Page 223 - DMGT524_TOTAL_QUALITY_MANAGEMENT

Total Quality Management

Notes 1
Mean range, R = (7.39) = 0.74
10
Control chart for Mean ( X )
Central line, CL = X = 7.95
X

UCL = X + A R = 7.95 + 0.58 × 0.74 = 8.38
X 2
UCL = X − A R = 7.95 − 0.58 × 0.74 = 7.52
X 2
Control chart for Range (R)

Central line = R
UCL = D R = 2.115 × 0.74 = 1.56
R 4
LCL R = D R = 0 × 0.74 = 0
2
Since all the sample means ( X ) and the sample ranges (R) lie within the corresponding control
limits, the production process is in a state of statistical control.

Illustration 4: In the following data are given the number of defective founded on 24 consecutive
production days in daily samples of 400 items. Draw (i) np-chart and (ii) p chart.

Production day: 1 2 3 4 5 6 7 8 9 10
No. of defections: 20 10 20 24 22 18 38 8 24 54
Production day: 11 12 13 14 15 16 17 18 19 20
No. of defections: 50 18 24 30 16 28 20 8 22 22

Production day: 21 22 23 24
No. of defections: 52 6 20 22

Solution: Total number of units examined.
= No. of samples × Sample size
= 24 × 400 = 9600
Total no. of units defective = 576.

No. of defectives
∴ Average proportion of defectives = P =
No. of units
576
= = 0.06
9600
∴ Average proportion of defectives = 0.06.
(i) np chart

CL = n p = 400 × 0.06 = 24.
np
LCL = np − 3 np ( 1 p− ) = 24 − 3 400 × 0.06 × 0.94 = 9.75
np
UCL = np + 3 np ( 1 p− ) = 24 + 3 400 × 0.06 × 0.94 = 38.25
np

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