Six Sigma is comprehensive approach to quality goal setting that involves statistics. An article in Aircraft Engineering and Aerospace Technology (Vol. 76, No. 6, 2004) demonstrate the use of the normal distribution in Six Sigma goal setting at Motorola Corporation. Motorola discovered that the average defect rate for parts produced on an assembly line varies from run-to-run and is approximately normally distributed with a mean equal to 3 defects per million. Assume that the goal at Motorola is for the average defect rate to vary no more than 1.5 standard deviations above or below the mean of 3. How like is it that the goal will be met?

Answer:

0.8664 = 86.64% probability that the goal will be met.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Assume that the goal at Motorola is for the average defect rate to vary no more than 1.5 standard deviations above or below the mean of 3. How like is it that the goal will be met?

It should be within 1.5 standard deviations of the mean, which means that the probability is the pvalue of Z = 1.5 subtracted by the pvalue of Z = -1.5.

Z = 1.5 has a pvalue of 0.9332

Z = -1.5 has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

0.8664 = 86.64% probability that the goal will be met.

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