Question

please help so lost...

Answer 1

The specification for the length of an aluminum tube is 780 mm ± 1 mm. The manufacturing process that produces the tubes is centered at 780 mm and has a standard deviation of 0.80 mm.

The process is in control and the process output is normally distributed.
Tubes that fail to meet the lower specification limit are scrapped at a loss of $12 per tube. Tubes that exceed the upper specification limit are reworked at a cost of $4.50 per tube.

Determine the expected monetary loss due to nonconforming tubes in a production run of 1,500 tubes. Assume that all process parameters remain as previously described during the production run.

Answer 2

this was hint that professor gave: Prob (X<=779) = Prob (Z<= (779-780)/0.8)

expected loss due to scraping

Prob(X<=779) * 1500 *12

Answer 3

hi

Answer 4

hi it is normally distributed, so gaussian distribution

Answer 5

not really sure how to set it up or start... =(

Answer 6

everything below 779 is scrapped, everything about 781 is reworked
STD 0.8 mm
1mm is how many standard deviaions away?

Answer 7

hey?

Answer 8

.2mm?

Answer 9

no okay so 0.8 mm is one std

Answer 10

and 1mm is more than one STD right

Answer 11

yes.

Answer 12

how many times more

Answer 13

0.8 * what = 1

Answer 14

1.25

Answer 15

yes

Answer 16

okay so you have to see what probability for more than z score of 1.25

Answer 17

hm. do i solve it or is there chart that i can look at?