Problem 1 · Probability distributions
Machined rod lengths are normally distributed with a mean of 250.0 mm and a standard deviation of 0.8 mm. The specification limits are 248.8 mm and 251.6 mm. For the standard normal distribution, Φ(1.50) = 0.9332 and Φ(2.00) = 0.9772. Enter the percentage of rods that fall within the specification, to one decimal place.
Fill in the blank: answer in %.
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Answer: 91 % (accepted 90.5 to 91.5)
z₁ = (248.8 − 250.0)/0.8 = −1.50 and z₂ = (251.6 − 250.0)/0.8 = 2.00. P = Φ(2.00) − Φ(−1.50) = 0.9772 − (1 − 0.9332) = 0.9104, or 91.0%. The limits are not symmetric about the mean, so a single ±z value does not apply.
Probability distributions · Reference Handbook: Engineering Probability and Statistics
