Amanufacturing process produces a critical part of average length 120 millimeters, with a standard deviation of 3 millimeters. all parts deviating by more than 5 millimeters from the mean must be rejected. what percentage of the parts must be rejected, on average? assume a normal distribution.

9.692%

Step-by-step explanation:

We have been given that a manufacturing process produces a critical part of average length 120 millimeters, with a standard deviation of 3 millimeters. All parts deviating by more than 5 millimeters from the mean must be rejected.

5 millimeters below mean would be and 5 millimeters above mean would be .

Corresponding z values for 115 and 125 would be:

Now, we need to find using normal distribution table.

We know that .

Now, we need to convert 0.09692 into percentage as:

Therefore, 9.692% of parts must be rejected on average.

step-by-step explanation:

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