Automated manufacturing operations are quite precise but still vary, often with distribution that are close to Normal. The width in inches of slots cut by a milling machine follows approximately the N(0.77,0.0012) distribution. The specifications allow slot widths between 0.7698 and 0.7702 . What proportion of slots meet these specifications?
Answer as a percent:
Answer as a percent:
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P(0.7698 < x < 0.7702) = P( (0.7698 - 0.77)/0.0012 < z < (0.7702 - 0.77)/0.0012) = P(-0.17 < z < 0.17)
Use the normal distribution table: 0.5675 - 0.4325 = 0.135 --> 13.5%
Use the normal distribution table: 0.5675 - 0.4325 = 0.135 --> 13.5%