7 points In Pullman one of the largest engineering firms is

(7 points) In Pullman, one of the largest engineering firms is Schweitzer Engineering Labs (SEL). One of the products they produce is 1000 ?m diameter fiber optic cables. The tolerance for the diameter is 0.0015 ?m. Assume the process is normally distributed with a mean of 1000 ?m and variance, = 0.00000075. Any cable with a diameter smaller than the lower spec limit is scrapped at a cost of $1. Any cable with a diameter above the upper spec limit is fixed for a cost of $0.75 per cable. Assuming that 100 cables are produced per day and that there are 250 working days per year, what is the annual cost of fixing plus scrapping?

Solution

m = 1000

var = 0.00000075

sd = sqrt(var) = 0.000866

tolerance = 0.0015

P(d < 999.9985) = P ( z < -0.0015 / 0.000866) = P(z<-1.73) = 0.0418

P(d > 1000.0015) = P(z>0.0015/0.000866) = P(z>1.73) = 0.0418

out of 100 cables,

lower spec. = 100 * 0.0418 = 4.18

higher spec = 100 * 0.0418 = 4.18

scrap cost = 4.18 * 250 * 1 = 1045 $

fixing cost = 4.18 * 250 * 0.75 = 783.75 $

Total cost = 1828.75 $