Question: 30 The specifications for the thickness of nonferrous washers are 1.0

Question

Question: 30 The specifications for the thickness of nonferrous washers are 1.0 +-0.04 mm. From process data, the distribution of the washer thickness is estimate to be normal with a mean of 0.98 mm and a standard deviation of 0.02 mm. The unit cost of rework is $0.10, and the unit cost of scrap is $0.15. For a daily production of 10,000 items:

(a) what proportion of the washers is conforming? What is the expected total daily cost of rework and scrap?

(b) In its study of constant improvement, the manufacturer changes the mean setting of the machine to 1.0 mm. If the standard deviation is the same as before, what is the expected total daily cost of rework and scrap?

(c) The manufacturer is trying to improve the process and reduces its standard deviation to 0.015 mm. If the process mean is maintained at 1.0 mm, what is the percent decrease in the expected total daily cost of rework and scrap compared to that of part (a)?

Explanation / Answer

The Specification for the thickness of nonferrous washers are 1+-0.04mm i.e ( 0.96mm, 1.04mm)

X : Thickness of nonferrous washers.

X ~ N( Mean= 0.98 ,S.D =0.02)

a) The propertion of the washer is conforming = P( 0.96 < X < 1.04)

= P( (0.96-0.98)/0.02 < ( X-mean)/S.D < (1.04-.98 ) /0.02)

= P( -1 < Z < 3) ( Z = (X-mean)/S.D. ~ N(0,1))

From normal probability table

  P( -1 < Z < 3) = 0.84

The propertion of conforming the washers is =0.84.

Hence the propertion of washer is non-conforming =1-0.84 =0.16

The expected number of washer which is rework and scrap =0.16 *10000 =1600

The expected total daily cost of rework and scrap = (0.10+0.15)*1600 = $400

b) The process mean shifting 0.98 mm to 1.00 mm

The propertion of the washer is conforming = P( 0.96 < X < 1.04)

= P( (0.96-1)/0.02 < ( X-mean)/S.D < (1.04-.1 ) /0.02)

= P( -2 < Z < 2) ( Z = (X-mean)/S.D. ~ N(0,1))

From normal probability table

  P( -2 < Z < 2) = 0.9545

The propertion of conforming the washers is =0.9545

Hence the propertion of washer is non-conforming =1-0.9545 = 0.455

The expected number of washer which is rework and scrap =0.0455 *10000 =455

The expected total daily cost of rework and scrap = (0.10+0.15)*455 = $113.75

c) The standard deviation is reduces to 0.015 and process mean maintained at 1.00 mm

The propertion of the washer is conforming = P( 0.96 < X < 1.04)

= P( (0.96-1)/0.015 < ( X-mean)/S.D < (1.04-.1 ) /0.015)

= P( -2.6666 < Z < 2.6666) ( Z = (X-mean)/S.D. ~ N(0,1))

From normal probability table

  P( -2.6666 < Z < 2.6666) = 0.9923

The propertion of conforming the washers is =0.9923

Hence the propertion of washer is non-conforming =1-0.9923 = 0.0077

The expected number of washer which is rework and scrap =0.0077 *10000 =77

The expected total daily cost of rework and scrap = (0.10+0.15)*77 = $19.25

The percent decrease in expected total daily cost of rework and scrap as copmpared to part a

= (400-19.25/400)*100 = 95.18%

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