ACCEPTABLE PINS
DATA INTERPRETATION IN UNCERTAINITY
PRODUCT OF COMPANY – MANUFACTURE & SUPPLY OF PINS
DATA ON HAND
MEAN, (µ )= 1.012
STANDARD DEVIATION, (σ) = 0.018
CUSTOMER DESIRED LENGTH = 1.00
ACCEPTABLE TOLERANCES = ± 0.02
NORMALLY DISTRIBUTED
# All the dimensions are in inches
ACCEPTABLE PINS
Đ1. WHAT PERCENTAGE OF THE PINS WILL BE ACCEPTABLE TO THE CONSUMER?
Now, X N(1.012, 0.018)
Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/0.018)) < (X- 1.012 )/ 0.018 < (1.02- 1.012)/0.018)]
= P[ -1.777 < Z < 0.444 ]
= P[ -1.777 < Z < 0 ] + P[ 0 < Z < 0.444 ]
= 0.4620 + 0.1720 ---------- (From Tables)
= 0.6340
# All the dimensions are in inches
ACCEPTABLE PINS
[Say, Z = (X- µ)/ σ]
U
# All the dimensions are in inches
ACCEPTABLE PINS
63.40% OF PINS ARE ACCEPTABLE TO THE CONSUMER
µ = 1.012
-1.777 0.444
Đ2. IF THE LATHE CAN BE ADJUSTED TO HAVE THE MEAN OF THE LENGTHS TO ANY DESIRED VALUE, WHAT SHOULD IT BE ADJUSTED TO? WHY?
SUPPOSE, µ=1.00 (ADJUSTED TO DESIRED VALUE) & σ = 0.018
Now, X N(1.00, 0.018)
Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/0.018)) < (X- 1.00)/ 0.018 < (1.02- 1.00)/0.018)]
= P[ -1.111 < Z < 1.111]
= P[ -1.111 < Z < 0 ] + P[ 0 < Z < 0.111 ]
= 2(0.3665) ---------- (From Tables)
= 0.7330 # All the dimensions are in inches
ACCEPTABLE PINS
[Say, Z = (X- µ)/ σ]U
# All the dimensions are in inches
ACCEPTABLE PINS
73.30% OF PINS ARE ACCEPTABLE TO THE CONSUMER
µ = 1.00
-1.111 1.111
Đ3. SUPPOSE THE MEAN CANNOT BE ADJUSTED BUT THE STANDARD DEVIATION CAN BE REDUCED. WHAT MAXIMUM VALUE OF THE STANDARD DEVIATION WOULD MAKE 90% OF THE PARTS ACCEPTABLE TO THE CONSUMER (ASSUME THE MEAN TO BE 1.012.)
SUPPOSE, µ=1.012 & σ (VALUE NOT KNOWN)
Now, X N(1.00, σ) given, P(0.98 < X < 1.02) = 0.90
Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/ σ) < (X- 1.00)/ σ < (1.02- 1.00)/ σ)]
0.90 = P[ -0.032/ σ < Z < 0.008/ σ]
0.5 + 0.40 = P[[ -0.032/ σ < Z < 0 ] + P[ 0 < Z < 0.008/ σ ]
0.5 + 0.40 = P[ a< Z < 0 ] + P[ 0 < Z < b ]# All the dimensions are in inches
ACCEPTABLE PINS
[Say, Z = (X- µ)/ σ]U
[ASSUME, P[[ -0.032/ σ < Z < 0 ] ] = 0.5
0.40 = P[ 0 < Z < b ]
Say, b = 0.008/ σ
Value of b = 1.28 @ 0.40 ---------- (From Tables)
1.28 = 0.008 / σ
Therefore, σ = 0.00625
# All the dimensions are in inches
ACCEPTABLE PINS
# All the dimensions are in inches
ACCEPTABLE PINS
µ = 1.012
-0.032/ σ 0.008/ σ
a b
“a” IS 4 TIMES THAT OF “b”
MAXIMUM VALUE OF σ = 0.00625 WILL MAKE 90% OF THE PARTS ACCEPTABLE TO THE CONSUMER
Đ4. REPEAT QUESTION 3, WITH 95% AND 99% OF THE PINS ACCEPTABLE?
SUPPOSE WITH 95% OF PINS ACCEPTANCE
µ=1.012 & σ (VALUE NOT KNOWN)
Now, X N(1.012, σ) given, P(0.98 < X < 1.02) = 0.95
Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/ σ) < (X- 1.00)/ σ < (1.02- 1.00)/ σ)]
0.95 = P[ -0.032/ σ < Z < 0.008/ σ]
0.5 + 0.45 = P[[ -0.032/ σ < Z < 0 ] + P[ 0 < Z < 0.008/ σ ]
0.5 + 0.45 = P[ a< Z < 0 ] + P[ 0 < Z < b ]
0.40 = P[ 0 < Z < b ] # All the dimensions are in inches
ACCEPTABLE PINS
[Say, Z = (X- µ)/ σ]U
Say, b = 0.008/ σ
Value of b = 1.645 @ 0.45 ---------- (From Tables)
1.645 = 0.008 / σ
Therefore, σ = 0.00486
# All the dimensions are in inches
ACCEPTABLE PINS
# All the dimensions are in inches
ACCEPTABLE PINS
µ = 1.012
-0.032/ σ 0.008/ σ
a b
“a” IS 4 TIMES THAT OF “b”
MAXIMUM VALUE OF σ = 0.00486 WILL MAKE 95% OF THE PARTS ACCEPTABLE TO THE CONSUMER
SUPPOSE WITH 99% OF PINS ACCEPTANCE
µ=1.012 & σ (VALUE NOT KNOWN)
Now, X N(1.012, σ) given, P(0.98 < X < 1.02) = 0.99
Then, P(0.98 < X < 1.02) = P[(0.98- 1.012/ σ) < (X- 1.00)/ σ < (1.02- 1.00)/ σ)]
0.99 = P[ -0.032/ σ < Z < 0.008/ σ]
0.5 + 0.49 = P[[ -0.032/ σ < Z < 0 ] + P[ 0 < Z < 0.008/ σ ]
0.5 + 0.49 = P[ a< Z < 0 ] + P[ 0 < Z < b ]
0.40 = P[ 0 < Z < b ]
Say, b = 0.008/ σ # All the dimensions are in inches
ACCEPTABLE PINS
[Say, Z = (X- µ)/ σ]
U
Value of b = 2.33 @ 0.49 ---------- (From Tables)
2.33 = 0.008 / σ
Therefore, σ = 0.00343
# All the dimensions are in inches
ACCEPTABLE PINS
# All the dimensions are in inches
ACCEPTABLE PINS
µ = 1.012
-0.032/ σ 0.008/ σ
a b
“a” IS 4 TIMES THAT OF “b”
MAXIMUM VALUE OF σ = 0.00343 WILL MAKE 99% OF THE PARTS ACCEPTABLE TO THE CONSUMER
Đ5. IN PRACTICE, WHICH ONE DO YOU THINK IS EASIER TO ADJUST, THE MEAN OR THE STANDARD DEVIATION? WHY?
THE SUMMARIZATION OF ABOVE RESULTS THINK US AS FOLLOWS:
# All the dimensions are in inches
ACCEPTABLE PINS
Q MEAN(µ)
STANDARD DEVIATION(σ)
LEVEL OF PINS ACCEPTABLE(% ) REMARKS
1 1.012 0.018 63.40 A minor change in “σ” results in comparatively
lesser impact on % acceptance
2 1.00 0.018 73.30
3 1.012 0.00625 90 A minor change in “σ” results in comparatively
lesser impact on % acceptance
4a 1.012 0.00486 95
4b 1.012 0.00343 99
AS PER THE PROBLEM, THE COST OF RESETTING THE MACHINE TO ADJUST THE MEAN (µ)INVOLVES THE ENGINEER’S TIME AND THE COST OF PRODUCTION TIME.
THE COST OF REDUCING THE STANDARD DEVIATION (σ) INVOLVES THE ENGINEER’S TIME, THE COST OF PRODUCTION TIME, THE COST OF OVERHAULING THE MACHINE AND REENGINEERING THE PROCESS. SO IN PRACTICE, THE MEAN IS EASIER TO ADJUST
Đ6. ASSUME IT COSTS $150 x2 TO DECREASE THE STANDARD DEVIATION (σ) BY (x/1000) INCH. FIND THE COST OF REDUCING THE STANDARD DEVIATION TO THE VALUES FOUND IN QUESTION 3 AND 4?
# All the dimensions are in inches
ACCEPTABLE PINS
From Q3, we know,
µ=1.012 & σd = 0.00625 & P(0.98 < X < 1.02) = 0.90
CHANGE IN σ, σc = (σa – σd)
= (0.018 – 0.00625)
= 0.01175
σc = 11.75/1000 i.e., x = 11.75
[Say, σa - ACTUAL STANDARD DEVIATION, σd - DERIVED STANDARD DEVIATION & σc – CHANGE IN STANDARD DEVIATION
# All the dimensions are in inches
ACCEPTABLE PINS
From given relation,
Cost = $150 x2
= $150 X (11.75) 2
Therefore, Cost = $20709.375
THE COST OF REDUCING THE STANDARD DEVIATION IS $20709.375, IF 90% OF THE PARTS ACCEPTABLE TO CONSUMER
From Q4a, we know,
µ=1.012 & σd = 0.00486 & P(0.98 < X < 1.02) = 0.95
CHANGE IN σ, σc = (σa – σd)
= (0.018 – 0.00486)
# All the dimensions are in inches
ACCEPTABLE PINS
σc = 0.01314
σc = 13.14/1000 i.e., x = 13.14
From given relation,
Cost = $150 x2
= $150 X (13.14) 2
Therefore, Cost = $25898.94
THE COST OF REDUCING THE STANDARD DEVIATION IS $25898.94, IF 95% OF THE PARTS ACCEPTABLE TO CONSUMER
# All the dimensions are in inches
ACCEPTABLE PINS
From Q4b, we know,
µ=1.012 & σd = 0.00343 & P(0.98 < X < 1.02) = 0.99
CHANGE IN σ, σc = (σa – σd)
= (0.018 – 0.00343)
= 0.01457
σc = 14.57/1000 i.e., x = 14.57
From given relation,
Cost = $150 x2
= $150 X (14.57) 2
Therefore, Cost = $31842.735
THE COST OF REDUCING THE STANDARD DEVIATION IS $31842.735, IF 99% OF THE PARTS ACCEPTABLE TO CONSUMER
Đ7. NOW ASSUME THAT THE MEAN HAS BEEN ADJUSTED TO THE BEST VALUE FOUND IN QUESTION 2 AT A COST OF $80. CALCULATE THE REDUCTION IN STANDARD DEVIATION NECESSARY TO HAVE 90%, 95% AND 99% OF THE PARTS ACCEPTABLE. CALCULATE THE RESPECTIVE COSTS, AS IN QUESTION 6?
# All the dimensions are in inches
ACCEPTABLE PINS
From Q2, we know,
µ=1.00 & σ (VALUE NOT KNOWN),WITH 90% OF PINS ACCEPTANCE
P(0.98 < X < 1.02) = 0.90
Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/µ)) < (X- 1.00)/ µ) < (1.02- 1.00)/µ)]
0.9 = P[-0.02/σ < Z < 0.02/σ ]
0.9 = P[-0.02/σ < Z < 0 ] + P[ 0 < Z < 0.02/σ]
# All the dimensions are in inches
ACCEPTABLE PINS
0.90 = 2 P[ 0 < Z < 0.02/σ]
0.45 = P[ 0 < Z < 0.02/σ]
0.45 = P[ 0 < Z < b]
Say, b = 0.02/ σ
Value of b = 1.645 @ 0.45 ---------- (From Tables)
1.645 = 0.02 / σ
Therefore, σd = 0.01216
# All the dimensions are in inches
ACCEPTABLE PINS
µ = 1.00
-0.02/σ 0.02/σ
# All the dimensions are in inches
ACCEPTABLE PINS
CHANGE IN σ, σc = (σa – σd)
= (0.018 – 0.01216)
= 0.00584
σc = 5.84/1000 i.e., x = 5.84
From given relation,
Cost = $150 x2 + $80 ($80 IS THE COST FOR ADJUSTING µ TO DESIRED VALUE)
= $150 X (5.84) 2 + $80
Therefore, Cost = $5195.84
THE COST OF REDUCING THE STANDARD DEVIATION IS $5195.84, IF 90% OF THE PARTS ACCEPTABLE TO CONSUMER
# All the dimensions are in inches
ACCEPTABLE PINS
From Q2, we know,
µ=1.00 & σ (VALUE NOT KNOWN),WITH 95% OF PINS ACCEPTANCE
P(0.98 < X < 1.02) = 0.95
Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/µ)) < (X- 1.00)/ µ) < (1.02- 1.00)/µ)]
0.95 = P[-0.02/σ < Z < 0.02/σ ]
0.95 = P[-0.02/σ < Z < 0 ] + P[ 0 < Z < 0.02/σ]
0.95 = 2 P[ 0 < Z < 0.02/σ]
0.475 = P[ 0 < Z < 0.02/σ]
# All the dimensions are in inches
ACCEPTABLE PINS
0.475 = P[ 0 < Z < 0.02/σ]
Say, b = 0.02/ σ
Value of b = 1.96 @ 0.475 ---------- (From Tables)
1.96 = 0.02 / σ
Therefore, σd = 0.01020
CHANGE IN σ, σc = (σa – σd)
= (0.018 – 0.01020)
= 0.00780
σc = 7.80/1000 i.e., x = 7.80
# All the dimensions are in inches
ACCEPTABLE PINS
From given relation,
Cost = $150 x2 + $80 ($80 IS THE COST FOR ADJUSTING µ TO DESIRED VALUE)
= $150 X (7.80) 2 + $80
Therefore, Cost = $9206
THE COST OF REDUCING THE STANDARD DEVIATION IS $9206, IF 95% OF THE PARTS ACCEPTABLE TO CONSUMER
# All the dimensions are in inches
ACCEPTABLE PINS
From Q2, we know,
µ=1.00 & σ (VALUE NOT KNOWN),WITH 99% OF PINS ACCEPTANCE
P(0.98 < X < 1.02) = 0.99
Then, P(0.98 < X < 1.02) = P[(0.98- 1.00/µ)) < (X- 1.00)/ µ) < (1.02- 1.00)/µ)]
0.99 = P[-0.02/σ < Z < 0.02/σ ]
0.99 = P[-0.02/σ < Z < 0 ] + P[ 0 < Z < 0.02/σ]
0.99 = 2 P[ 0 < Z < 0.02/σ]
0.495 = P[ 0 < Z < 0.02/σ]
# All the dimensions are in inches
ACCEPTABLE PINS
0.495 = P[ 0 < Z < 0.02/σ]
Say, b = 0.02/ σ
Value of b = 2.57 @ 0.495 ---------- (From Tables)
2.57 = 0.02 / σ
Therefore, σd = 0.00778
CHANGE IN σ, σc = (σa – σd)
= (0.018 – 0.00778)
= 0.01022
σc = 10.22/1000 i.e., x = 10.22
# All the dimensions are in inches
ACCEPTABLE PINS
From given relation,
Cost = $150 x2 + $80 ($80 IS THE COST FOR ADJUSTING µ TO DESIRED VALUE)
= $150 X (10.22) 2 + $80
Therefore, Cost = $15747.26
THE COST OF REDUCING THE STANDARD DEVIATION IS $15747.26, IF 99% OF THE PARTS ACCEPTABLE TO CONSUMER
Đ8. BASED ON YOUR ANSWERS TO QUESTION 6 AND 7, WHAT ARE YOUR RECOMMENDED MEAN AND STANDARD DEVIATION?
# All the dimensions are in inches
ACCEPTABLE PINS
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