Answer :
Answer:
Part a
The value of cp and cpk for Armand is 0.94 and 0.77 respectively. As both values are less than 1 so Armand is not capable.
The value of cp and cpk for Jerry is 1.42 and 1.33 respectively. As both values are more than 1 so Jerry is capable.
The value of cp and cpk for Melissa is 0.98 and 0.98 respectively. As both values are less than 1 so Melissa is not capable.
Part b
The value of cpk cannot be more than cp. It can at maximum equal to the value of the cp.
Explanation:
Part a
As per the given data
Lower Tolerance Limit=LTL=30 min
Upper Tolerance Limit=UTL=47 min
The process capability ratio is given as
[tex]c_p=\frac{UTL-LTL}{6\sigma}[/tex]
where σ is the standard deviation .
The process capability index is given as
[tex]c_p_k=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})[/tex]
where μ is the mean.
Armand
Now for Armand, μ is 37.0 , σ is 3.0.
[tex]c_p_{A}=\frac{UTL-LTL}{6\sigma}\\c_p_{A}=\frac{47-30}{6\times 3}\\c_p_{A}=0.94[/tex]
[tex]c_p_k_{A}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{A}=min(\frac{47-37}{3\times 3},\frac{37-30}{3\times 3})\\c_p_k_{A}=min(1.1,0.77)\\c_p_k_{A}=0.77[/tex]
The value of cp and cpk for Armand is 0.94 and 0.77 respectively. As both values are less than 1 so Armand is not capable.
Jerry
Now for Jerry, μ is 38.0 , σ is 2.0.
[tex]c_p_{J}=\frac{UTL-LTL}{6\sigma}\\c_p_{J}=\frac{47-30}{6\times 2}\\c_p_{J}=1.42[/tex]
[tex]c_p_k_{J}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{J}=min(\frac{47-38}{3\times 2},\frac{38-30}{2\times 3})\\c_p_k_{J}=min(1.5,1.33)\\c_p_k_{J}=1.33[/tex]
The value of cp and cpk for Jerry is 1.42 and 1.33 respectively. As both values are more than 1 so Jerry is capable.
Melissa
Now for Jerry, μ is 38.5 , σ is 2.9.
[tex]c_p_{M}=\frac{UTL-LTL}{6\sigma}\\c_p_{M}=\frac{47-30}{6\times 2.9}\\c_p_{M}=0.98[/tex]
[tex]c_p_k_{M}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{M}=min(\frac{47-38.5}{3\times 2.9},\frac{38.5-30}{3\times 2.9})\\c_p_k_{M}=min(0.977,0.977)\\c_p_k_{M}=0.98[/tex]
The value of cp and cpk for Melissa is 0.98 and 0.98 respectively. As both values are less than 1 so Melissa is not capable.
Part b
The value of cpk cannot be more than cp. It can at maximum equal to the value of the cp.
The standard deviation shows that the participant that will be capable is Jerry.
How to explain the standard deviation?
From the information given, the cp of Armand will be:
= (47 - 30) / (6 × 3.0)
= 0.94
The cp of Jerry will be:
= (47 - 30) / (6 × 2)
= 1.42
The cp of Melissa will be:
= (47 - 30) / (6 × 2.9)
= 17/17.4
= 0.98
Therefore, Jerry is capable since the cp is higher than 1.
In conclusion, the value of the Cpk exceed the value of Cp for a given participant.
Learn more about standard deviation on:
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