A process manufactures ball bearings whose diameters are normally distributed wi

Question

A process manufactures ball bearings whose diameters are normally distributed with mean 2.105 cm and standard deviation 0.010 cm. Specifications call for the diameter to be in the interval 2.1 plusminus 0.02 cm. Suppose the process can be recalibrated so that the mean will equal 2.1 cm, the center of the specification interval. The standard deviation of the process remains 0.010 cm. What proportion of the diameters will meet the specifications? Recalibrating will increase the proportion of diameters that meet the specification to %.

Explanation / Answer

Acceptance range of the diameter is 2.08 to 2.12

Before recaliberation:
mean = 2.105 and std. dev. =0.01

Proportion of the diameters that meet the specifications are
P(2.08<=X<=2.12) = P((2.08-2.105)/0.01 < Z < (2.12-2.105)/0.01) = P(-2.5 < z < 1.5) = 0.927

Hence before recaliberation 92.7% of the diameters meet the specifictions.

After recaliberation:
mean = 2.1 and std. dev. =0.01

Proportion of the diameters that meet the specifications are
P(2.08<=X<=2.12) = P((2.08-2.1)/0.01 < Z < (2.12-2.1)/0.01) = P(-2 < z < 2) = 0.9545

Hence after recaliberation 95.45% of the diameters meet the specifictions.

Recaliberation increased the proportion of diameters that meet the spcification to 95.45%

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