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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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(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

Step-by-step explanation:

We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.

Let Probability that item are defective = P(D) = 0.20

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Probability of items being repaired from the given defective items = P(R/D) = 0.60

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(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.

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           p = probability of success which in our question is % of randomly

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<em>LET X = Number of items that are defective and cannot be repaired</em>

So, it means X ~ Binom(n=20, p=0.08)

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                 = 0.2711

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