Page 181 - DMGT501_OPERATIONS_MANAGEMENT

Unit 6: Quality Assurance and Control

Solution: Notes
k = Number of samples = 10
n = Number of observations in each sample = 5
X = X/10 = 34.545/10 = 3.4545
R = 0.441/10 = 0.0441
For X chart

UCL = X + A X R = 3.4545 + 0.577 × 0.0441
2
= 3.48

LCL = X – A X R = 3.4545 – 0.577 × 0.0441
2
= 3.43
For R chart
UCL = D × R = 2.115 × 0.0441
4
= 0.013

LCL = D × R = 0 × 0.0441
3
= 0

UCL

X
1 2 3 4 5 6 7 8 9 10 Sample No.
LCL

Example: Draw the control charts for X (mean) and R (range) for the above example with
the following information:

For n = 5, d = 2.326
2
d = 0.864
3
Solution:

X = X / 10 = 34.545/10
= 3.4545
= 0.441/10 = 0.0441

For X chart

UCL = X + 3 × R /(d / n )
2
= 3.4545 + 3 × 0.0441/( 2.326 / 5 )
= 3.4545 + 0.1268
= 3.5813
LCL = X – 3 × R / (d / n )
2
= 3.4545 – 3 × 0.0441 / (2.326/ 5 )
= 3.3277

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