Motorola used the normal distribution to determine the probability of defects an

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Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of ounces.

a. The process standard deviation is , and the process control is set at plus or minus standard deviation . Units with weights less than or greater than ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of parts, how many defects would be found (round to the nearest whole number)?

b. Through process design improvements, the process standard deviation can be reduced to . Assume the process control remains the same, with weights less than or greater than ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)?

In a production run of parts, how many defects would be found (to the nearest whole number)?

c. What is the advantage of reducing process variation, thereby causing a problem limits to be at a greater number of standard deviations from the mean?

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Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 13 ounces.

a) The process standard deviation is 0.1, and the process control is set at plus or minus 1.5 standard deviations. Units with weights less than 12.85 or greater than 13.15 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

b)In a production run of 1000 parts, how many defects would be found (to 0 decimals)?

c) Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 12.85 or greater than 13.15 ounces being classified as defects. What is the probability of a defect (rounded to 4 decimals; getting the exact answer, although not necessary, will require Excel)?

d)In a production run of 1000 parts, how many defects would be found (to 0 decimals)?

Ans:

population mean,u = 13
standard deviation,sigma = 0.1

a.
P( 12.85 <X< 13.15 )
= P( ((12.85-13) / 0.1) <Z< ((13.15-13) / 0.1) )
= P( -1.50 <Z< 1.50 )
= P(Z<1.5) - P(Z<-1.5)
= P(Z<1.5) - (1 - P(Z<1.5))
= 0.9332 - ( 1 - 0.9332)
= 0.9332 - 0.0668
= 0.8664

the probability of a defect
= P(X<12.85) + P(X>13.15)
= 1 - P( 12.85 <X< 13.15 )
= 1 - 0.8664
= 0.1336
_______________________________________________________

1000*0.1336
= 133.6 or 134 defects would be found in a production run of 1000 parts.
_______________________________________________________
b
standard deviation,sigma = 0.05

P( 12.85 <X< 13.15 )
= P( ((12.85-13) / 0.05) <Z< ((13.15-13) / 0.05) )
= P( -3.00 <Z< 3.00 )
= P(Z<3) - P(Z<-3)
= P(Z<3) - (1 - P(Z<3))
= 0.9987 - ( 1 - 0.9987)
= 0.9987 - 0.0013
= 0.9974

the probability of a defect
= P(X<12.85) + P(X>13.15)
= 1 - P( 12.85 <X< 13.15 )
= 1 - 0.9974
= 0.0026
_______________________________________________________
1000*0.0026
= 2.6 or 3 defects would be found in a production run of 1000 parts.

c) As sample SD is decreases, the process control lmits increases and probability of producing the defective items is reduced.

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