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Question
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
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Solution
| Sample number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
| `bar"X"` | 49 | 45 | 48 | 53 | 39 | 47 | 46 | 39 | 51 | 45 | 462 |
| R | 7 | 5 | 7 | 9 | 5 | 8 | 8 | 6 | 7 | 6 | 68 |
The control limit for `bar"X"` chart is
`\overset{==}{"X"} = (sumbar"X")/"No. of samples" = 462/10` = 46.2
`bar"R" = (sum"R")/"No. of samples" = 68/10` = 6.8
UCL = `\overset{==}{"X"} + "A"_2 bar"R"`
= 46.2 + (0.58)(6.8)
= 46.2 + 3.944
= 50.144
= 50.14
CL = `\overset{==}{"X"}` = 46.2
LCL = `\overset{==}{"X"} + "A"_2 bar"R"`
= 46.2 – (0.58)(6.8)
= 46.2 – 3.944
= 42.256
= 42.26
The control limits for Range chart is
UCL = `"D"_4 bar"R"`
= (2.115)(6.8)
= 14.382
= 14.38
CL = `bar"R"` = 6.8
LCL = `"D"_3bar"R"` = (0)(6.8) = 0
Since all individual sample range points fall within the control limits, the process variability is in a state of statistical control.
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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)
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How many causes of variation will affect the quality of a product?
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The assignable causes can occur due to
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A typical control charts consists of
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R is calculated using
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The upper control limit for `bar"X"` chart is given by
