The production department of the newspaper has embarked on a quality improvement

Question

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 indicates that the blackness is normally distributed with an average of 1.005 and a standard deviation of 0.10.


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 ? Explain.

Explanation / Answer

mu = 1.005, sigma = 0.10

z = (x - mu)/sigma

(a) z = (1.0 - 1.005)/0.10 = -0.05

P(x < 1.0) = P(z < -0.05) = 0.4801

(b) z = (0.95 - 1.005)/0.10 = -0.55

P(0.95 < x < 1.0) = P(-0.55 < z < -0.05) = 0.1889

(c) z = (1.05 - 1.005)/0.10 = 0.45

(d) P(x < 0.95) = P(z < -0.55) = 0.2912 and P(x > 1.05) = P(z > 0.45) = 0.3264

So, P(x < 0.95 or x > 1.05) = 0.2912 + 0.3264 = 0.6176

Those were for Q1

Now, Q2 ...

If the mean is 1.0, z = (0.95 - 1.0)/0.10 = -0.5 and z = (1.05 - 1.0)/0.10 = 0.5

P(x < 0.95 or x > 1.05) = P(z < -0.5) + P(z > 0.5) = 0.3085 + 0.3085 = 0.6170

If the SD is 0.075, z = (0.95 - 1.005)/0.075= -0.7333 and z = (1.05 - 1.005)/0.075 = 0.6

P(x < 0.95 or x > 1.05) = P(z < -0.7333) + P(z > 0.6) = 0.2317 + 0.2743 = 0.5060

Since 0.5060 < 0.6170, the remedial action should be to reduce the SD to 0.075

DONE

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