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Sagot :
Answer:
b. 23.16 and 16.84
Explanation:
The computation is shown below:
As we know that
The Mean of each and every sample (X-bar) is
= Sum of observations Number of observations
And,
Range (R) = (Highest observation - Lowest observation)
For Machine 1:
X-bar = (17 + 15 + 15 + 17) ÷ 4
= 16
R = (17 - 15)
= 2
For Machine 2:
X-bar = (16 + 25 + 18 + 25) ÷ 4
= 21
R = (25 - 16)
= 9
For Machine 3:
X-bar = (23 + 24 + 23 + 22) ÷ 4
= 23
R = (24 - 22)
= 2
Now, the Mean of means (X-double bar) is determined as follows
X-double bar = Sum of X-bar ÷ Number of samples
= (16 + 21 + 23) ÷ 3
= 20
The Mean of ranges (R-bar) is determined as follows
R-bar = Sum of R ÷ Number of samples
= (2 + 9 + 2) ÷ 3
= 4.33
with reference to the table of constants for determining the 3-sigma upper and lower control limits,
n = 4 that represent the Number of units sampled
A2 = 0.729,
Now
UCL = X-double bar + (A2 × R-bar)
= 20 + (0.729 × 4.33)
= 23.16
LCL = X-double bar - (A2 × R-bar)
= 20 - (0.729 × 4.33)
= 16.84
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