For a certain process, R and X control charts based on subgourps of size 5, have
ID: 3178677 • Letter: F
Question
For a certain process, R and X control charts based on subgourps of size 5, have centerlines of 1.163 and 14.4, respectively. Given that the process has specification limits of 12 and 16,
a. Process standard deviation
b. Compute CPU
c. Computer CPL
d. Compute CPK
e. What percentage of products will be out of specifications?
6. For a certain process, R and X control charts based on subgroups of size 5, have centerlines of 1.163and 14.4, respectively. Given that the process has specification limits of 12 and 16, a. Process standard deviation. b. Compute CPU. c. Compute CPI. d. Compute CPK. e. What percentage of products will be out of specification?Explanation / Answer
Given, sub-group size, n = 5; Central Line for Xbar-Chart (i.e., Xdouble bar) = 14.4;Central Line for R- Chart (i.e., Rbar) = 1.163; Lower Specification Limit (LSL) = 12 and Upper Specification Limit (USL) = 16
Part (a)
Process standard deviation, = Rbar/d2, where d2 is a constant that can be obtained from Standard Control Chart Constants, which is 2.326 for n = 5.
So, = 1.163/2.326 = 0.5 ANSWER
Part (b)
CPU measures how close is the process centering to the Upper Specification Limit (USL)
= 16 – 14.4 = 1.6 ANSWER
Part (c)
CPL measures how close is the process centering to the Lower Specification Limit (USL)
= 14.4 – 12 = 2.4 ANSWER
Part (d)
CPK = Min {CPL, CPU} = Min(1.6, 2.4) = 1.6 ANSWER
Part (e)
If X represents the measurement of study, which is assumed to be Normally distributed, proportion of products within specification = P(L X U)
= P[{(L – process average)/Process SD} Z {(U – process average)/Process SD}], where Z is the Standard Normal Variate
= P[(- 2.4/0.5 ) Z (1.6/0.5)] = P(- 4.8 Z 3.2) = P(Z 3.2) - P(Z - 4.8)
= 0.999313 – 0.0000007936 [using Excel Function]
= 0.9993
So, percentage of products out of specifications = 100(1 – 0.9993) = 0.07% ANSWER