Question

Components used in a cellular telephone are manufactured with nominal dimension of .3 mm and lower and upper specification limits of .295 mm and .305 mm respectively. the x bar 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 bar chart at .3015 mm and the center line of the R chart at .00154 mm.

a) Estimate the mean and standard deviation of this process.
b) Suppose that parts below the LSL can be reworked, but parts above the USL 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? Estimate the proportion of scrap and rework produced by the process at this new mean.

Answer 1

Kindly check explanation

Explanation:

Xbar = 0.3015 ; Rbar = 0.00154

USL = 0.295 ; LSL = 0.305

For a subgroup size of 3 ; k = 3 ; d2 = 1.693

The mean = xbar = 0.3015 mm

The standard deviation = Rbar / d2 = 0.00154 / 1.693 = 0.0009096

b) Suppose that parts below the LSL can be reworked, but parts above the USL must be scrapped. Estimate the proportion of scrap and rework produced by this process.

P(X < 0.295)

(x - mean) / standard deviation

(0.295 - 0.3015) / 0.0009096 = 0.00003

0.00003 should be reworked.

P(X > 0.305)

1 - P(X < 0.305)

(x - mean) / standard deviation

(0.305 - 0.3015) / 0.0009096 = 3.847

1 - P(Z < 3.847)

1 - 0.99994

= 0.00006

0.00006 should be reworked.

c) Suppose that the mean of this process can be reset by fairly simple adjustments. What value of the process mean would you recommend? Estimate the proportion of scrap and rework produced by the process at this new mean.

Recommended mean = 0.3000 mm

P(X < 0.295)

(x - mean) / standard deviation

(0.295 - 0.3000) / 0.0009096 = 0.00003

0.00003 should be reworked.

P(X > 0.305)

1 - P(X < 0.305)

(x - mean) / standard deviation

(0.305 - 0.3000) / 0.0009096 = 5.497

1 - P(Z < 5.497)

1 - 0.99997

= 0.00003

0.00003 should be reworked.