Question

Aspen Plastics produces plastic bottles to customer order. To
monitor the

process, statistical process control charts are used. The central
line of the chart for

the sample means is set at 8.50 and the range at 0.31in. Assume
that the sample

size is 6 and the specification for the bottle neck diameter is
8.50 ± 0.25.

a. Calculate the control limits for the mean and range charts. (3
points)

b. Suppose that the standard deviation of the process distribution
is 0.13 in.

If the firm is seeking three-sigma performance, is the process
capable of

producing the bottle? (4 points)

c. If the process is not capable, what percent of the output will
fall outside the

specification limits? (3 points

Answer #1

**Solution:**

Let,

(X=) = Overall Mean = 8.50

R- = Range Bar = 0.31

A2 = Constant value derived from the control chart constant value table = 0.483

D3 = Constant value derived from the control chart constant value table = 0

D4 = Constant value derived from the control chart constant value table = 2.004

U = Upper Specification Limit = 8.50 + 0.25 = 8.75

L = Lower Specification Limit = 8.50 - 0.25 = 8.25

**Answer a:**

**Control Limits for Mean
Chart:**

CL = Average Mean (X=) = 8.50

UCL = (X=) + (A2 * R-)

= 8.50 + (0.483 * 0.31)

= 8.6497 (Rounded to 4 decimal places)

LCL = (X=) - (A2 * R-)

= 8.50 - (0.483 * 0.31)

= 8.3503 (Rounded to 4 decimal places)

**Control Limits for Range
Chart:**

CL = Average Range (R-) = 0.31

UCL = D4 X R- = 2.004 X 0.31 = 0.6212 (Rounded to 4 decimal places)

LCL = D3 X R- = 0 X 0.31 = 0

**Answer b:**

σ = Std Dev. = 0.13

Here, we will calculate Cp and Cpk as mentioned below:

Where,

So,

Cpk = Min (0.641,0.641) = 0.641

As Cpk < 1, The process is not capable of meeting the desired specifications.

**Answer c:**

i) % of output falling outside the Upper Specification Limit:

= (8.75 - 8.50) / 0.13

= 1.9231

So,

% of units above USL = Probability value derived from the standard normal table for Z_USL = 0.0272 or 2.72%

ii) % of output falling outside the Lower Specification Limit:

= (8.50 - 8.25) / 0.13

= 1.9231

% of units below LSL = Probability value derived from the standard normal table for Z_LSL = 0.0272 or 2.72%

Thus, % of total units out of the desired specification limits = 2.72 + 2.72 = 5.44 % (Rounded to the two decimal places)

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this answer, if you found it useful)**

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