Answer:
The mean of sampling distribution of the proportion of the students who prefer later start times for school
μₓ = p = 0.63
Step-by-step explanation:
Explanation:-
Given sample size 'n' =200
given data Of the 200 students surveyed at the local high school, 126 say that they prefer earlier start times for school
sample proportion
[tex]p = \frac{x}{n} = \frac{126}{200} = 0.63[/tex]
The mean of sampling distribution of the proportion of the students who prefer later start times for school
μₓ = p = 0.63
Standard deviation of the proportion
[tex]S.D = \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.63(1-0.63)}{200} } = 0.034[/tex]
