Motorola used the normal distribution to determine the probability of defects and the number
of defects expected in a production process. Assume a production process produces
items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected
number of defects for a 1000-unit production run in the following situations.
a. The process standard deviation is .15, and the process control is set at plus or minus
one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces
will be classified as defects.
b. Through process design improvements, the process standard deviation can be reduced
to .05. Assume the process control remains the same, with weights less than 9.85 or
greater than 10.15 ounces being classified as defects.
c. What is the advantage of reducing process variation, thereby causing process control
limits to be at a greater number of standard deviations from the mean?

[tex]\mu=10,\sigma=0.15\\Therefore:-\\z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.15}\approx-1.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.15}\approx+1.0\\P(x<9.85/x>10.15)=P(Z<-1.0/Z>+1.0)\\=2P(Z<-1.0)=2(0.1587)=0.3174\\[/tex]

Total defects=Production*Probability

=0.3174*1000

[tex]=317.4[/tex]

b. [tex]\mu=10,\sigma=0.05[/tex]

therefore:-

[tex]z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.05}\approx-3.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.05}\approx+3.0\\\\=P(x<9.85/x>10.15)=P(Z<-3.0/Z>+3.0)\\=2P(Z<--3.0)=2(0.0013)\\=0.0026\\[/tex]

Defects=Probability*Production

=0.0026*1000

=2.6

c.Reducing process variation results in a significant reduction in the number of unit defects.