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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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View Full Document(c) To assess normality we can look at a histogram of the data.
From Minitab we get
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 Spring '10
 BillNotz
 Statistics

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