PLEASE HELP! A shop in a manufacturing plant makes three gears that fit together

Question

PLEASE HELP!

A shop in a manufacturing plant makes three gears that fit together to become part of the gear shift mechanism for a forklift.  Each part must be a specific diameter within specific tolerances.  The three parts, when put together to slip into the gearbox, must also meet an overall diameter within a specific tolerance. To solve this problem, you must understand the relationship between standard deviation and process sigma (1 sigma = 1 standard deviation). You must also depict the relationship between standard deviation and % distribution.

Part

Nominal (inches)

Tolerances (inches)

1

1.750

+ 0.042

2

2.000

+ 0.060

3

1.250

+ 0.030

"Nominal" is the intended or target diameter of the part. "Tolerance" is the amount of variation that is considered acceptable. For example, Part 1 should have a diameter of 1.750 inches but the part will be considered acceptable if its diameter is as small as 1.705 inches or as large as 1.795 inches.

Samples for the three parts were taken:

Part 1

Part 2

Part 3

1.74831

2.01144

1.25426

1.75740

2.00448

1.24775

1.75134

2.01492

1.24558

1.73316

2.00448

1.24992

1.75134

2.00448

1.23907

1.71498

2.00100

1.25860

1.75437

2.00796

1.25643

1.73619

2.00100

1.25209

1.73922

1.98708

1.25426

1.73619

1.99752

1.24341

1.74528

1.99752

1.25643

1.74831

2.00100

1.24775

1.76346

1.99404

1.24341

1.74528

2.00796

1.24558

1.76952

2.00448

1.23907

1.70892

2.00448

1.24124

1.75134

2.00100

1.24558

1.71195

1.99752

1.24992

1.74831

2.00100

1.25209

1.77861

2.00100

1.24992

1.75740

1.98708

1.24992

1.73619

1.99404

1.24558

1.74225

1.98708

1.24992

1.74528

1.98360

1.24775

1.73922

2.00100

1.24775

Calculate the mean and standard deviations for Part 1, 2, and 3 and compare them to the specification (tolerance) limits above.The machine shop manufactures hundreds of the parts each day. The production manager has decided to randomly pull 25 of each part from the production line to verify whether the parts are being manufactured within the tolerance limits.

1.) Will the production process permit an acceptable fit of Part 1 at least 99.73 % of the time?

2.) Will the production process permit an acceptable fit of Part 2 at least 99.73 % of the time?

3.) Will the production process permit an acceptable fit of Part 3 at least 99.73 % of the time?

4.) Will the production process permit an acceptable fit of all three parts combined into a slot with a specification of 5 + 0.081 inches at least 99.73% of the time?

Part

Nominal (inches)

Tolerances (inches)

1

1.750

+ 0.042

2

2.000

+ 0.060

3

1.250

+ 0.030

Explanation / Answer

we have given data three parts of gear and for each of part we have different tolerence limits.

For part 1, we have values for mean and sd from sample as below

Mean=1.744553

Standard Deviation=0.016277

For part 2, we have values for mean and sd from sample as below

Mean=1.999886

Standard Deviation=0.007763

For part 3, we have values for mean and sd from sample as below

Mean=1.248531

Standard Deviation=0.005199

By using given tolerence limit table we get upper limit and lower limit for

part lower limit upper limit

1 1.750 1.792

2 2.000 2.06

3 1.250 1.28

...............................................................................................................

for part-1 we get 9/25 sampling unit acceptable

for part-2 we get 19/25 sampling unit acceptable

for part-3 we get 7/25 sampling unit acceptable

.....................................................................................................

by using acceptance proportion for given sample we get

part value from sample value given

1 36 99.73

2 64 99.73

3 28 99.73

From above table we conclude as

1) The production process will not permit an acceptable fit of part 1 at least 99.73% of the time

2) The production process will not permit an acceptable fit of part 2 at least 99.73% of the time

3) The production process will not permit an acceptable fit of part 3 at least 99.73% of the time   

4) We have lower limit as 5.00 and upper limit as 5.081

Therefore by oberving three parts of gear simultaniously, we come into know that all the observations are belongs to interval (5.000 5.081)

Therefore we conclude that the production will permit an acceptable fit of all three parts combined into a slot with a specification of 5+0.081 inches at least 99.73% of the time.

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