Question

The diameter of a ball bearings produced by a machine is random variable having a normal distribution with mean 6.00 mm and standard deviation 0.025 mm. If the diameter tolerance ±1%, find the proportion of ball bearings produced that are out of tolerance.

Answer 1

Answer on Question #41626 - Math - Statistics and Probability

The diameter of a ball bearings produced by a machine is random variable having a normal distribution with mean $6.00 \mathrm{~mm}$ and standard deviation $0.025 \mathrm{~mm}$. If the diameter tolerance $\pm 1\%$, find the proportion of ball bearings produced that are out of tolerance.

Solution

$$X _ {m i n} = 0. 9 9 \mu = 0. 9 9 \cdot 6. 0 0 = 5. 9 4 \mathrm {m m}.$$ $$X _ {m a x} = 1. 0 1 \mu = 6. 0 6 \mathrm {m m}.$$ $$P (X > X _ {m a x}) = 1 - P (X < 6. 0 6) = 1 - \Phi \left(\frac {6 . 0 6 - 6 . 0 0}{0 . 0 2 5}\right) = 1 - \Phi (2. 4) = 1 - 0. 9 9 1 8 = 0. 0 0 8 2.$$ $$P (X < X _ {m i n}) = P (X < 5. 9 4) = \Phi \left(\frac {5 . 9 4 - 6 . 0 0}{0 . 0 2 5}\right) = \Phi (- 2. 4) = 0. 0 0 8 2.$$

The proportion of ball bearings produced that are out of tolerance is

$$P (X < X _ {m i n}) + P (X > X _ {m a x}) = 0. 0 0 8 2 + 0. 0 0 8 2 = 0. 0 1 6 4 = 1. 6 4 \%$$

Answer: $1.64\%$.

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