Gary\'s drivers have been producing computer drivers with an average life of 290
ID: 442813 • Letter: G
Question
Gary's drivers have been producing computer drivers with an average life of 290 hours and a standard deviation of 24 hours. The customer specification are 300+ or -100 for the upper and lower specification limits respectively.
A. Gary has the opportunity to gain a big order from Vic's computers if he can produce the driver with the specification limits of 300+ or -100 hours. If the minimum acceptable process capability is 1.33 (a 4-sigma process), can Gary meet the customers specification demands at this time? if he cannot explain if it is due to a drifting of the mean or too much variability. - Explai,
b. Suppose that Gary has introduced some changes in his operations. The customer has also agreed to use Gary's if he can provide a process capability of 1.33 or greater. After the first 20 days of production under the new procsses, it finds that the average life is 295 hours but it has not determined the standard deviation. What is the maximum acceptable value of the standard deviation for Gary to be selected? The customer spec limits are still 300+ or -100
Explanation / Answer
S.D=24 hrs
Customer requirement =300-100 to 300+100=200 hours to 400 hours
A)
Minimum acceptable process capability=1.33 S.D
Minimum Acceptable deviation =1.33*24=31.9 hrs
Maximum Deviation (4 sigma) =4*S.D=4*24=96hrs
So, If Gary computer targets average life of 290 hrs, the range of hours it will be able to produce is
Lower Range=290-32.9 to 290-96=257.1 to 194 hrs
Upper range=290+32.9 to 290+96 =312.9 to 386 hrs
However lower range 194 hrs falls outside of the range of customer requirement of 200 hrs to 400 hrs.
Thus Gary cannot meet the customer requirement at lower range.
However the maximum deviation of Gary process is +/- 96 hours (4*S.D) will is less the customer specification limit of +/- hours. Thus Gary process variability lies well within the customer limit of variability.
It is the lower mean of 290 hours because of which Gary is not able to meet the customer requirement at lower range limit of 200 hrs.
Thus Gary cannot meet the customer specification demand at this time because of the drift of the mean.
B)
Mean=295 hrs
Customer requirement =300-100 to 300+100=200 hours to 400 hours
Maximum Deviation (4 sigma) =4*S.D
Maximum Deviation =Min((Mean-lower limit),(Upper limit-Mean))
Maximum Deviation =Min(295-200,400-295)=Min(95,105)=95
4*S.D=95
S.D=95/4=23.75 hrs
Thus maximum acceptable value of S.D=23.75 hrs
C)
Mean+4*S.D<=400
Mean-4*S.D>=200
Or,
Mean+4*24<=400, Mean-4*24>=200
Mean<=304 and Mean>=296
Lower limit of mean=296 hrs
Upper limit of mean=304 hours
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.