Question

A manufacturer of a certain type of large machine wishes to buy rivets from one of two manufacturers. It is important that the breaking strength of each rivet exceed 10,000 psi. Two manufacturers (A and B) offer this type of rivet and both have rivets whose breaking strength is normally distributed. The mean breaking strengths for manufacturers A and B are 14,000 psi and 13,000 psi, respectively.Which manufacturer will produce, on the average, the fewest number of defective rivets?

Answer 1

Answer:

Not solvable.

Step-by-step explanation:

The normal way to answer this question would be to use the following formula formula for Standard Normal Distribution

$P(X) = P (\frac{X-u}{a})$

Where

$X$ = Minimum Breaking Strength = 10,000 psi

$u$ = Mean Breaking Strength = 14,000 psi / 13,000 psi

$a$ = Standard Deviation of each manufacturer

We put in the values, get the values of Z and use the Standard Normal Distribution Table to get the values for each manufacturer and compare. The manufacturer with the lower value will produce the lower amount of defective sprints.

Since standard deviation values are not given in this question we cannot solve this equation.