BIA 654 STATISTICS HW

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Stevens Institute Of Technology *

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654

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Industrial Engineering

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Dec 6, 2023

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1. Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situations.  Motorola defect probability and expected number of defects a. The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. a. Probability of defects with a standard deviation of 0.15: Z-score for 9.85 ounces: Z = (9.85 - 10) / 0.15 approximately -1.00 Z-score for 10.15 ounces: Z = (10.15 - 10) / 0.15 approximately 1.00 Using a standard normal distribution table or calculator: The probability for Z < -1.00 is approximately 0.1587. The probability for Z > 1.00 is also approximately 0.1587. The total probability of defects is 2 * 0.1587 = 0.3174 Expected number of defects = P(defect) = 0.3174*1000 = 317 b. Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects . z_lower = (9.85 - 10) / 0.05 = -3 z_upper = (10.15 - 10) / 0.05 = 3 P(defect) = 0.9987 - 0.0013 = 0.9974 The probability of a defect after reduced standard deviation is 0.9974 Expected number of defects with reduced process variation: Expected number of defects = P(defect) * sample size = 0.9974 * 1000 = 997 c. What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

Reducing process variance has the benefit of enabling more items to meet specification limitations, which lowers the number of faults. Reducing the standard deviation will reduce probability of defects in cases (a) and (b). 2. After deducting grants based on need, the average cost to attend the University of Southern California (USC) is $27,175 (U.S. News & World Report, America’s Best Colleges, 2009 ed.). Assume the population standard deviation is $7400. Suppose that a random sample of 60 USC students will be taken from this population. a. What is the value of the standard error of the mean? a. Standard Error (SE): SE = σ / √n SE = $7,400 / √60 approximately $955.75 b. What is the probability that the sample mean will be more than $27,175? b. Calculate the Z-score: Z = (27175 - 27175) / SEM = 0 P(Z > calculated Z) = 0.5 c. What is the probability that the sample mean will be within $1000 of the population mean? c. Calculate the Z-scores for the lower and upper limits: Lower limit: 27175 - 27175 - 1000 = -1.046 Upper limit: 27175 - 27175 + 1000 = 1.046 P(lower limit < X < upper limit) = 0.704 d. How would the probability in part (c) change if the sample size were increased to 100? With a larger sample size (n = 100), SE = $7,400 / √100 = $740. A smaller SE means a smaller margin of error and a more precise estimate of the sample mean.

3. A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weights 𝑥̅ . If test results over a long period of time show that 5% of the 𝑥̅ values are over 2.1 pounds and 5% are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process? Given n=30 To calculate mean which is the average of the two given boundaries M = 2.1+2.9/2 = 2.0 Z-score with a probability of 5% or 0.05 Z-score approximately -1.645 Solving the standard error using Z-score z = (x-μ)/σ -1.645 = (1.9-2.0)/ σ σ approximately equals to 0.0608 Standard deviation is the product of standard error and the square root of the sample size σ = 0.0608 √30 approximately 0.3330 4. A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $33.77 (AAII Journal, February 2006). The survey is conducted annually. With the historical data available, assume a known population standard deviation of $15. Given data: (x̄) = $33.77 population standard deviation (σ) = $15 (n) = 54 Confidence level = 95% a. Using the sample data, what is the margin of error associated with a 95% confidence interval? Calculate the Margin of Error (ME):

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