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:
18
Step-by-step explanation:
48-12 = 36
36/2
18
Answer:
46
Step-by-step explanation:
Let number is x greater than 21 and x less than 71.
Now, according to question,
21+x=71−x
2x=50
x=25
So, required number is 21+x=21+25=46.
Alternatively,
The required number would be the average or mean of 21 and 71.
So, required number =
2
(21+71)
=46.
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