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Sagot :
This question is incomplete, the missing table is uploaded along this answer below.
Answer:
percentage of age reduction is 40
Option B) 40% is the correct answer
Step-by-step explanation:
Given the data in the question;
from the table in the image below;
sample x = 25.69 , 25.58 , 24.78 , 24.94 , 24.87 , 24.63 , 25.70 , 25.52 , 24.95 , 24.91
mean = ∑x/n
where n is sample size
= (25.69 + 25.58 +24.78 + 24.94 + 24.87 + 24.63 + 25.70 + 25.52 +24.95 ) / 10
= 251.57 / 10
mean u = 25.157
so;
S/N Sample (X) Mean (u) (X-U)²
1 25.69 25.157 0.284089
2 25.58 25.157 0.178929
3 24.78 25.157 0.142129
4 24.94 25.157 0.047089
5 24.87 25.157 0.082369
6 24.63 25.157 0.277729
7 25.7 25.157 0.294849
8 25.52 25.157 0.131769
9 24.95 25.157 0.042849
10 24.91 25.157 0.061009
Total 251.57 1.54281
Mean 25.157
Sample Variance = ∑(x-u)²/n-1 = 1.54281 / (10 - 1) = 1.54281 / 9 = 0.1714
Standard deviation STD1 = √ (variance) = √0.1714 = 0.414
now, for 6-sigma capability, Cpk = 2
Cpk = min of ( USL - mean ) / 3×STD), (Mean - LSL) / 3×STD
since our mean is centered between USL and LSL;
Cpk = ( 26.5 - 25 ) / 3×STD2
we substitute
2 = ( 26.5 - 25 ) / 3×STD2
6STD2 = 1.5
STD2 = 1.5 / 6
STD2 = 0.25
so current standard deviation STD1 = 0.414
required standard deviation for 6-sigma process STD2 = 0.25
so percentage of age reduction will be;
⇒ ( 0.414 - 0.25 ) 0.414 = 0.164 / 0.414 = 0.396 ≈ 0.4
⇒ 0.4 × 100 = 40%
Therefore, percentage of age reduction is 40
Option B) 40% is the correct answer
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