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The lengths of nails produced in a factory are normally distributed with a mean of 6.13 centimeters and a standard deviation of 0.06 centimeters. Find the two lengths that separate the top 7% and the bottom 7%. These lengths could serve as limits used to identify which nails should be rejected.

User Abhi V
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Answer:

A value of 6.0415 centimeters separates the bottom 7%, while a value of 6.2185 centimeters separates the top 7%.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the z-score of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 6.13 centimeters and a standard deviation of 0.06 centimeters.

This means that
\mu = 6.13, \sigma = 0.06

Value that separated the top 7%:

The 100 - 7 = 93rd percentile, which is X when Z has a p-value of 0.93, so X when Z = 1.475.


Z = (X - \mu)/(\sigma)


1.475 = (X - 6.13)/(0.06)


X - 6.13 = 1.475*0.06


X = 6.2185

Value that separates the bottom 7%:

The 7th percentile, which is X when Z has a p-value of 0.07, so X when Z = -1.475.


Z = (X - \mu)/(\sigma)


-1.475 = (X - 6.13)/(0.06)


X - 6.13 = -1.475*0.06


X = 6.0415

A value of 6.0415 centimeters separates the bottom 7%, while a value of 6.2185 centimeters separates the top 7%.

User Konrud
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