Industrial Statistics with Minitab

12.2 Rivets

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 ...