An in-control, centered process that follows a normal distribution has a $C _ { p }$ of 3.0. How many standard deviations away from the process mean is the upper specification limit?
A) 9
B) 3
C) 18
D) 6

Question

Quizplus · Accepted Answer

The $C _ { p }$ is defined as:$$C _ { p } = \frac { USL - LSL } { 6 \sigma }$$where $USL$ is the upper specification limit, $LSL$ is the lower specification limit, and $\sigma$ is the standard deviation.In this case, $C _ { p } = 3.0$, so:$$3.0 = \frac { USL - LSL } { 6 \sigma }$$$$USL - LSL = 18 \sigma$$Since the process is centered, $LSL = -USL$. Therefore:$$USL = 9 \sigma$$So the upper specification limit is 9 standard deviations away from the process mean.

An In-Control, Centered Process That Follows a Normal Distribution Has CpC _ { p }Cp​

An In-Control, Centered Process That Follows a Normal Distribution Has $C _ { p }$