Question

the specification for a plastic handle calls for a length of 6.0 inches +- .2 inches. The standard deviation of the process is estimated to be 0.05 Inches. what are the upper and lower specification limits for this product.

Answer 1

Answer:

a)

USL = 6.2 inches

LSL = 5.8 inches

b) Cp = 1.33

Cpk = 0.67

c)

Yes it meets all specifications

Step-by-step explanation:

The specification for a plastic handle calls for a length of 6.0 inches ± .2 inches. The standard deviation of the process is estimated to be 0.05 inches. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.1 inches. What is the Cp and Cpk for this process?   Is this process capable of producing the desired part?

Given that:

Mean (μ) = 6.1 inches, Standard deviation (σ) = 0.05 inches and the length of the plastic handle is 6.0 inches ± .2

a) Since the length of the plastic handle is 6.0 inches ± .2  = (6 - 0.2, 6 + 0.2)

The Upper specification limits (USL) = 6 inches + 0.2 inches = 6.2 inches

The lower specification limits (LSL) = 6 inches - 0.2 inches = 5.8 inches

b) The Cp is given by the formula:

$Cp=\frac{(USL-LSL)}{6\sigma} =\frac{(6.2-5.8)}{6*0.05} =1.33$

The Cpk is given by the formula:

c)

The upper specification limit lies about 3 standard deviations from the centerline, and the lower specification limit is further away, so practically all units will meet specifications

$Cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma})=min(\frac{6.2-6.1}{3*0.05},\frac{6.1-5.8}{3*0.05})=min(0.67,2)=0.67$