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The lengths of nails produced in a factory are normally distributed with a mean of 4.95 centimeters and a standard deviation of 0.05 centimeters. Find the two lengths that separate the top 5% and the bottom 5%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.

Sagot :

Answer:

The length that separates the bottom 5% is of 4.87 centimeters.

The length that separates the top 5% is of 5.03 centimeters.

Step-by-step explanation:

Normal Probability Distribution:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 4.95 centimeters and a standard deviation of 0.05 centimeters.

This means that [tex]\mu = 4.95, \sigma = 0.05[/tex]

Bottom 5%:

The 5th percentile, which is X when Z has a pvalue of 0.05. So X when Z = -1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.645 = \frac{X - 4.95}{0.05}[/tex]

[tex]X - 4.95 = -1.645*0.05[/tex]

[tex]X = 4.87[/tex]

The length that separates the bottom 5% is of 4.87 centimeters.

Top 5%:

The 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.645 = \frac{X - 4.95}{0.05}[/tex]

[tex]X - 4.95 = 1.645*0.05[/tex]

[tex]X = 5.03[/tex]

The length that separates the top 5% is of 5.03 centimeters.

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