Question 1209556
Here's how to solve this problem:

**Understanding the Concept**

The tolerance represents the acceptable variation in pencil length.  A smaller tolerance means more consistent lengths, but also potentially more rejects.  We're looking for the tolerance that, when combined with the original mean, would result in the same *proportion* of acceptable pencils as the *new* tolerance.

**Steps**

1. **Determine the Z-score:**  The tolerance is related to the standard deviation (a measure of how spread out the lengths are) and a Z-score.  The Z-score tells us how many standard deviations away from the mean a particular length is.  We need to find the Z-score that corresponds to the desired percentage of pencils within the tolerance.  Since the problem implies we are looking at a 6-sigma process, we can assume that the percentage of pencils within the tolerance is 99.73%

2. **Calculate the Original Standard Deviation:** We can use the original tolerance to calculate the standard deviation of the process, using the Z-score we calculated above.

3. **Calculate the new standard deviation:** We can use the new tolerance to calculate the standard deviation of the process, using the Z-score we calculated above.

4. **Calculate the Equivalent Tolerance:** Now that we have the standard deviation of the new distribution, we can calculate the tolerance that would produce the same percentage of pencils under the original mean, using the Z-score we calculated above.

**Calculation**

1. **Z-score:** For a 99.73% confidence interval, the Z-score is approximately 3.

2. **Original Standard Deviation:** original_std = original_tolerance / 3

3. **New Standard Deviation:** new_std = new_tolerance / 3

4. **Equivalent Tolerance:** equivalent_tolerance = 3 * original_std

**Result**

The tolerance used to produce the same percentage of pencils before the increase is 0.05 in.