Answer:
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
We are given
Since, we have to solve for F
so, we will isolate F on anyone side
step-1:
Multiply both sides by 60
step-2:
Divide both sides by 11
step-3:
Add both sides by 32
so, we get
................Answer
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