SOLUTION: I was looking for some assistance, i've been struggling with Z-scores and would like to see if these are correct. Thanks. The production department of the newspaper has embar

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Question 909240: I was looking for some assistance, i've been struggling with Z-scores and would like to see if these are correct.
Thanks.
The production department of the newspaper has embarked on a quality improvement effort and has chosen as its first project an issue that relates to the blackness of the newspaper print. Each day a determination needs to be made concerning how “black” the newspaper is printed. This is measured on a standard scale in which the target value is 1.0. Data collected over the past year indicate that the blackness is normally distributed with an average of 1.005 and a standard deviation of 0.10.
Each day, one spot on the first newspaper printed is chosen and the blackness of the spot is measured. Suppose the blackness of the newspaper is considered acceptable if the blackness of the spot is between 0.95 and 1.05. Assuming that the distribution has not changed from what it was in the past year, what is the probability that the blackness of the is
Z=x-u/o U=Mean, O= Standard Deviation X= Sample
U= a,b,c, or d, O=.10 X=1.005
Once I found Z, I used the Z-score table to solve my answers
a. less than 1.0? (Z-score=.05) (.5199) = or 51.99%
b. between 0.95 and 1.0? (.95 Z-score=.55 , 1.0 Z-score =.05) .1899 or 18.99%
c. between 1.0 and 1.05? (1.0 Z-score =.05, 1.05 Z-score =.-.45) -.1537 or 15.37%
d. less than 0.95 or greater than 1.05? (.95 Z-score=.55, 1.05 Z-score =.-.45) .3824 or 38.24%




If the objective of the production team is to reduce the probability that the blackness is below 0.95 or above 1.05, would it be better off focusing on process improvement that lowered the blackness to the target value of 1.0 or on process improvement that reduced the standard deviation to 0.075? Please explain.

Z=x-u/o U=Mean, O= Standard Deviation X= Sample

U= .95 and 1.0, x=1.005, O=.075

(1.05 Z-Score=-.6, .95 Z-score=.733)
.
7637-.2743= 48.94 or 48.9%
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi

 = 1.005, = .10

*Note: 

The idea (the x-value - the mean) and then divide by sd to find z

a) P(X< 1) = P(z < (1 - 1.005)/.10) = P(z < -.05)  = 48.1%

b) P(.95 < x < 1) = P(x<1) - P(x < .95) = P(z <-.05) - P(z < -.55) =  48.1% - 29.12%= 19.26%

19.26% the Area Under the Standard Normal curve between the Green Lines



c) P(1 < x < 1.05) = P(x <1.05) - P(x <1 ) = P(z<.45) - p(z <-.05)  = 67.36 - 48.1 = 

19.26% the Area Under the Standard Normal curve between the Green Lines



d) P(x<.95 0r x>1.05) = P(z<-.55) + [1 - P(z < .45)] = 29.12 + (1-.6736) = .2912 + .3264  = ..6176 0r 61.76%

61.76% the Area Under the Standard Normal curve Outside the Green Lines

 

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