Syllabus-BA 2300-180 × \\. Cengage x / Mindlap-Cengage Lea x \\ e Chegg Study |
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Syllabus-BA 2300-180 × . Cengage x / Mindlap-Cengage Lea x e Chegg Study | Guided S: X -Google Not secure ngcengage.com/static/nb/ui/index.html?nbId=740021&nbNodeld;:279396983&eISBN;:9781 3371153774|&parentld;:279397314 d : Apps D Bootyquak Bookmarks D Point Reduction and The ideal growing cc D CAKE BIRTH Shroomery PF TEK Shroomery-casing D How to Legally Acce Other bookmark MINDTAP James lyons- Chapter 6 Assignment Due on Mar 19 at 11-59 PM EDT Questions Exercise 06.35 Algorithmic Question 7 of 10 5. Check My Work 5. eBook 8. 9. 10. 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 13 ounces. a. The process standard deviation is 0.10, and the process control is set at plus or minus 1.25 standard deviation s. Units with weights less than 12.875 or greater than 13.125 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found (round to the nearest whole number)? b. Through process design improvements, the process standard deviation can be reduced to 0.08. Assume the process control remains the same, with weights less than 12.875 or greater than 13.125 ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)? In a production run of 1000 parts, how many defects would be found (to the nearest whole number)? c. What is the advantage of reducing process variation, thereby causing a problem limits to be at a greater number of standard deviations from the mean? It can substantially reduce the number of defects 7:03 PM Type here to searchExplanation / Answer
a)
hence probability of defect =1-P(12.875<X<13.125)=1-0.7887 =0.2113
number of defects =np =1000*0.2113 =211
b)
hence probability of defect =1-P(12.875<X<13.125)=1-0.8818 =0.1182
number of defects =np =1000*0.1182 =118
for normal distribution z score =(X-)/ here mean= = 13 std deviation == 0.1000Related Questions
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