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प्रश्न
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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उत्तर
We first find the sample mean and range for each of the 8 given samples.
| Sample Number | Observations | `bar"X"` | R | ||
| 1 | 32 | 36 | 42 | 36.67 | 10 |
| 2 | 28 | 32 | 40 | 33.33 | 12 |
| 3 | 39 | 52 | 28 | 39.67 | 24 |
| 4 | 50 | 42 | 31 | 41 | 19 |
| 5 | 42 | 45 | 34 | 40.33 | 11 |
| 6 | 50 | 29 | 21 | 33.33 | 29 |
| 7 | 44 | 52 | 35 | 43.67 | 17 |
| 8 | 22 | 35 | 44 | 33.67 | 22 |
| Total | 301.67 | 144 | |||
The control limits for `bar"X"` chart is
`\overset{==}{"X"} = (sumbar"X")/"Number of samples" = 301.67/8` = 37.71
`bar"R" = 144/8` = 18
UCL = `\overset{==}{"X"} + "A"_2 bar"R"`
= 37.71 + (0.58)(18)
= 37.71 + 10.44
= 48.15
CL = `\overset{==}{"X"}` = 37.71
LCL = `\overset{==}{"X"} - "A"_2 bar"R"`
= 37.71 – (0.58)(18)
= 37.71 – 10.44
= 27.27
The control limits for Range chart is
UCL = `"D"_4 bar"R"` = 2.115(18) = 38.07
CL = `bar"R"` = 18
LCL = `"D"_3 bar"R"` = 0(18) = 0
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