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Alex Ar [27]
3 years ago
9

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 ? No/Yes

Jerry: Cpk _____ Capable ? Yes/No

Melissa Cp ________ No/Yes

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

yes or no
Mathematics
1 answer:
hammer [34]3 years ago
3 0

Answer:

1) cpk < 1.33, therefore it is not capable

b) cpk = 1.33, therefore it is capable

c) cpk < 1.33, therefore it is not capable

2) Cpk can never be greater than the Cp, but can be equal to it

Step-by-step explanation:

Upper limit (USL) = 47 minutes and Lower limit (LSL) = 30 minutes

1)

a) mean (μ) = 37 minutes, standard deviation (σ) = 3 minutes

cpk=min(\frac{USL-\mu}{3\sigma}, \frac{\mu - LSL}{3\sigma})=min(\frac{47-37}{3*3},\frac{37-30}{3*3}  )=min(1.11,0.78)=0.78

cp=(\frac{USL-LSL}{6\sigma})=\frac{47-30}{6*3}=0.94

cpk < 1.33, therefore it is not capable

b) mean (μ) = 38 minutes, standard deviation (σ) = 2 minutes

cpk=min(\frac{USL-\mu}{3\sigma}, \frac{\mu - LSL}{3\sigma})=min(\frac{47-38}{3*2},\frac{38-30}{3*2}  )=min(1.5,1.33)=1.33

cp=(\frac{USL-LSL}{6\sigma})=\frac{47-30}{6*2}=1.42

cpk = 1.33, therefore it is capable

c) a) mean (μ) = 38.5 minutes, standard deviation (σ) = 2.9 minutes

cpk=min(\frac{USL-\mu}{3\sigma}, \frac{\mu - LSL}{3\sigma})=min(\frac{47-38.5}{3*2.9},\frac{38.5-30}{3*2.9}  )=min(0.98,0.98)=0.98

cp=(\frac{USL-LSL}{6\sigma})=\frac{47-30}{6*2.9}=0.98

cpk < 1.33, therefore it is not capable

2) Cpk can never be greater than the Cp, but can be equal to it

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The missing options are:

A.\ f(20) =5,000

B.\ f(5000) = 20

C. In the xy plane, point (20, 5,000) is on the graph y = f(x).

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<em />

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