Question

Motorola used the normal distribution to determine the probability of defects and the number

Answer 1

Answer:

a.P<x<9.85 orx>10.15)=0.3174, Total defects=317.4

b.p=0.0026,total defects=2.6

c.Less of the items produced will be classified as defects.

Step-by-step explanation:

a.The standard score,z, is esentially x reduced by process mean then divided my process standard deviation.

$\mu=10,\sigma=0.15\\Therefore:-\\z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.15}\approx-1.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.15}\approx+1.0\\P(x10.15)=P(Z+1.0)\\=2P(Z$

Total defects=Production*Probability

                      =0.3174*1000

                       $=317.4$

b. $\mu=10,\sigma=0.05$

therefore:-

$z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.05}\approx-3.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.05}\approx+3.0\\\\=P(x10.15)=P(Z+3.0)\\=2P(Z$

Defects=Probability*Production

            =0.0026*1000

             =2.6

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