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Floyd Electric is fabricating flanges for a new electrical motor. The flange is to be 25 millimeters in diameter with an upper tolerance limit of 26.5 millimeters and a lower tolerance limit of 23.5 millimeters. Bill Floyd took a sample consisting of ten flanges, measured the diameter and recorded the observations in the table. He has heard discussion of 6 sigma in the business news recently and would like to join the club. What percentage decrease in process standard deviation would be required for Floyd Electric to achieve 6-sigma process capability? Assume that his process mean remains where it is.
A) 36%
B) 40%
C) 46%
D) 58%

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nuhulawal20

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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