Assume the machine shifts and is filling the vials with a mean...

Question

Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams and a standard deviation of 0.05 milligram. You select five vials and find the mean amount of compound added.

a. What is the probability that you select a sample of five vials that has a mean that is within the acceptable range? (See figure.)

Accepted Answer

Hello everyone. Let's take a look at this question together. A company produces small bottles of essential oil with an average fill amount of 12.02 mL and a standard deviation of 0.04 mL. You randomly select a sample of 5 bottles and measure the mean fill amount. What is the probability that the sample mean is within the acceptable range of 11.95 mL to 12.05 mL? Is it answer choice A 0.9531, answer choice B, 0.3595, answer choice C 0.3559, or answer choice D 0.9355. So in order to solve this question, we have to recall how to determine a probability to determine the probability that the sample mean of our randomly selected five bottles has a mean within the acceptable range of 11.95 mL to 12.05 mL, given that these small bottles of essential oil have an average fill amount of 12.02 mL and a standard. of 0.04 mL. And so the first step in solving this problem and determining our probability is identifying the sampling distribution, which we know that since the sample size is N equals 5, we find the standard error of the mean using the following formula where the sample mean is equal to the standard deviation divided by the square root of our sample size N. So then the standard error of the mean is equal to 0.04 divided by the square root of 5, which is approximately 0.0179. And then we can calculate the Z scores for the acceptable range using the formula for the Z score where Z is equal to the sample mean subtracted by the population mean divided by the standard error of the mean. And so for the lower bound, which is 11.95 mL, our Z score of the lower bound is equal to 11.95, subtracted by 12.02, divided by 0.0179, which is approximately 3.910. 6. And then for the upper bound, which is 12.05 mL, our Z score for the upper bound is 12.05, subtracted by 12.02, divided by 0.0179, which is approximately 1.6760. So we have our Z score for the lower bound of -3.9106, and our Z score for the upper bound, which is approximately 1.6760. And now we can determine the probability using the standard normal distribution, starting with our Z score 1, which is -3. 9106, we know that this value is deep in the left tail of the normal curve, and so the area under the curve below our Z equaling -3.91 is extremely close to zero, so the probability that Z is less than -3.9106 is approximately 0. Since we know we can treat it as effectively zero for practical purposes, and then for our second Z score, which is equal to 1.6760, we interpolate between the two values, which are the probability that Z is less than 1.67, which is equal to 0.9525, and then the probability that Z is less than 1.68, which is equal to 0.9535. And since 1.6760 is 60% of the way from 1.67 to 1.68, we know that the probability of Z being less than 1.6760 is equal to 0.9525 plus 0.6 multiplied by in parentheses 0.9535 subtracted by 0.9525, which is equal to 0.953. want. So then we can find the probability that the sample mean is within the range, given that the probability of the sample mean is between 11.95 and 12.05 is equal to the probability of Z being between the lower bound and the upper bound, which we know is the same as the probability of Z being less than the upper bound, subtracted by the probability of Z being less than the lower. Bound, which is equal to 0.9531, subtracted by zero, so the probability is equal to 0.9531. Therefore, the probability that the sample mean of the five bottles is within the acceptable range is 0.9531. And looking at our answer choices, we can identify that answer choice A is 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

Sampling Distribution of the Mean

Standard Error

Probability and Normal Distribution

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