Question

Design specifications require that a key dimension on a product measure 101 ± 16 units. A process being considered for producing this product has a standard deviation of five units.

a. What can you say (quantitatively) regarding the process capability? Assume that the process is centered with respect to specifications. (Round your answer to 4 decimal places.)

Process capability index  ????

b. Suppose the process average shifts to 96. Calculate the new process capability. (Round your answer to 4 decimal places.)

New process capability index ????

c. What is the probability of defective output after the process shift? (Use Excel's NORM.S.DIST() function to find the correct probability. Round "z" values to 2 decimal places. Round probabilities to 4 decimal places (0.####).)

Probability of defective output ​​​​​​​ ???

Answer 1

Given Informations:-

USL =101+16 =117 Units

LSL =101-16 =85 Units

Process is centered with specification , means that the actual mean is equal to specified mean.

Therefore , Xbar=101 Units

Standard Deviation($\sigma$) = 5 units

(a)

We have process capability index

Cp =(USL -LSL) /6$\sigma$ = (117-85)/6*5 = 32/30 =1.07

Cp=1.06

And, Cpk = Min of ( (USL-Xbar)/3$\sigma$ or (Xbar-LSL)/3$\sigma$=Min of (117-101)/3*5 or (101-85)/3*5)

Cpk= Min of ( 1.07 or 1.07)

Cp=1.07

Cpk=1.07

The process Capability index shows that the process is well within specification limits with 99.73% conformance.

(b)

If the average shifts to 96 , Cp will be the same but Cpk will change.

New Cpk = Min of ( ( USL-Xbar)/3$\sigma$ or (Xbar - LSL)/3$\sigma$ ) =Min of ( 117-96)/3*5 or (96-85)/3*5 )

Cpk = Min of ( 1.4 , 0.73 )

Therefore Cpk =0.73

(c)

We have Z= (x - Xbar)/$\sigma$= (101-96)/5 =1

NORMDIST(x, mean, standard deviation, z) is the actual form of NORM.S.DIST()

X= 101

Mean=96

Standard Deviation ($\sigma$) =5

z=1

Therefore return of NORMDIST(101,96,5,1)= 0.8413

NORMDIST(X,MEAN,S.D,Z)=NORMDIST(101,96,5,1)=					0.8413