Components used in a cellular telephone are manu- factured with nominal dimension of 0.3 mm and lower and upper specification limits of 0.295 mm and 0.305 mm respectively. The x and R control charts for this process are based on subgroups of size 3 and they exhibit statistical control, with the center line on the x chart at 0.3015 mm and the cen- ter line on the R chart at 0.00154 mm. Required:
a. Estimate the mean and standard deviation of this process.
b. Suppose that parts below the lower specifica- tion limits can be reworked, but parts above the upper specification limit must be scrapped. Estimate the proportion of scrap and rework produced by this process.
c. Suppose that the mean of this process can be reset by fairly simple adjustments. What value of the process mean would you recommend

1. Mean = r bar = 0.0154mm

Standard deviation = 0.0009096

2. Proportion of rework produced = 30ppm or 0.0003%

Proportion of scrap = 60ppm or 0.0006%

3 recommended value = 0.3000

Step-by-step explanation:

The mean = x bar = 0.3015mm

Standard deviation = r bar/d2

R bar = 0.0154 mm

d2 = 1.693mm

= 0.0154mm/1.693mm

= 0.0009096mm

B. Lower specification 0.295

P(x<0.295)

= ϕ(0.295-0.3015/0.0009096)

This gives us 0.00003

So we conclude that 30 ppm or 0.0003% of the parts produced by the process would be reworked

For scrap,

X>0.305 upper specifications

1-ϕ(0.305-0.3015/0.0009096)

= 1-0.99994

= 0.00006

So we conclude that 60 ppm or 0.0006% of the parts produced by the process would be scrapped.

c. to get the recommended value of the process mean

= (0.305+0.295)/2

= 0.6/2

= 0.3000

the recommended value would be 0.3000

i not know

step-by-step explanation:

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: