Using the z-distribution, it is found that a sample size of 3,385 is required.
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:
The margin of error is:
In which:
is the sample proportion.
In this problem, we have a 98% confidence level, hence
, z is the value of Z that has a p-value of 
, so the critical value is z = 2.327.
We have no prior estimate, hence 
is used, which is when the largest sample size is needed. To find the sample size, we solve the margin of error expression for n when M = 0.02, hence:
n = 3,385.
A sample size of 3,385 is required.
More can be learned about the z-distribution at brainly.com/question/25890103
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16/25 is 64%. Hope this helps
Using the mean concept, the average collection for a student in this school was of $16.13.
<h3>What is the mean?</h3>
The mean of a data-set is given by the <u>sum of all observations in the data-set divided by the number of observations</u>.
The number of students is given as follows:
82 + 74 + 96 + 99 = 351.
The sum is:
82 x 26.75 + 74 x 12.25 + 96 x 15.50 + 99 x 10.85 = $5662.15
Hence the mean is:
M = $5662.15/351 = $16.13.
More can be learned about the mean of a data-set at brainly.com/question/24628525
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