acceptable pins

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


ACCEPTABLE PINS

[Say, Z = (X- µ)/ σ]

U


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


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


ACCEPTABLE PINS


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


1.645 = 0.008 / σ



ACCEPTABLE PINS


ACCEPTABLE PINS

µ = 1.012

-0.032/ σ 0.008/ σ

a b



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


2.33 = 0.008 / σ



ACCEPTABLE PINS


ACCEPTABLE PINS

µ = 1.012

-0.032/ σ 0.008/ σ

a b



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:


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?


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


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


= (0.018 – 0.00486)


ACCEPTABLE PINS

σc = 0.01314

σc = 13.14/1000 i.e., x = 13.14


Cost = $150 x2

= $150 X (13.14) 2




ACCEPTABLE PINS

From Q4b, we know,

µ=1.012 & σd = 0.00343 & P(0.98 < X < 1.02) = 0.99


= (0.018 – 0.00343)

= 0.01457

σc = 14.57/1000 i.e., x = 14.57


Cost = $150 x2

= $150 X (14.57) 2



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?


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]


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


1.645 = 0.02 / σ

Therefore, σd = 0.01216


ACCEPTABLE PINS

µ = 1.00

-0.02/σ 0.02/σ


ACCEPTABLE PINS


= (0.018 – 0.01216)

= 0.00584

σc = 5.84/1000 i.e., x = 5.84


Cost = $150 x2 + $80 ($80 IS THE COST FOR ADJUSTING µ TO DESIRED VALUE)

= $150 X (5.84) 2 + $80




ACCEPTABLE PINS

From Q2, we know,


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]


ACCEPTABLE PINS

0.475 = P[ 0 < Z < 0.02/σ]

Say, b = 0.02/ σ


1.96 = 0.02 / σ



= (0.018 – 0.01020)

= 0.00780

σc = 7.80/1000 i.e., x = 7.80


ACCEPTABLE PINS



= $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


ACCEPTABLE PINS

From Q2, we know,


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]


ACCEPTABLE PINS

0.495 = P[ 0 < Z < 0.02/σ]

Say, b = 0.02/ σ


2.57 = 0.02 / σ



= (0.018 – 0.00778)

= 0.01022

σc = 10.22/1000 i.e., x = 10.22


ACCEPTABLE PINS



= $150 X (10.22) 2 + $80



Đ8. BASED ON YOUR ANSWERS TO QUESTION 6 AND 7, WHAT ARE YOUR RECOMMENDED MEAN AND STANDARD DEVIATION?


ACCEPTABLE PINS

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