# Baria Case Study

Topics: Extrusion, Standard deviation Pages: 3 (739 words) Published: October 11, 2012
Six Sigma At Flyrock Homework 28 Sept 2012

1.  If the extruder setting is accurate, what proportion of the rubber extruded will be within specifications? m = 400
σ = 4
P(x<410) = P(z<410-400/4) = P(z<2.5) = 0.9938
P(x<390) = P(z<390-400/4) = P(z<-2.5) = 0.0062
P(-2.5<z<2.5) = 0.9938 – 0.0062 = 0.9876
Therefore, 98.76% of the rubber extruded will be within specifications.
2.  Douglas has asked to take a sample of 10 sheets of rubber each hour from the extruder and measure the thickness of each sheet.  Based on the average thickness of the sample, operators will decide whether the extrusion process is in control or not.  Given that Douglas plans Three Sigma control limits, what upper and lower control limits should she specify to the operators? σx = 4/100.5 = 1.265

UCL = 400 + 3*1.265 = 403.795
LCL = 400 - 3*1.265 = 396.205

3.  If a bearing is worn out, the extruder produces a mean thickness of 403 thou when the setting is 400 thou.  a)      Under this condition, what proportion of defective sheets will the extruder produce?  P(x<410) = P(z<410-403/4) = P(z<1.75) = 0.9599

P(x<390) = P(z<390-403/4) = P(z<-3.25) = 0.0006
P(-3.25<z<1.75) = 0.9599 – 0.0006 = 0.9593             Proportion of defective sheets =  1 – 0.9593 = 0.0407 = 4.07%
b)      Assuming the control limits in question 2, what is the probability that a sample taken from the extruder with the worn bearings will be out of control?  P(In Control) = P(396.205<x<403.795) = P(396.205-403/1.265<x<403.795-403/1.265)                       = P(-5.37<z<0.628) = 0.7357 P(Out of Control) = 1 – 0.7357 = 0.2643 = 26.43%

c)      On average, how many hours are likely to go by before the worn bearing is detected? Proportion of defective sheets = 0.0407
~ 1 in 25 sheets is defective.
=> 2.5 Hours may go...