Standard Deviation and Production Team

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Final Project

2.a.) Assuming no change in the distribution the probability that the blackness of the spot is less than 1.0 is 48.01%. (1.0-1.005/.10= z-score of -.05 = .4801 or 48.01%)
b.) The probability that the blackness of the spot is between .95 and 1.0 is 18.89% (.95-1.005/.10=z score of -0.55 = 29.12%. take answer from #1 and subtract from this answer 48.01-29.12= 18.89%) c.) The probability that the blackness of the spot is between 1.0 and 1.05 is 19.35%. (1.05-1.005/.10= z-score of .45 = 67.36%, then take area from 1.0= 48.01% and subtract it from 67.36% and you get 19.35%) d.) The probability that the blackness of the spot would be less than .95 is 29.12%. The probability that the blackness of the spot would be greater than 1.05 is 32.64%. (Already solved for .95. Already solved for 1.05, therefore just took 1-the area for 1.05 and got 32.64%)

If the objective of the production team is to reduce the probability that the blackness is below 0.95 or above 1.05, I feel that the production team would be better off focusing on the process of improving the blackness to the target value of 1.0 rather than trying to reduce the standard further. The standard deviation is already relatively low at 0.10 which indicates little deviation from the mean value of 1.0 therefore reducing the standard deviation any further is just a waste.

3.e.) The probability that the average blackness of the 25 spots is less than 1.0 is 40.13%. (1.0-1.005/.10/square root of 25= standard error of .02= z-score of -0.25= 40.13%) f.) The probability that the average blackness of the 25 spots is between 0.95 and 1.0 is 39.83%. (Already solved for 1.0=40.13%, therefore .95-1.005/.02= z-score= -0.25= 0.3%. Then subtract the two= 39.83%) g.) The probability that the blackness of the spot is between 1.0 and 1.05 is 58.65%. (Already solved for 1.0=40.13%, therefore 1.05-1.005/.02= z-score= 2.25= 98.78%. Then 9878= 1.22%) i.) If the average blackness of today’s sample of 25...
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