Shafts manufactured for use in optical storage devices have diameters that are n

Question

Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean = 0.652 cm and standard deviation = 0.003 cm. The specification for the shaft diameter is 0.650 ± 0.005 cm a. What proportion of the shafts manufactured by this process meet the specifications? [Select] b. The process mean can be adjusted through calibration. If the mean is set to 0.650 cm, what proportion of the shafts will meet specifications? [ Select ] c. If the mean is set to 0.650 cm, what is the greatest standard deviation that will allow 99% of the shafts to meet specifications? [ Select ]

Explanation / Answer

Ans:

a)specification ranges from 0.645 to 0.655

Use Normalcdf(0.645,0.655,0.652,0.003) for TI-84

z(0.645)=(0.645-0.652)/0.003=-0.007/0.003=-2.33

z(0.655)=(0.655-0.652)/0.003=1

P(-2.33<=z<=1)=P(z<=1)-P(z<=-2.33)=0.8413-0.0099=0.8314

hence,83.14% will meet the specification.

b) use Normalcdf(0.645,0.655,0.650,0.003) for TI-84

z(0.645)=(0.645-0.650)/0.003=-0.005/0.003=-1.67

z(0.655)=(0.655-0.650)/0.003=1.67

P(-1.67<=z<=1.67)=P(z<=1.67)-P(z<=-1.67)=0.9525-0.0475=0.9051

hence,90.51% will meet the specification.

c)99% means that z=+/-2.58

2.58=(0.655-0.650)/std dev

standard deviation=0.005/2.58=0.00194

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