1. The width of a slot of a duralumin forging is (in inches) normally distributed with m = 0.9000
and s= 0.0030. The specification limits were given as 0.9000
±
0.0050.
(a)
What percentage of forgings will be defective?
(b)
What is the maximum allowable value of
σ
that will permit no more than 1 in 100
defectives when the widths are normally distributed with
μ
=0.9000 and
σ
?
1. The weight of anodized reciprocating pistons produced by Brown Company follows a normal
distribution with mean 10 lb and standard deviation 0.2lb.
a. The heaviest 2.5% of the pistons produced are rejected as overweight. What weight, in
pounds, determines the overweight classification? Give your arguments.
Let Y= weight (in lb)
Find the z such that 0.975 = Prob ( Z < (Y10)/0.2) .
This z is 1.96 (from the table). So the weight is Y=
1.96x0.2 + 10 = 10.392. So the weight that
determines the overweight classification is 10.392 lbs.
Prob(Y > 10.392)
10
10.392
Weight
Prob(z > 1.96)
1.96
z
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
b. Suppose Brown Company can sell only those pistons weighing between 9.8 and 10.4 lbs.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '09
 Normal Distribution, Standard Deviation, pistons, PROB

Click to edit the document details