Use you brain. Please use it.
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
Answer:
A. 30000
B. 12000
Step-by-step explanation:
Find out how much spectators leave in 1 minute.
1 second = 5
1 minute = 5 x 60 = 300 (60 seconds)
20 minutes = 300 x 20 = 6000
Since there were only 80% of the full capacity of spectators and after minutes it became 60%...
80% - 60% = 20%
20% = 6000
1% = 6000 ÷ 20 = 300
100% = 300 x 100 = 30000
1 minute = 300 spectators leaving
1 hour = 300 x 60 = 18000 (60 minutes)
30000 - 18000 = 12000
