An industry produces rivets for metal sheets used in the building sector. An essential feature, besides the dimensions of the rivet, is its shear strength, which is a measure of the strength that the rivet can resist in the direction of a cross section.

The intended use of a certain type of rivet and market conditions require having a minimum strength of 2500 psi (pounds per square inch). The manufacturing process gives this magnitude a variability that can be characterized by a normal distribution with σ = 20 psi.

The questions we pose are:

**1.** What should be the average value of strength if we accept 2.5% of the rivets with a value below the required minimum?

**2.** Some assembly requires the placement of 20 rivets. For the correct behaviour of the installation, at least 15 rivets must have a shear strength above the required minimum. If the rivets are manufactured with the process centred in 2525 psi, what is the probability that the assembly behaves correctly?

**1.** We have to find the value of μ in the following drawing:
As the tail area is 2.5%, the value 2500 will be approximately 2σ far from μ, so μ will be around 2540. We can use Minitab to, by trial and error, find the exact value. Go to Calc > Probability Distributions > Normal:
and the result is 0.0228. As it does not reach 2.5% we can try again reducing the average. But if we replace 2540 by 2539 ...