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a)
| Sample | x1 | x2 | x3 | x4 | x5 | X-bar | Range |
| 1 | 27 | 18 | 20 | 23 | 19 | 21.4 | 9.0 |
| 2 | 22 | 25 | 31 | 40 | 17 | 27.0 | 23.0 |
| 3 | 16 | 15 | 22 | 19 | 23 | 19.0 | 8.0 |
| 4 | 35 | 27 | 16 | 20 | 24 | 24.4 | 19.0 |
| 5 | 21 | 33 | 45 | 12 | 22 | 26.6 | 33.0 |
| 6 | 17 | 15 | 22 | 20 | 30 | 20.8 | 15.0 |
| 7 | 25 | 21 | 26 | 33 | 19 | 24.8 | 14.0 |
| 8 | 15 | 38 | 23 | 25 | 31 | 26.4 | 23.0 |
| 9 | 31 | 26 | 24 | 35 | 32 | 29.6 | 11.0 |
| 10 | 28 | 23 | 29 | 20 | 27 | 25.4 | 9.0 |
| Avg. | 24.54 | 16.40 | |||||
| X_double-bar | R-bar | ||||||
| A2 | 0.577 | ||||||
| D3 | 0 | ||||||
| D4 | 2.114 | ||||||
| UCLx | 34.003 | ||||||
| LCLx | 15.077 | ||||||
| UCLr | 34.670 | ||||||
| LCLr | 0 | ||||||



The process seems to be in control in that none of the estimates of mean or ranges overshoot the limits.
(b)
R-bar = 16.40
Constant d2 for n=5 is 2.326
Estimated process standard deviation, σ = R-bar / d2 = 16.40 / 2.326 = 7.05
The process mean is estimated by the X_double bar, so, μ = 24.54
USL = 25+5 = 30 minutes
LSL = 25 - 5 = 20 minutes
Since the process mean is shifted towards the LSL,
Cpk = (μ - LSL) / 3.σ = (24.54 - 20)/(3*7.05) = 0.215
Cp = (USL - LSL) / 6σ = (30 - 20)/(6*7.05) = 0.236
So, the process in incapacble of producing even 3-sigma quality in that both the parameters are less than 1.0.
To, improve, the major step should be to reduce the variability i.e. the standard deviation.
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