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aksik [14]
3 years ago
13

Suppose that a process is currently operating at a 3.5-sigma quality level, and it is planned to use improvement projects to mov

e this process to a 6-sigma level? What project improvement rate would be necessary to achieve that new performance in 2 years?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

11,373 PPM

Step-by-step explanation:

The following  formula is used to compute the Proportion of error-free units given the sigma level in Excel.

<em>  The proportion of error-free units = NORMSDIST(Sigma Level - 1.5) </em>

Suppose that a process is currently operating at a 3.5-sigma quality level.

The proportion of error-free units for this process is

                           NORMSDIST(3.5 - 1.5) = 0.977250

Therefore, the rejection rate is equal to

                  1 - 0.977250 = 0.022750

or 22750 PPM.

Now, consider a 6-sigma level process.

The proportion of error-free units equals

                     NORMSDIST(6.0 - 1.5) = 0.9999966

and the rejection rate is

                             1 - 0.9999966 = 3.4 PPM

Hence, reduction is PPM required per year , that is

                              \dfrac{22750 - 3.4}{2} = 11,373 PPM

reduction per year

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=================================================

Explanation:

JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.

See the diagram below.

The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)

We know that JN = 60

Let x = JQ and y = QN

The ratio of x to y is x/y and this is 2/1

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Now use the segment addition postulate

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JQ = 2*y = 2*QN = 2*20 = 40

--------------

We have

JQ = 40 and QN = 20

We see that JQ is twice as larger as QN and that JQ + QN is equal to 60.

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