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 $sqrt{\frac{pq}{n}}$. the probability is $square$. (round to four decimal places as needed.)

Explanation:

Step1: Identify values

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

Step2: Calculate standard deviation of sample proportion

The standard deviation 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 z - score

The z - score is calculated as $z=\frac{\hat{p}-p}{\sigma_{\hat{p}}}$. Substituting the values, we get $z=\frac{0.79 - 0.74}{0.0439}=\frac{0.05}{0.0439}\approx1.14$.

Step4: Find 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 $P(Z > 1.14)=1 - 0.8729 = 0.1271$.

Answer:

$0.1271$