Question

If a process has a variance of 4 units and a specification of 96 ± 4, what is the process performance index (Pp)?

Option 1: 0.33
Option 2: 0.66
Option 3: 1.00
Option 4: 1.50

Answer 1

The process performance index (Pp) is calculated using the formula Pp = (USL - LSL) / (6 √ variance). For a variance of 4 and a specification of 96 ± 4, Pp is approximately 0.67. Thus, the correct answer from the provided options is 0.66.

Step-by-step explanation:

To answer the question about the process performance index (Pp), we need to use the formula for Pp which is Pp = (USL - LSL) / (6 √ variance). Here, USL stands for upper specification limit, LSL stands for lower specification limit, and the variance is the square of the standard deviation (σ^2). Given that the variance is 4 units (σ^2=4, hence σ=2), and the specification is 96 ± 4, we can calculate USL as 96 + 4 = 100 and LSL as 96 - 4 = 92.

Now we can substitute the values into the Pp formula:

Pp = (100 - 92) / (6 × 2) = 8 / 12 = 0.6667

After calculating, we find that the process performance index (Pp) is 0.6667, which can be approximately considered as 0.67 when rounded to two decimal places. Therefore, out of the given options, the closest to our calculated value is Option 2: 0.66.