A process manufactures ball bearings whose diameters are normally distributed with mean 2.505 cm and standard deviation 0.008 cm. Specifications call for the diameter to be in the interval 2.5±0.01 cm. What proportion of the ball bearings will meet the specification? Let X represent the diameter of a randomly chosen ball bearing. Then X N(2.505, 0.0082). We need to determine P(2.49 < x < 2.51), which is the proportion of ball bearings that meet the specification. We will show below that about 70.56% of ball bearings will meet the specification. 0.7056 Exercise: Ball Bearings = P(Z<0.63)-P(Z<-1.88) 0.7357-0.0301 0.7056 2.49 z=-1.88 2.505 2.51 z=0.63 For next class The process can be re-calibrated so that the mean will equal 2.5 cm, the center of the specification interval. The standard deviation of the process remains 0.008 cm. What proportion of the diameters will meet the specifications? (Ans: 78.88%) 2 Assume that the process has been re-calibrated so that the mean diameter is now 2.5 cm. To what value must the standard deviation be lowered so that 95% of the diameters will meet the specification? (Ans: a 0.0051) Read: Section 1.4-1.6 of the text. 2 Do the following problems on pp 44-46: 30, 31, 32, 36, 40