The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.585.58 millimeters and a standard deviation of 0.040.04 millimeters. Find the two diameters that separate the top 5%5% and the bottom 5%5%. These diameters could serve as limits used to identify which bolts should be rejected. Round your answer to the nearest hundredth, if necessary.

We are given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.58 millimeters and a standard deviation of 0.04 millimeters.

Let X = diameters of bolts produced in a machine shop

So, X ~ N([tex]\mu=5.58,\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] = mean diameter = 5.58 millimeter

[tex]\sigma[/tex] = standard deviation = 0.04 millimeter

Now, we have to find the two diameters that separate the top 5% and the bottom 5%.

Firstly, Probability that the diameter separate the top 5% is given by;

P(X > x) = 0.05

P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-5.58}{0.04}[/tex] ) = 0.05

P(Z > [tex]\frac{x-5.58}{0.04}[/tex] ) = 0.05

So, the critical value of x in z table which separate the top 5% is given as 1.6449, which means;

[tex]\frac{x-5.58}{0.04}[/tex] = 1.6449

[tex]{x-5.58} = 1.6449 \times {0.04}[/tex]

[tex]x[/tex] = 5.58 + 0.065796 = 5.65

Secondly, Probability that the diameter separate the bottom 5% is given by;

P(X < x) = 0.05

P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-5.58}{0.04}[/tex] ) = 0.05

P(Z < [tex]\frac{x-5.58}{0.04}[/tex] ) = 0.05

So, the critical value of x in z table which separate the bottom 5% is given as -1.6449, which means;

[tex]\frac{x-5.58}{0.04}[/tex] = -1.6449

[tex]{x-5.58} = -1.6449 \times {0.04}[/tex]

[tex]x[/tex] = 5.58 - 0.065796 = 5.51

Therefore, the two diameters that separate the top 5% and the bottom 5% are 5.51 and 5.65 respectively.