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Product specifications call for a part at Vaidy​ Jayaraman's Metalworks to have a length of 1.200 " plus or minus . 070 ". ​Currently, the process is performing at a grand average of 1.2​00" with a standard deviation of 0.010​". Calculate the capability index of this process. The Upper C Subscript pk of this process is nothing ​(round your response to two decimal​ places).

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Answer:

Step-by-step explanation:

Upper specification limit (USL) = 1.200+0.070 = 1.270

Lower specification limit (LSL) = 1.200-0.070 = 1.130

Mean = 1.200

Standard deviation (SD) = 0.010

Cpl = (Mean - LSL) / (3 x SD) = (1.200-1.130)/(3x0.010) = 0.070/0.03 = 2.33

Cpu = (USL - Mean) / (3 x SD) = (1.270-1.200)/(3x0.010) = 0.070/0.03 = 2.33

Process capability index (Cpk) = Minimum of (Cpu, Cpl)

= Minimum of (2.33, 2.33)

= 2.33

The capability index of this process is 2.33.

Given

Product specifications call for a part at Vaidy​ Jayaraman's Metalworks to have a length of 1.200 " plus or minus . 070 ".

Capability index process

Process capability index (Cpk) is a statistical tool, to measure the ability of a process to produce output within customer’s specification limits.

The upper specification limit (USL) = 1.200 + 0.070 = 1.270

And the lower specification limit (LSL) = 1.200 - 0.070 = 1.130

The value of CPL is;

[tex]\rm CPL=\dfrac{Mean - Lower \ specification \ limit }{3 \times Standard \ deviation}\\\\CPL= \dfrac{1.200-1.130}{3 \times 0.010}\\\\CPL=\dfrac{0.070}{0.03}\\\\CPL=2.33[/tex]

Hence, the capability index of this process is 2.33.

To know more about the capability index process click the link given below.

https://brainly.com/question/7052971

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