Question

about 74% of the residents in a town say that they are making an effort to conserve water or electricity. one hundred residents are randomly selected. what is the probability that the sample proportion making an effort to conserve water or electricity is greater than 79%? 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 $sqrt{\frac{pq}{n}}$. 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 79% of a sample of 100 residents are making an effort to conserve water or electricity is about %. b. the probability that more than 74% of a sample of 100 residents are making an effort to conserve water or electricity is about %. c. the probability that less than 74% of a sample of 100 residents are making an effort to conserve water or electricity is about %. d. the probability that more than 79% of a sample of 100 residents are making an effort to conserve water or electricity is about %.

Explanation:

Step1: Identify population and sample - related values

Let $p = 0.74$ (population proportion), $q=1 - p=1 - 0.74 = 0.26$, and $n = 100$ (sample size). We want to find $P(\hat{p}>0.79)$ where $\hat{p}$ is the sample - proportion.

Step2: Calculate the standard deviation of the sampling distribution of the sample proportion

The standard deviation of the sampling distribution of the sample proportion $\sigma_{\hat{p}}=\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.74\times0.26}{100}}=\sqrt{\frac{0.1924}{100}}=\sqrt{0.001924}\approx0.0439$.

Step3: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\hat{p}-p}{\sigma_{\hat{p}}}$. Substitute $\hat{p}=0.79$, $p = 0.74$, and $\sigma_{\hat{p}}\approx0.0439$ into the formula: $z=\frac{0.79 - 0.74}{0.0439}=\frac{0.05}{0.0439}\approx1.14$.

Step4: Find the probability

We want $P(\hat{p}>0.79)$, which is equivalent to $P(Z > 1.14)$ in the standard normal distribution. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard - normal table $P(Z\leq1.14) = 0.8729$, then $P(Z > 1.14)=1 - 0.8729 = 0.1271\approx12.71\%$.

Answer:

D. The probability that more than 79% of a sample of 100 residents are making an effort to conserve water or electricity is about 12.71%.