Question

about 75% of the residents in a town say that they are making an effort to conserve water or electricity. one hundred ten residents are randomly selected. what is the probability that the sample proportion making an effort to conserve water or electricity is greater than 82%? interpret your result. assume the sampling distribution of sample proportions is a normal distribution. the mean of the sample proportion is equal to the population proportion and the standard deviation is equal to \\( \sqrt { \frac { p q } { n } } \\). the probability is \\( \square \\). (round to four decimal places as needed.) interpret the result. select the correct choice and fill in the answer box to complete your choice. (round to two decimal places as needed.) a. the probability that less than 82% of a sample of 110 residents are making an effort to conserve water or electricity is about \\( \square \\% \\). b. the probability that less than 75% of a sample of 110 residents are making an effort to conserve water or electricity is about \\( \square \\% \\). c. the probability that more than 75% of a sample of 110 residents are making an effort to conserve water or electricity is about \\( \square \\% \\).

Explanation:

Step1: Calculate the standard deviation

Given \(p = 0.75\), \(q=1 - p=0.25\), \(n = 110\).
The standard deviation \(\sigma_{\hat{p}}=\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.75\times0.25}{110}}=\sqrt{\frac{0.1875}{110}}\approx\sqrt{0.0017045}\approx0.0413\)

Step2: Calculate the z - score

The sample proportion \(\hat{p}=0.82\), the mean of the sampling distribution \(\mu_{\hat{p}} = p=0.75\)
The z - score \(z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.82 - 0.75}{0.0413}=\frac{0.07}{0.0413}\approx1.69\)

Step3: Find the probability

We want to find \(P(\hat{p}>0.82)\), which is equivalent to \(P(Z > 1.69)\)
Since \(P(Z>z)=1 - P(Z\leq z)\)
From the standard normal table, \(P(Z\leq1.69)=0.9545\)
So \(P(Z > 1.69)=1 - 0.9545=0.0455\)

Answer:

The probability is \(0.0455\)
For the interpretation:
We found \(P(\hat{p}>0.82) = 0.0455\), which means \(P(\hat{p}<0.82)=1 - 0.0455 = 0.9545\approx95.45\%\)
So the answer is A. The probability that less than \(82\%\) of a sample of \(110\) residents are making an effort to conserve water or electricity is about \(95.45\%\)