{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework6-key-2011

homework6-key-2011 - 1 The width of a slot of a duralumin...

This preview shows pages 1–3. Sign up to view the full content.

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 < (Y-10)/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.

{[ snackBarMessage ]}

Page1 / 3

homework6-key-2011 - 1 The width of a slot of a duralumin...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online