Bridies' Bearing Works manufactures bearing shafts whose diameters are normally distributed with parameters [math]\mu = 1[/math], [math]\sigma = .002[/math]. The buyer's specifications require these diameters to be [math]1.000 \pm .003[/math] cm. If the manufacturer improves her quality control, she can reduce the value of [math]\sigma[/math]. What value of [math]\sigma[/math] will ensure that no more than 1 percent of her shafts are likely to be rejected?

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: B

We need to find [math]\sigma [/math] small enough so that the range of values outside [math]1.000 \pm .003[/math] cm has probability less than or equal to 1%. If [math]X[/math] has a normal distribution with mean 1 and standard deviation [math]\sigma [/math] then

[[math]] P(X \in 1.000 \pm .003 ) = P(Z \in \pm 0.003/\sigma ) \geq 0.99 = P(Z \in \pm 2.576) [[/math]]

where [math]Z [/math] is a standard normal. Hence we need [math]\sigma \leq 0.001328[/math].