Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so 
Now, find the margin of error M as such

In which
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when
. So






Rouding up
We need a sample size of at least 719
Answer:
7986
Step-by-step explanation:
66x0.5x(140+102)=7986
6hrs = school
9hrs = sleeps = 18hrs // 24hrs ----6hrs left
3hrs = plays
6hrs = other things
playing = 12.5% -- 3/24 3/24*100
other things = 25% -- 6/24 6/24*100
It's 10. I hope this helps!