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Question
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.
Solution
| Sample No | Sample Mean `bar"X"` |
Sample Range R |
| 1 | 1080 | 410 |
| 2 | 1390 | 670 |
| 3 | 1460 | 180 |
| 4 | 1380 | 320 |
| 5 | 1230 | 690 |
| 6 | 1370 | 450 |
| 7 | 1310 | 380 |
| 8 | 1630 | 350 |
| 9 | 1580 | 270 |
| 10 | 1510 | 660 |
| 11 | 1270 | 440 |
| 12 | 1200 | 310 |
| Total | `sum"X"` = 16410 | `sum"R"` = 5130 |
The control limits for `bar"X"` chart is
`\overset{==}{"X"} = (sum"X")/"Number of sample" = 16410/12` = 1367.5
`bar"R" = (sum"R")/"n" = 5130/12` = 427.5
UCL = `\overset{==}{"X"} + "A"_2 bar"R"`
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.6675
= 1614.1675
= 1614.17
CL = `\overset{==}{"X"} ` = 1367.5
LCL = `\overset{==}{"X"} + "A"_2 bar"R"`
= 1367.5 – 0.577(427.5)
= 1367.5 – 246.6675
= 1120.8325
= 1120.83
The control limits for Range chart is
UCL = `"D"_4bar"R"`
= 2.115(427.5)
= 904.1625
= 904.16
CL = `bar"R"` = 427.5
LCL = `"D"_3bar"R"`
= 0(427.5)
= 0
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RELATED QUESTIONS
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Ten samples each of size five are drawn at regular intervals from a manufacturing process. The sample means `(bar"X")` and their ranges (R) are given below:
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| `bar"X"` | 49 | 45 | 48 | 53 | 39 | 47 | 46 | 39 | 51 | 45 |
| R | 7 | 5 | 7 | 9 | 5 | 8 | 8 | 6 | 7 | 6 |
Calculate the control limits in respect of `bar"X"` chart. (Given A2 = 0.58, D3 = 0 and D4 = 2.115) Comment on the state of control
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)
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)
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:
Variations due to natural disorder is known as
Choose the correct alternative:
A typical control charts consists of
Choose the correct alternative:
R is calculated using
