Question

Broody Plastics produces plastic bottles for the customer orders. Statistical process control charts is being used to monitor all the process. assume that the sample size is 4 and the specification for the bottle neck diameter is 6.00+-0.35.

(Note that the central line of chart for the sample means is set at 6.00 and the range at 0.4 in.)

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

(b) If the firm is looking for the three-sigma performance, is the process capable of producing the bottle?

(The standard deviation of process distribution is 0.25 in.)

(c) If the process is not capable, what percent of the output will fall outside the specification limits?

Answer 1

(1.)

Mean () = 6

Range () = 0.4

Given sample size (n) = 4

CONTROL LIMITS FOR MEAN CHART

*Use mean factor table to find value of A2 at n=4

Value of A2 = 0.729

UCL = 6 + (0.729 × 0.4)

UCL = 6.2916

LCL = 6 - (0.729 × 0.4)

LCL = 5.7084

CONTROL LIMITS FOR RANGE CHART

*Use mean factor table to find value of D3 & D4 at n=4

Value of D3 = 0

Value of D4 = 2.282

UCL = 2.282 × 0.4

UCL = 0.9128

LCL = 0 × 0.4

LCL = 0

(2.)

Upper Specification Limit (USL) = 6 + 0.35 = 6.35

Lower Specification Limit (LSL) = 6 - 0.35 = 5.65

Standard deviation of process () = 0.25

Let Process Capability = Cp

Cp = (6.35 - 5.65)/(6 × 0.25)

Cp = 0.467

Since Cp < 1, therefore the process is not capable.

(3.)

Z = (6 - 5.65)/0.25

Z = 1.4

* Use one sided Z table to find value of probability at Z = 1.4.

Probability = 0.4192

For one sided, the defective percentage is given by:

Defective percentage = 100 × (0.5 - Probability)

Defective percentage = 100 × (0.5 - 0.4192)

Defective percentage = 8.08%

For two tailed::

​​​​​​Total defective percentage = 2 × 8.08%

Total defective percentage = 16.16%

Therefore, 16.16% of the output falls outside the specification limits.