Page 221 - DMGT524_TOTAL_QUALITY_MANAGEMENT

Total Quality Management

Notes For the Range Chart
The center line = The average of all the sample ranges (R-bar) = 0.425.
UCL = D × (R-bar) = 2.283 × 0.425 = 0.97
4
LCL = D × (R-bar) = 0 × 0.425 = 0
3
For X-bar or Average Chart

The center line (X-double bar) = is the average of all the averages of the 25 sample points = 1.043
UCL = X-double bar + A × (R-bar).
2
= 1.043 + (0.729 × 0.425) = 1.486
LCL = X-double bar – A × (R-bar)
2
= 1.05 – (0.729 × 0.425) = 0.60
Where the constants D , D , and A are based on the sample size of each sample. Any text on the
3 4 2
subject will have a table for these values with respect to the sample size n.
Illustration 1: Measurements on averages ( X ) and ranges (R) from 20 samples each of size 5
7.0. Determine the values of the control limits for
gave the following results. X = 99.6 , R =
drawing a mean chart.
Given that for n = 5, mean range = 2.32 × population S.D.

Solution: Here we are given: X = 99.6, R = 7.0

R 7
Now, R = 2.32σ ⇒ ˆ σ = = = 3.0172
2.32 2.32
3s control limits for mean chart.
⎛ ˆ σ ⎞ ⎛ 3.0172 ⎞
UCL = X + 3 ⎜ ⎟ = 99.6 + ⎜ 3 × ⎟
X ⎝ n ⎠ ⎝ 5 ⎠

⎛ 9.0516 ⎞
= 99.6 + ⎜ ⎟ = 99.6 + 4.0479 = 103.6479
⎝ 2.2361 ⎠
⎛ ˆ σ ⎞
LCL = X − ⎜ ⎟ = 99.6 − 4.0479 = 95.5521
X ⎝ n ⎠

CL = X = 99.6
X
Illustration 2: The following figures give the number of defectives in 20 samples. Containing
2000 items. 425, 430, 216, 341, 225, 322, 280, 306, 337, 305, 356, 402, 216, 264, 126, 409, 193, 280, 389.
Calculate the values for central line and the control limits for P-chart (Fraction defectives chart).
Solution: Total number of defectives out of 40,000 items in sample is:
Σd = 425 + 430 + …. + 389 = 6,148

P = 6,148 ÷ 40,000 = 0.1537
For p-chart
CL = P = 0.1537.
P
LCL = P − 3 P (1 − P)ln
P

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