A town council wants to estimate the proportion of residents who are in favor of a proposal to upgrade the computers in the town library. A random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal. Is it appropriate to assume that the sampling distribution of the sample proportion is approximately normal?

A. No, because the sample is not large enough to satisfy the normality conditions.
B. No, because the size of the population is not known.
C. Yes, because the sample was selected at random.
D. Yes, because sampling distributions of proportions are modeled with a normal model.
E. Yes, because the sample is large enough to satisfy the normality conditions.

E.Yes, because the sample is large enough to satisfy the normality conditions

Step-by-step explanation:

Given that  a random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal then;

The percent in favor of the proposal is 97/100 *100 =97%

Thus from this information ;

p=0.97 and q=1-p =1-0.97 =0.03 where p=population proportion

Finding mean and standard deviation will be;

μp =mean of sample proportion

μp=p= 0.97

δp= standard deviation of sample proportion

δp=√{pq/n }= √{(0.97*0.03)/100} where n=100 (sample size)

δp= 0.0171

3δp = 3*0.0171 =0.051

check using ;

{p- 3δp, p+3δp} = { 0.97-0.05,0.97+0.05} ={0.92,1.02}

compare if {0.92,1.02} lies in {0,1)

A sample is large enough if the interval {p- 3δp, p+3δp} lies wholly within the interval {0,1}

This sample wholly lie in the interval of {0,1} thus it is safe to assume p' is approximately normally distributed

only 10/60 isn't equivalent

hope this : )

answer: you need the equation in the question so i can answer it

step-by-step explanation: