Answer:
Given:
Sample size (n) = 90
Proportion of residents conserving water or electricity (p) = 0.72
To find:
Mean and standard deviation of the proportion of sampled residents who are making an effort to conserve water or electricity.
Probability that the proportion of sampled residents who are making an effort to conserve water or electricity is greater than 80%.
Solution:
Mean and Standard Deviation
The mean of the sampling distribution of sample proportions is given by:
μp = p = 0.72
The standard deviation of the sampling distribution of sample proportions is given by:
σp = sqrt((p*(1-p))/n) = sqrt((0.72*0.28)/90) = 0.055
Therefore, the mean of the proportion of sampled residents who are making an effort to conserve water or electricity is 0.72, and the standard deviation is 0.055.
Probability
To calculate the probability that the proportion of sampled residents who are making an effort to conserve water or electricity is greater than 80%, we need to standardize the sample proportion using the z-score formula:
z = (x - μ) / σ
where x is the sample proportion, μ is the mean of the sampling distribution, and σ is the standard deviation of the sampling distribution.
Substituting the given values, we get:
z = (0.8 - 0.72) / 0.055 = 1.45
Using a standard normal table or calculator, we can find that the probability of obtaining a z-score greater than 1.45 is 0.0735, or approximately 7.35%.
Therefore, the probability that the proportion of sampled residents who are making an effort to conserve water or electricity is greater than 80% is 0.0735 or 7.35%.
Explanation: