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