Normal Distribution Problem: A certain component for the newly developed electronic diesel engine is considered to be defective if its diameter is less than 8 mm or greater than 10.5 mm. The distributions of the diameters of these parts is known to be normal with a mean of 9.0 mm and a standard deviation of 1.5 mm. If a component is randomly selected, what is the probability that it will be defective?Q: Do I add the Right and Left tails and that's my total probability of defectiveness? Or, Do I take the Normal.Dist (from Excel) to the left of 10.5 minus the N.D left of 8 for a total of .588852 probability that it will be defective (that's a lot)?Please help! And, thank you in advance!

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Asked Nov 11, 2019
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Normal Distribution Problem: A certain component for the newly developed electronic diesel engine is considered to be defective if its diameter is less than 8 mm or greater than 10.5 mm. The distributions of the diameters of these parts is known to be normal with a mean of 9.0 mm and a standard deviation of 1.5 mm. If a component is randomly selected, what is the probability that it will be defective?

Q: Do I add the Right and Left tails and that's my total probability of defectiveness? Or, Do I take the Normal.Dist (from Excel) to the left of 10.5 minus the N.D left of 8 for a total of .588852 probability that it will be defective (that's a lot)?

Please help! And, thank you in advance!

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Expert Answer

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Step 1

Given information:

Mean = 9.0 mm

Standard deviation = 1.5 mm

Step 2

Let x is random variable.

Now, the probability that it will be defective is calculated as follows:

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P(X<8or X10.5) P(X<8)+P(X>10.5) = P(X <8)+1-P(X<10.5) X-10.5-u X -и 8-u 1 P = P 8-9 +1-P Z 1.5 10.5-9 = P Z < 1.5

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Step 3

Simplify further a...

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P(X <8 or X10.5) P(Z <-0.67)+1-P(Z<1) = 0.2514 +1-0.8413 = 0.25 14 +0.1587 = 0.4101

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