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so as to reduce the uncertainty while leaving the mean ﬁll weight (and the advertised
average net weight) unchanged.
b. If the ﬁrst method is adopted, ﬁnd an acceptable mean ﬁll weight.
SOLUTION:
From part a., one sees that for µ = 8.2, the AUC of the speciﬁcation interval is less that
what will be asked for (0.865). First, one realizes that there could be multiple possible
options for µ. Since one of them will fulﬁll the requirement, we ask which value of µ will
maximize the AUC for the speciﬁcation interval ([8,8.3]). The obvious choice based on the geometric characteristic of normal curve will be the one that is symmetric about µ. So, if
µ = 8.15, the speciﬁcation interval will be symmetric about the mean, yielding the max of
AUC.
.
P(8 < X < 8.3) = P( 8−8115 ≤ Z ≤ 8.3−8.15 )
0.
0 .1
= P(Z ≤ 1.5) − P(Z ≤ −1.5) = 2P(Z ≤ 1.5) − 1 = 2(0.9332) − 1 = 0.8664 > 0.865.
Actually ∀µ ∈ [8.142, 8.158], AUC([8, 8.3]) > 0.865.
c. If the second method is adopted, ﬁnd an acceptable standard deviation of ﬁll weights.
SOLUTION:
To increase the AUC of speciﬁcation interval ([8,8.3]) without shifting the curve, one can
dec...
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 Fall '11
 DANIELCONUS
 Statistics, Standard Deviation

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