12/50=24% of student bring lunch so 24%-524=398 student buy lunch out of 524
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insta
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txt and ill give u the deatails
Answer:
A sample of 1068 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error is:
95% confidence level
So , z is the value of Z that has a pvalue of , so .
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?
We need a sample of n.
n is found when M = 0.03.
We have no prior estimate of , so we use the worst case scenario, which is
Then
Rounding up
A sample of 1068 is needed.
Answer:
X = 4
Y = 30
i think.
Step-by-step explanation: