Mastorola ured the ormal dsaribation to aiehe probables nf dfects s peced n prod

Question

Mastorola ured the ormal dsaribation to aiehe probables nf dfects s peced n productin sroces. Asumrrocess roduce with a mean weight of 14 ounces. a. The precess standard deviation is 0.10, and the precess control is set at plus or minus 1.75 standard deviation s. Units with weights less will be classified as defects. What is the probability of a defect (to 4 decimals)? 0.0455 In a production run of 1000 parts, how many defects would be found (round to the nearest whiole number)?) 0.0244 3 75 standard deviation s. Units with weights less than 13.825 or greater than 14.175 ounces CengegeBci-Logn or Regster ess design improvements, the process standard deviation can be reduced to 0.06. Assume the process control remains the same, with weights less than 13.825 or greater than 14 175 ounces being dassifled as defects. What is the probuability of a defect (Cround to 4 decimals if necessary)? 46 ie a production run of 1000 parts, how many defects would be found (to the nearest iwhole number) 25 3 c. What is the advantage of reducing process variatien, thereby causing a probilem limits to be at a greater number of standard deviations from the mean? It can substantially reduce the number of defects

Explanation / Answer

a)P(defects) =1-P(13.825<X<14.175)=1-P(-1.75<Z<1.75)

=1-(0.9599-0.0401)=0.0802 ( please try 0.0801 if this comes wrong)

number of defects =1000*0.0802 ~ 80

b)P(defets)=1-P(13.825<X<14.175)=1-(-2.92<Z<2.92)

=1-(0.9982-0.0018)=0.0036 ( please try 0.0035 if this comes wrong)

number of defects =1000*0.0036=4

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