A quality analyst wants to construct a control chart for determining whether thr

Question

A quality analyst wants to construct a control chart for determining whether three machines, all producing the same product, are under control with regard to a particular quality variable. Accordingly, he sampled four units of output from each machine, with the following results?   

What are the Mean chart three-sigma upper and lower control limits?

a) 22 and 18

b) 23.16 and 16.84

c) 22.29 and 16.71

d) 23.5 and 16.5

e) 24 and 16

Machine 1 Measirement 17 15 15 17 Machine 2 Measurement 16 25 18 25 Machine 3 Measurement 23 24 23 22

Explanation / Answer

Option b is correct option

UCL = Average(X) + 3*Sigma(X)

LCL = Average(X) - 3*Sigma(X)

Mean for Machine 1 =17+15+15+17/4=64/4=16

Mean of Machine 2 = 16+25+18+25=84/4=21

Mean of Machine 3= 23+24+23+22=92/4=23

We have to find Standard Deviation for all the three machines

For Standard Deviation we have to follow the steps as under

Therefore for Machine 1 UCL= 16+3(1) =19

                                        LCL= 16-3(1) =13

Similarly we can find the Standard Deviations for Machine 12 and 3 as follows

Standard Deviation for Machine 2

Therefore for Machine 2 UCL= 21+3(4.06)=21+12.18=33.18

LCL= 21-3(4.06) =8.82

Standard Deviation for Machine3

Therefore for Machine 3 UCL= 23+3(0.70)=23+2.1=25.1

LCL= 23-3(0.70) =20.9

Therefore Mean chart UCL= 19+33.18+25.1/3= 25.76

Mean chart LCL= 13+8.82+20.9/3= 14.24

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