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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)
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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संबंधित प्रश्न
Define chance cause
Name the control charts for variables
Write the control limits for the R chart
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)
In a certain bottling industry the quality control inspector recorded the weight of each of the 5 bottles selected at random during each hour of four hours in the morning.
Time | Weight in ml | ||||
8:00 AM | 43 | 41 | 42 | 43 | 41 |
9:00 AM | 40 | 39 | 40 | 39 | 44 |
10:00 AM | 42 | 42 | 43 | 38 | 40 |
11:00 AM | 39 | 43 | 40 | 39 | 42 |
Choose the correct alternative:
`bar"X"` chart is a
Choose the correct alternative:
R is calculated using
From the following data, calculate the control limits for the mean and range chart.
Sample No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Sample Observations |
50 | 21 | 50 | 48 | 46 | 55 | 45 | 50 | 47 | 56 |
55 | 50 | 53 | 53 | 50 | 51 | 48 | 56 | 53 | 53 | |
52 | 53 | 48 | 50 | 44 | 56 | 53 | 54 | 549 | 55 | |
49 | 50 | 52 | 51 | 48 | 47 | 48 | 53 | 52 | 54 | |
54 | 46 | 47 | 53 | 47 | 51 | 51 | 47 | 54 | 52 |
The following data gives the average life(in hours) and range of 12 samples of 5lamps each. The data are
Sample No | 1 | 2 | 3 | 4 | 5 | 6 |
Sample Mean | 1080 | 1390 | 1460 | 1380 | 1230 | 1370 |
Sample Range | 410 | 670 | 180 | 320 | 690 | 450 |
Sample No | 7 | 8 | 9 | 10 | 11 | 12 |
Sample Mean | 1310 | 1630 | 1580 | 1510 | 1270 | 1200 |
Sample Range | 380 | 350 | 270 | 660 | 440 | 310 |
Construct control charts for mean and range. Comment on the control limits.
The following are the sample means and I ranges for 10 samples, each of size 5. Calculate; the control limits for the mean chart and range chart and state whether the process is in control or not.
Sample Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Mean | 5.10 | 4.98 | 5.02 | 4.96 | 4.96 | 5.04 | 4.94 | 4.92 | 4.92 | 4.98 |
Range | 0.3 | 0.4 | 0.2 | 0.4 | 0.1 | 0.1 | 0.8 | 0.5 | 0.3 | 0.5 |