. Plugging in our values, we get a standard deviation of approximately 9.1 (rounded to the nearest tenth).
Question:
A college has over 5,000 students, 10, percent of which are left-handed. A new lecture hall is being planned that will seat 300 students, and college officials want to be sure there are enough left-handed desks. Suppose we randomly select groups of 300 students from this college. What are the mean and standard deviation of the number of left-handed students in each group of 300? You may round your answers to the nearest tenth.