Question

As part of an insurance company’s training program, participants learn how to conduct an analysis of clients’ insurability. The goal is to have participants achieve a time in the range of 30 to 47 minutes. Test results for three participants were: Armand, a mean of 37.0 minutes and a standard deviation of 3.0 minutes; Jerry, a mean of 38.0 minutes and a standard deviation of 2.0 minutes; and Melissa, a mean of 38.5 minutes and a standard deviation of 2.9 minutes.
a.Which of the participants would you judge to be capable? (Do not round intermediate calculations. Round your answers to 2 decimal places.)

Participants :

Armand: Cpk _____ Cp Capable ? (Click to select)NoYes

Jerry: Cpk _____ Capable ? (Click to select)YesNo

Melissa Cp ________ (Click to select)NoYes

b.Can the value of the Cpk exceed the value of Cp for a given participant?

yes or no

Answer 1

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

$c_p=\frac{UTL-LTL}{6\sigma}$

where σ is  the standard deviation .

The process capability index is given as

$c_p_k=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})$

where μ is the mean.

Armand

Now for Armand, μ is 37.0 , σ is 3.0.

$c_p_{A}=\frac{UTL-LTL}{6\sigma}\\c_p_{A}=\frac{47-30}{6\times 3}\\c_p_{A}=0.94$

$c_p_k_{A}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{A}=min(\frac{47-37}{3\times 3},\frac{37-30}{3\times 3})\\c_p_k_{A}=min(1.1,0.77)\\c_p_k_{A}=0.77$

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.

$c_p_{J}=\frac{UTL-LTL}{6\sigma}\\c_p_{J}=\frac{47-30}{6\times 2}\\c_p_{J}=1.42$

$c_p_k_{J}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{J}=min(\frac{47-38}{3\times 2},\frac{38-30}{2\times 3})\\c_p_k_{J}=min(1.5,1.33)\\c_p_k_{J}=1.33$

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.

$c_p_{M}=\frac{UTL-LTL}{6\sigma}\\c_p_{M}=\frac{47-30}{6\times 2.9}\\c_p_{M}=0.98$

$c_p_k_{M}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{M}=min(\frac{47-38.5}{3\times 2.9},\frac{38.5-30}{3\times 2.9})\\c_p_k_{M}=min(0.977,0.977)\\c_p_k_{M}=0.98$

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.