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.

Graph showing distributions of sample means with mean values and acceptable range limits for vial masses in milligrams.

b. You randomly select three samples of five vials. What is the probability that you select at least one sample of five vials that has a mean that is within the acceptable range?

Accepted Answer

Hello, everyone. Let's take a look at this question together. A company's filling machine dispenses essential oil with a mean of 11.96 mL and a standard deviation of 0.05 mL. A quality inspector randomly selects 3 separate samples, each containing 5 bottles, and measures the sample mean fill amount for each. What is the probability that at least 1 of these 3 samples has a sample mean within the acceptable range of 11.90 mL to 12 mL? Is it answer choice A 0.8899? Answer choice B, 0.9999, answer choice C 0.1675, or answer choice D 0.7692. So, in order to solve this question, we have to recall how to determine a probability, so that we can determine what is the probability that at least one of the three samples has a sample mean within the acceptable range of 11.90 mL to 12 mL, given that the company's filling machine dispenses the oil with a mean of 11.96 mL and a standard deviation of 0.05 mL. Leads. And so the first step in solving this problem is identifying the sampling distribution of the sample mean, which we know can be done using the formula for the standard error, where the standard error of the sample mean is equal to the standard deviation divided by the square root of N. So we plug in the values which gives us 0.04 divided by the square root of 5. So the standard error of the same. mean is approximately 0.0179, and then we can calculate the Z scores for the acceptable range using the formula for the Z score, which is Z is equal to the sample mean subtracted by the population mean divided by our standard deviation of the sample mean, and then we know for the lower bound, we have 11.90 mL, so the Z score for the lower bound is 11.90. Subtract. By 11.96 divided by 0.0224, which is approximately 2.6786. And then for the upper bound, which is 12 mL, the Z score of the upper bound is 12 minus 11.96, divided by 0.0224, which is approximately 1.7857. So we have our Z score of the lower bound, which is approximately. -2.6786, and our Z score of the upper bound, which is approximately 1.7857. And then the next step is to interpolate to find the probabilities starting with our Z1, which is -2.6786, as this is our lower bound Z score. So we interpolate between our Z equaling -2.67 and Z equaling -2.68. So Z equals -2.67 is the probability that Z is less than -2.67, which using a standard normal distribution table, we have 0.0038. And then our Z equals -2.68 is the probability that Z is less than -2.68, which is 0.0037. And since -2.6786 is 86% of the way from -2.67 to -2.68, the probability that Z is less than -2.6786. is equal to 0.0038 subtracted by 0.86 multiplied by in parentheses 0.0038 subtracted by 0.0037. So the probability that Z is less than -2.6786 is equal to 0.003714. And then for our second Z score, which is 1. 7857, we interpolate between the following, which is Z equals 1.78 and Z equals 1.79. So for Z equals 1.78, we have the probability of Z being less than 1.78, which is equal to 0.9625, and the probability that Z is less than 1.79 is 0.9633. And since 1.7857 is 57% of the way from 1.78 to 1.79, the probability that Z is less than 1.7857 is equal to 0.9625 plus 0.57 multiplied by in parentheses 0.9633, subtracted by 0.9625, so the probability. of our Z being less than 1.7857 is 0.962956. And now we can find the probability that one sample mean is within the range, so the probability that our sample mean is between 11.90 and 12 is equal to the probability that Z is between our Z1 and our Z2. which we can substitute for the values that we found for both Z1 and Z2. So this is written as the probability that Z is less than 1.7857, subtracted by the probability that Z is less than -2.6786. So this is equal to 0.962956, subtracted by 0.003714. So the probability that one sample mean is within our acceptable range is approximately 0.9592. And then we can use the complement rule for at least one, and since we are selecting three independent samples and we want to find the probability that at least one sample has its mean in the acceptable range, this is given as the probability that at least one is in range. is equal to 1 minus the probability that none are in range, and we know that the probability that one sample is not within the range is given as Q is equal to 1 minus P, so this is equal to 1 minus 0.9592, as we found the probability that one sample mean is within range is 0.9592, so the probability. That one sample is not within the range is 0.0408. And so the probability that none of the three samples are within the range is given as the probability that none are in range is equal to Q to the power of 3, so 0.04 08 to the power of 3 is approximately 6.79 times 10 to -5. So, the probability that at least 1 is in range is equal to 1 minus 6.79 times 10-5, which is approximately 0.9999. So we know the probability that at least one of the three samples has a mean within the acceptable range is 0.9999. And looking at our answer choices, we can identify this is answer choice B, so answer choice B 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 Acceptable Range

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