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A process that is in control has a mean of 12.5 and a standard deviation of 1.1.
a. Construct the control chart for this process if samples of size are to be used (to 1 decimal).
For N=4
Find the UCL
Find the LCL
b. Repeat part (a) for samples of size and (to 2 decimals).
For N=10
Find the UCL
Find the LCL
For N=15
Find the UCL
Find the LCL
c. What happens to the limits of the control chart as the sample size is increased? Discuss why this is reasonable. and become - Select your answer (closerfarther) together as n increases. If the process is in control, the larger samples should have less variance and should fall - Select your answer (closerfarther) to 12.5

Answer :

batolisis

Answer:

1a) UCL = 14.2

    LCL = 10.9

b) UCL = 15.63

    LCL = 9.37

c) UCL = 16.48

   LCL = 8.52

2) The difference between the limits falls close together as n increases

and the difference between the limits  falls farther away from 12.5

Step-by-step explanation:

mean ( μ ) = 12.5

std ( б ) = 1.1

UCL = μ + ( n - 1 ) б / √n

LCL = μ - ( n - 1 ) б / √n

1) a) Given  n = 4

UCL = 12.5 + ( 3 ) * 1.1 / 2

       = 12.5 + 3.3/2 = 14.15 ≈ 14.2

LCL = 12.5 - 3.3/2 = 10.85 ≈ 10.9

b) Given n = 10

UCL = 12.5 + ( 9 ) * 1.1 /√10

       = 12.5 + 3.13 = 15.63

LCL = 12.5 - 3.13 = 9.37

c) Given n = 15

UCL = 12.5 + 14 * 1.1 / √15

       = 12.5 + 3.98 = 16.48

LCL = 12.5 - 3.98 = 8.52

2) As the sample size increases the difference between the limits of the control chart decreases

Hence the difference falls close together as n increases

and the difference between the limits  falls farther away from 12.5

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