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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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संबंधित प्रश्न
Define assignable cause
Define the mean chart
A machine is set to deliver packets of a given weight. Ten samples of size five each were recorded. Below are given relevant data:
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| `bar"X"` | 15 | 17 | 15 | 18 | 17 | 14 | 18 | 15 | 1 | 16 |
| R | 7 | 7 | 4 | 9 | 8 | 7 | 12 | 4 | 11 | 5 |
Calculate the control limits for mean chart and the range chart and then comment on the state of control, (conversion factors for n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115)
A quality control inspector has taken ten samples of size four packets each from a potato chips company. The contents of the sample are given below, Calculate the control limits for mean and range chart.
| Sample Number | Observations | |||
| 1 | 2 | 3 | 4 | |
| 1 | 12.5 | 12.3 | 12.6 | 12.7 |
| 2 | 12.8 | 12.4 | 12.4 | 12.8 |
| 3 | 12.1 | 12.6 | 12.5 | 12.4 |
| 4 | 12.2 | 12.6 | 12.5 | 12.3 |
| 5 | 12.4 | 12.5 | 12.5 | 12.5 |
| 6 | 12.3 | 12.4 | 12.6 | 12.6 |
| 7 | 12.6 | 12.7 | 12.5 | 12.8 |
| 8 | 12.4 | 12.3 | 12.6 | 12.5 |
| 9 | 12.6 | 12.5 | 12.3 | 12.6 |
| 10 | 12.1 | 12.7 | 12.5 | 12.8 |
(Given for n = 5, A2 = 0.58, D3 = 0 and D4 = 2.115)
The following data show the values of sample means and the ranges for ten samples of size 4 each. Construct the control chart for mean and range chart and determine whether the process is in control.
| Sample Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| `bar"X"` | 29 | 26 | 37 | 34 | 14 | 45 | 39 | 20 | 34 | 23 |
| R | 39 | 10 | 39 | 17 | 12 | 20 | 05 | 21 | 23 | 15 |
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:
The quantities that can be numerically measured can be plotted on a
Choose the correct alternative:
Variations due to natural disorder is known as
Choose the correct alternative:
The assignable causes can occur due to
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.
