In Exercises 39 and 40, determine whether the finite correctio...

Question

In Exercises 39 and 40, determine whether the finite correction factor should be used. If so, use it in your calculations when you find the probability.

Parking Infractions In a sample of 1000 fines issued by the City of Toronto for parking infractions in September of 2020, the mean fine was $49.83 and the standard deviation was $52.15. A random sample of size 60 is selected from this population. What is the probability that the mean fine is less than $40?

Accepted Answer

Hello everyone. Let's take a look at this question together. A university library issued 800 late return fines in October of 2022. The average fine was $35.50 with a standard deviation of $40.20. If a random sample of 50 fines is selected, what is the probability that the sample mean fine is less than $30? Should the finite population correction factor be applied? So, in order to solve this question, we have to recall how to find the probability, so that we can find the probability that the sample mean fine is less than $30 as well as if the finite population correction factor should be applied. And so the first step in solving this problem is to check if the finite population correction or FPC should be applied, which we know that the finite population correction is recommended when the sample size N is more than 5% of the population size N, and using the information provided in the question, we know that. Our sample size N is 50 and our population size N is 800, so 50 divided by 800 is 0.0625 or 6.25%, which is more than 5%. So since this is greater than 5%, the finite population correction factor should be applied. So then we can calculate the standard error with the finite population correction, and we know that the formula for the standard error of the mean with finite population correction is equal to SE, which is equal to the standard deviation divided by the square root of the sample size multiplied by the square root of the population size minus the sample size divided by. The population size minus 1, and then we can plug in the values into the formula for the standard error of the mean, which gives us 40 multiplied by 20 divided by the square root of 50, all multiplied by the square root of 800 minus 50 divided by 800 minus 1, which is approximately 5.510. Therefore, the standard error of the mean with the finite. The correction applied is approximately 5.510, and then we standardize the value, meaning we compute the Z score using the formula Z equals the sample mean subtracted by the population mean divided by the standard error, and then substituting the values, we have 30 minus 35.50 divided by 5.510, which is approximately -1. And then we can find the probability that our Z is less than -1, which is approximately 0.1587. And so we have determined that the probability that the sample mean fine is less than $30 is 0.1587, and the finite population correction factor should be applied because the sample is more than 5% of the population. Which makes answer choice A the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Finite Correction Factor

Sampling Distribution of the Mean

Z-Score

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