664 HW5 Solns

664 HW5 Solns - Statistics 664 Autumn 2010 Homework 5...

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Statistics 664 – Autumn 2010 Homework 5 Solutions 6.2 (b) From part (a) (previous homework) we saw that x = 10.32 so that the mean of all the actual voltages is 350 + 10.32/10 = 350 + 1.032 = 351.032 and we would estimate the process mean to be ö μ = 351.032. We also saw that R = 6.25 Transforming back to the original units, the range of the voltages would be 6.25/10 = 0.625 Assuming the process was in control (from the previous homework there was no evidence that it was not in control), an estimate of the standard deviation in the voltages is (using the value of d 2 given in Appendix VI for a sample size of n = 4) ö σ = 0.625/d 2 = 0.625/2.059 = 0.3035 Thus, assuming the measurements follow a normal distribution, the voltages of the power supplies would appear to follow a normal distribution with mean 351.032 and standard deviation 0.3035. The probability that the process produces power supplies with voltages between 345 and 355 is P(345 ≤ normal r.v. with mean 351.032 and std. dev. 0.3035 ≤ 355) = P( (345 – 351.032)/0.3035 ≤ std. normal r.v. ≤ (355 – 351.032)/0.3035) = P( –19.875 ≤ std. normal r.v. ≤ 13.074) 2245 1 So the process is almost certain to meet specifications. (This probability can also be computed using software, but should be similar to the above value.)
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(c) To assess normality we can look at a histogram of the data. From Minitab we get
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664 HW5 Solns - Statistics 664 Autumn 2010 Homework 5...

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