Question

The lengths of nails produced in a factory are normally distributed with a mean of 5.02 centimeters and a standard deviation of 0.05 centimeters. Find the two lengths that separate the top 6% and the bottom 6%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.

Answer 1

The length that separates the top 6% is 5.1 centimeters.

The length that separates the bottom 6% is 4.94 centimeters.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 5.02, \sigma = 0.05`

Find the two lengths that separate the top 6% and the bottom 6%.

Top 6%:

The 100-6 = 94th percentile, which is X when Z has a pvalue of 0.94. So X when Z = 1.555.

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

`1.555 = (X - 5.02)/(0.05)`

`X - 5.02 = 1.555*0.05`

`X = 5.1`

So the length that separates the top 6% is 5.1 centimeters.

Bottom 6%:

The 6th percentile, which is X when Z has a pvalue of 0.06. So X when Z = -1.555.

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

`-1.555 = (X - 5.02)/(0.05)`

`X - 5.02 = -1.555*0.05`

`X = 4.94`

The length that separates the bottom 6% is 4.94 centimeters.