Calculate the theoretical process yield using the normal distribution. Is this a reasonable calculation? If yes, why? If no, why not? Answer: ZLSL= (5.95 - 6.024)/ 0.01139 = -6.4969, hence, P( X <…

Question

Options

• AZLSL= (5.95 - 6.024)/ 0.01139 = -6.4969, hence, P( X < -6.4969) = 0.0000 ZUSL= (6.05 - 6.024)/ 0.01139 = 2.2827, hence, P( X > -2.2827) = 0.0112 P (5.95 ≤ X ≤ 6.05) = 0.0000 + 0.0112 = 0.0112 The above calculation is not reasonable because the process is not in statistical control. The above calculation is not reasonable because the process is not in stati

• A
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Explanation

Option A is correct because it accurately performs the Z-score calculations using the process mean (6.024) and standard deviation (0.01139) relative to both specification limits, and correctly identifies the fatal flaw: the calculation is unreasonable because the process is not in statistical control. When a process is out of control, it exhibits special-cause variation, meaning its behavior is unpredictable and the historical mean/standard deviation are not valid descriptors of future output - rendering any normal-distribution-based yield prediction meaningless.

Since only one option is presented here, the focus is on understanding the two-part logic: (1) the arithmetic is mechanically correct, and (2) the interpretation is invalid. A common mistake is accepting the calculation at face value just because the math is sound - but statistical validity requires both correct arithmetic and a stable, controlled process as a prerequisite.

Memory tip: Think "Control before Capability." You must confirm a process is in statistical control (via control charts showing no special causes) before using the normal distribution to estimate yield or process capability (Cp, Cpk). If the process is unstable, capability indices and yield predictions are meaningless - garbage in, garbage out.

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