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प्रश्न
The following data show the values of sample mean `(bar"X")` and its range (R) for the samples of size five each. Calculate the values for control limits for mean, range chart and determine whether the process is in control.
| Sample Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Mean | 11.2 | 11.8 | 10.8 | 11.6 | 11.0 | 9.6 | 10.4 | 9.6 | 10.6 | 10.0 |
| Range | 7 | 4 | 8 | 5 | 7 | 4 | 8 | 4 | 7 | 9 |
(conversion factors for n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115)
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उत्तर
| Sample Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
| Mean `bar"X"` | 11.2 | 11.8 | 10.8 | 11.6 | 11.0 | 9.6 | 10.4 | 9.6 | 10.6 | 10.0 | 106.6 |
| Range (R) | 7 | 4 | 8 | 5 | 7 | 4 | 8 | 4 | 7 | 9 | 63 |
The control limits for `bar"X"` chart is
`\overset{==}{"X"} = (sumbar"X")/"Number of samoles" = 106.6/10` = 10.66
`bar"R" = (sum"R")/"R" = 63/10` = 6.3
UCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 10.66 + (0.58)(6.3)
= 10.66 + 3.654 = 14.314
= 14.31
CL = `\overset{==}{"X"}` = 10.66
LCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 10.66 – (0.58)(6.3)
= 10.66 – 3.654
= 7.006
The control limits for Range chart is
UCL = `"D"_4 bar"R"` = 2.115(6.3)`
= 13.3245
= 13.32
CL = `bar"R"` = 6.3
LCL = `"D"_3 bar"R"` = 0(6.3) = 0
Conclusion: Since all the points of sample range is within UCL of R chart, the process is in control.
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संबंधित प्रश्न
Define chance cause
Define assignable cause
What do you mean by process control?
Define a control chart
Write the control limits for the mean chart
Construct `bar"X"` and R charts for the following data:
| Sample Number | Observations | ||
| 1 | 32 | 36 | 42 |
| 2 | 28 | 32 | 40 |
| 3 | 39 | 52 | 28 |
| 4 | 50 | 42 | 31 |
| 5 | 42 | 45 | 34 |
| 6 | 50 | 29 | 21 |
| 7 | 44 | 52 | 35 |
| 8 | 22 | 35 | 44 |
(Given for n = 3, A2 = 1.023, D3 = 0 and D4 = 2.574)
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