Calculates the inverse of the NORMSDIST function.
NORMSINV(p)
NORMSINV returns the value x, such that NORMSDIST(x) is p.
It is equivalent to NORMINV(p, 0, 1).
NORMSINV(0.5)
returns 0. Half of the standard normal distribution is below 0.
NORMINV(0.5, 0, 1)
also returns 0.
Setting Production Quality Control Limits
Scenario: A company produces machine parts, and the diameter of these parts is normally distributed. The company wants to set up a quality control system to identify parts that are too large or too small. They have determined that any part falling outside the middle 99.7% of the distribution is considered defective and needs to be inspected.
Goal: The quality control manager needs to find the Z-scores that correspond to the cutoff points for the top 0.15% and bottom 0.15% of the production. This will help them set up automated sensors that flag any part with a diameter corresponding to these Z-scores.
Calculations:
Table of Results:
Cumulative Probability (p) | Z-Score (NORMSINV(p)) | Interpretation | ||
|---|---|---|---|---|
A | B | C | ||
1 | 0.0015 | -2.968 | This is the Z-score for the lower quality control limit. Any part with a diameter at or below this Z-score is in the bottom 0.15% and considered defective. | |
2 | 0.9985 | 2.968 | This is the Z-score for the upper quality control limit. Any part with a diameter at or above this Z-score is in the top 0.15% and considered defective. |
Conclusion:
Based on the NORMSINV function, the quality control manager knows that any part with a diameter that results in a Z-score less than -2.968 or greater than 2.968 is a defect. They can now set up their automated inspection equipment to flag any part that falls outside these limits, ensuring that only high-quality parts are approved for sale.
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