Assume the machine shifts and the distribution of the amount o...

Question

Assume the machine shifts and the distribution of the amount of the compound added now has a mean of 9.96 milligrams and a standard deviation of 0.05 milligram. You select one vial and determine how much of the compound was added.

b. You randomly select 15 vials. What is the probability that you select at least one vial that is within the acceptable range?

b. You randomly select 15 vials. What is the probability that you select at least one vial that is within the acceptable range?

Accepted Answer

Hello, everyone. Let's take a look at this question together. A factory packages bags of flour with an average weight of 2.02 kg and a standard deviation of 0.03 kg. The acceptable weight range for each bag is 1.95 kg to 2.05 kg. You randomly select 10 bags of flour. What is the probability that you select at least one bag with a weight within the acceptable range? Is it answer choice A 0.9, answer choice B, 1, Answer choice C 0.5, or answer choice D 0.3. So, in order to solve this question, we have to recall how to find a probability so that we can find the probability that we select at least one bag with a weight within the acceptable range, given that the bags of flour have an average weight of 2.02 kg, and a standard deviation of 0.03 kg. And the acceptable weight range for each bag is 1.95 to 2.05 kg. And so the first step in determining our probability is to calculate the Z scores for the acceptable range, which can be done so using the formula for the Z scores, which is Z is equal to X subtracted by our population mean divided by the standard deviation. So then we plug. In values into our formula for the Z score starting with the lower bound, we know that the Z score for the lower bound is equal to 1.95, subtracted by 2.02, divided by 0.03, which is equal to 0.07 divided by 0.03, giving us our value of the Z score for the lower bound as approximately 2.3333. And then we do the same for the upper bound so that our Z score for the upper bound is 2.05, subtracted by 2.02, divided by 0.03, which is equal to 1. So we have our Z scores for the lower and the upper bounds, and then we can determine the probability of the weight being within the range where the probability within the range is the probability of our Acceptable value is equal to Z being between the lower and the upper bounds, which is the probability that Z is between -2.3333, which can be rewritten as the probability of Z being less than 1, subtracted by the probability of Z being less than -2.3333. And then we can use interpolation to find the area under the standard normal curve, which We know the Z table usually gives values for Z rounded to two decimal places. And since our Z score is -2.3333, which lies between -2.33 and -2.34, we need to interpolate to get a more accurate probability. And then using the standard normal distribution table, the probability of Z being less than 1 is 0.8413. The probability of Z being less than -2.33 is 0.0099, and the probability of Z being less than -2.34 is 0.0096. And since -2.3333 is one third of the way from -2.33 to -2.34, we use linear. so that the probability of Z being less than -2.3333 is equal to 0.0099, subtracted by 1/3 multiplied by in parentheses 0.0099 subtracted by 0.0096, which is equal to 0.0098. And then we can calculate the probability that a single The bag is within the acceptable range, so the probability of X being between 1.95 and 2.05, which is our acceptable range, is equal to the probability of Z being between the lower and the upper bounds, which we know is also the same as saying the probability of Z being less than 1, subtracted by the probability of Z being Less than -2.3333, which we know is equal to 0.8413, subtracted by 0.0098, so the probability that a single bag is within the acceptable range is 0.8315. Therefore, E is equal to 0.8315, which is the probability a single bag is within the acceptable range. And then we can use the complement rule to calculate the probability of at least 1 success in 10 trials, as the probability of no successes is calculated as the probability of no successes is equal to 1 minus P in parentheses to the power of N, so the probability of no successes is equal to 1 minus 0.8315. in parentheses to the power of 10, as we know that there are 10 trials, and so this is equal to 0.1685 to the power of 10, which is approximately 1.85 multiplied by 10 to the power of -8. So the probability of no successes is approximately 1.85 multiplied by 10 to the power of -8. And then the probability of at least one success is the probability of at least 1, which is equal to 1 minus the probability of no successes. So we substitute that value for the probability of no successes. So the probability of at least 1 success is equal to 1 minus 1.85 multiplied by 10 to the power of -8, which is approximately 1. Therefore, the probability that at least 1 of the 10 bags is within the acceptable weight range is approximately 1. And so looking at our answer choices, we can identify that 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

Normal Distribution

Standard Deviation

Probability of Events

Watch next