The lengths of nails produced in a factory are normally distributed with a mean of 5.095.09 centimeters and a standard deviation of 0.040.04 centimeters. Find the two lengths that separate the top 7%7% and the bottom 7%7%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.

We are given that the lengths of nails produced in a factory are normally distributed with a mean of 5.09 centimeters and a standard deviation of 0.04 centimeters.

Let X = lengths of nails produced in a factory

So, X ~ N([tex]\mu=5.09,\sigma^{2} = 0.04^{2}[/tex])

The z score probability distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean

[tex]\sigma[/tex] = standard deviation

Now, we have to find the two lengths that separate the top 7% and the bottom 7%.

Firstly, Probability that the diameter separate the top 7% is given by;

P(X > x) = 0.07

P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-5.09}{0.04}[/tex] ) = 0.07

P(Z > [tex]\frac{x-5.09}{0.04}[/tex] ) = 0.07

So, the critical value of x in z table which separate the top 9% is given as 1.4996, which means;

[tex]\frac{x-5.09}{0.04}[/tex] = 1.4996

[tex]x-5.09 = 0.04 \times 1.4996[/tex]

[tex]x[/tex] = 5.09 + 0.059984 = 5.15

Secondly, Probability that the diameter separate the bottom 7% is given by;

P(X < x) = 0.07

P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-5.09}{0.04}[/tex] ) = 0.07

P(Z < [tex]\frac{x-5.09}{0.04}[/tex] ) = 0.07

So, the critical value of x in z table which separate the bottom 9% is given as -1.4996, which means;

[tex]\frac{x-5.09}{0.04}[/tex] = -1.4996

[tex]x-5.09 = 0.04 \times (-1.4996)[/tex]

[tex]x[/tex] = 5.09 - 0.059984 = 5.03

Therefore, the two diameters that separate the top 7% and the bottom 7% are 5.15 and 5.03 respectively.