HW assignment 3.1
1.
a) Tyler Plant = 133/200 = .665 = 66.50%
b) Mechanical problem = 108/200 = .54 = 54.00%
c) Lincoln and electrical problem = (67/200) x (92/200) = .1541 = 15.41%

2.
a) both color machines = (.20) x (.20) = .04 = 4.0%
b) 1 - .04 = .96 = 96%
c) yes it is sufficient (.1)x(.1)x(.1)x(.2)x(.2) = .00004 1-.00004 = .99996 = 99.996%
d)(.9)x(.9)x(.9)x(.8)x(8) = .4666 = 46.66% (.9)x(.9)x(.9)x(.9)x(.8)x(8) = .4199 = 41.99%

3.
a) 100/299 = .3344 = 33.44%
b) 30/299 = .1003 = 10.03%
c) (100/299) + (75/299) = .5853 = 58.53%
d) (31/75) = .4133 = 41.33%

4.
a).4420= 44.20%
b).1215 = 12.15%
c) I would suggest increasing the population size to a larger amount. I think that would help create a more accurate representation of the results and help tell production managers when they need to stop production.

5.
a) .988 = 98.8%
b) .929 = 92.9%
c) It favors the supplier. It means that 98.8% of the time the shipment will be kept if only 5% of the shipment is defected. But Ziteck will keep the shipment 92.9% of the time if 10% of the shipment is defected. I think that Ziteck needs to change there sampling size and create a lower cutoff point to make sure they are not receiving extra defects in a shipment.

6.
a) .7161 - .2838 = .4323 = 43.23%
b) 12.03 would be the average. Setting the average to above 12.00 will decrease your chances of getting fined but also increases your chances of over filling your soda cans, which can cause your company to lose money.

7
a).691 - .158 = .533 = 53.3%
b) .001