The width of a slot of a duralumin forging is (in inches) normally distributed with μ = .9000 and σ = .0030. The specification limits were given as .9000 ± .0050. What percentage of forgings will be defective? What is the maximum allowable value of σ that will permit no more than 1 in 100 defectives when the widths are normally distributed with μ = .9000 and σ = .0030?
Answers
Given : The width of a slot of a duralumin forging is (in inches) normally distributed with μ = .9000 and σ = .0030. The specification limits were given as .9000 ± .0050.
To Find : What percentage of forgings will be defective?
What is the maximum allowable value of σ that will permit no more than 1 in 100 defectives
Solution:
mean = .9000
sd = .0030
z score = (value - mean)/sd
The specification limits were given as .9000 ± .0050.
z score = ( .9000 ± .0050. - .9000 )/.0030
= ±5/3
4.78 % to 95.22% are with in the range
Hence Defectives = 2 * 4.78 = 9.56 %
1 in 100 defectives = 1 %
0.5% to 99.5%
z score between -2.575 2.575
± 2.575 = ± .0050 / σ
=> σ = 0.00194
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