Please include a complete step-by-step explanation. I shall be grateful and will
ID: 357602 • Letter: P
Question
Please include a complete step-by-step explanation. I shall be grateful and will give thumbs up!
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Please include a complete step-by-step explanation. I shall be grateful and will give thumbs up!
Regulations on milk packaging is such that a gallon of milk cannot have less than 1 gallon in the container. If this is violated, the company is fined. The company fills milk containers with a standard deviation of 0.5 ounces regardless of the mean setting. To be sure that regulations are met, the operator decides to set the mean at 1 gallon (128 ounces)+1.5 ounces, or 129.5 ounces a). To check if the process is in control, the operator plans to take samples where each sample consists of 4 gallons containers. The average volume of this sample of 4 containers is used to determined if the process is in control. Following industry practice, what control limits should the operator use? b). If the process is in control, approximately what fraction of contrainers will have a volume under 1 gallon (this is the fraction that would violate government regulation)? c). Putting more than 1 gallon in a container to prevent regulation fines is costly in terms of milk given away. Management wants to reduce the average excess volume to 1 ounce while staying in line with regulations "practically always," which means 99.87 % of the time. In what sense will this require improved process technology? Give an explanation in words as well as a specific numeric answerExplanation / Answer
Lower Specification Limit, LSL = 1 gallons (128 ounces)
Std dev, s = 0.5 ounces
Mean, X?? = 129.5 ounces
a) For sample size, n = 4, the control chart constants are A3 = 1.628, B3 = 0, B4 = 2.266 (these values are taken for the control chart constants table for X?-s chart)
UCL for X?-chart = X?? + A3*s = 129.5 + 1.628*0.5 = 130.314
LCL for X?-chart = X?? - A3*s = 129.5 - 1.628*0.5 = 128.686
b) z = (128-129.5)/0.5 = -3
P(z) = NORMSDIST(-3) = 0.00135
Fraction of containers having volume under 1 gallon (128 ounces) = 0.00135
c)
Required upper specification limit, USL = 128 + 1 = 129 ounces (mgmt wants to reduce the excess volume to 1 ounce)
z = (USL-X??)/s = (129-129.5)/0.5 = -1
P(z) = NORMSDIST(-1) = 0.1587
Fraction of samples having volume less than or equal to 129 ounces is 0.1587. So this will require improved process technology to reduce the process variability.
And fraction of samples having volume violating regulations (128 gallon) is 0.00135 (as determined in part b)