Exhibit 18-2 Students in statistics classes were asked whether they preferred a 10-minute break or to get out of class 10 minutes early. In a sample of 150 students, 40 preferred a 10-minute break, 80 preferred to get out 10 minutes early, and 30 had no preference. We want to determine if there is a difference in students' preferences. Refer to Exhibit 18-2. The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are

The mean and standard are 0.533 and 0.82 respectively

Step-by-step explanation:

We are given the following data

Total number of students = 150

Students preferred to get out of 10mins = 80

Our mean of the sampling distribution of the number of students who preferred to get out early will be

= 80:150

= 0.533

The formula for the standard deviation is as follows

= mean - pⁿ/√ (pⁿ (1-p) / )

Substituting the values we have

= 0.533-0.5/√ (0.50*0.5/15))

=0.82

The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are 0.533 and 0.82

Step-by-step explanation:

According to the given data we have the following:

Total sample of students = 150

80 students preferred to get out 10 minutes early

Therefore, the mean of the sampling distribution of the number of students who preferred to get out early is = 80/150 = 0.533

Therefore, standard deviation of the sampling distribution of the number of students who preferred to get out early = phat - p0/sqrt (p0 (1-p) / )

= 0.533-0.5/sqrt (0.5*0.5/15))

= 0.816 = 0.82