Conservation About 74% of the residents in a town say that the...

Question

Conservation About 74% 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 80%? Interpret your result.

Accepted Answer

All right, hello, everyone. So this question says, a poll indicates that 59% of employees at a company are satisfied with their work-life balance. If 80 employees are randomly chosen, what is the probability that the sample proportion satisfied is less than 50%? And here we have 4 different answer choices labeled A through D. So first, what information do we know? Well, we already know that P is equal to 0.59, because that's the number of employees that indicated they were satisfied. And the sample size is equal to 80. And pea hat or. Rather, our goal is for PHAT to be less than 0.50, referring to the sample proportion. From here we calculate the meat. Now In this case, mew. Of Phat would just be equal to P, which is 0.59. And now we can calculate the standard deviation. So, sigma of Phat, that's the standard deviation of the sample proportion. Is equal to the square root of N. Multiplied excuse me, P. It's P multiplied by 1 subtracted by P. Divided by and. And you would take the square root of this term. So, sigma of Phat is equal to the square root of. 0.59. Multiplied by 1 subtracted by 0.59. And divided by 80. Evaluating this expression gives you approximately 0.055. And now you would find the Z score for Phat equals 0.50. So Z Is equal to Phat subtracted by P. And divided by sigma of Phat. So because we are. Solving for Phat equals 0.50, excuse me, Z is equal to 0.50, subtracted by 0.59. Divided by 0.055 And this equals approximately 1.64. And now using the. Standard normal table. You would find P of Z less than -1.64. Which, if you recall, would correspond to P of Phat being less than 0.50. So from the standard normal table. The probability of Z less than -1.64 is 0.505. Or excuse me, 0.0505. So ultimately, there is about a 5.05% chance that less than 50% of a random sample of 80 employees will be satisfied with their work-life balance. This corresponds to option A in the multiple choice, making that your correct answer. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

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Key Concepts

Sample Proportion

Normal Approximation to the Binomial Distribution

Hypothesis Testing

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