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Question
Define the mean chart
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Solution
The mean chart `(bar"X" "chart")` is used to show, the quality average of the samples taken from the given process.
The `bar"X"` charts are usually required for decision making to accept or reject the process.
Procedure for `bar"X"`
i. Let X1, X2, X3, etc. be the samples selected each containing “n” observations usually (n = 4.5 or 6)
ii. Calculate mean for each samples `bar"X"_1, bar"X"_2, bar"X"_3`, ......... by using `bar"X"_"i" = (sum"X"_"i")/"n"`, i = 1, 2, 3, 4, ….
Where `sum"X"_"i"` = Total f “n” values included in the sample X1.
iii. Find the mean `(bar"X")` of the sample means
`bar"X" = (sumbar"X")/"Number of samples"`
Where `sumbar"X"` = Total of all the sample means.
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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)
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)
Choose the correct alternative:
Variations due to natural disorder is known as
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A typical control charts consists of
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`bar"X"` chart is a
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R is calculated using
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The upper control limit for `bar"X"` chart is given by
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The LCL for R chart is given by
