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

Answer:

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

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.555 = \frac{X - 5.02}{0.05}$

$X - 5.02 = 1.555*0.05$

$X = 5.1$

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.

$Z = \frac{X - \mu}{\sigma}$

$-1.555 = \frac{X - 5.02}{0.05}$

$X - 5.02 = -1.555*0.05$

$X = 4.94$

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