Question

Your manufacturing company sells metal nuts and bolts that come together in a package. The diameter of the bolts coming off the factory line follows a normal distribution with mean 18mm and standard deviation 0.1mm. The diameters of the holes in the nuts follow a normal distribution with mean 18.7mm and standard deviation of 0.08mm. Assume that in order for a nut and bolt to fit together the diameter of the hole in the nut has to be between .5 and .9 mm larger than the bolt. Find the z score for bolt of 18.12 mm.

Answer 1

The z score for bolt of diameter 18.12 mm is 1.20.

Explanation:

Let X = diameter of bolts.

It is provided that the random variable X follows a Normal distribution with mean, μ = 18 mm and standard deviation, σ = 0.10 mm.

A z-score is a standardized score, a numerical, that defines how far a data value from the mean.

The distribution of z-scores is defined by the Standard Normal distribution.

`Z\sim N(0, 1)`

The formula to compute the z-score is:

`z=(x-\mu)/(\sigma)`

The value of the diameter of a bolt is, x = 18.12 mm.

Compute the z-score for this value as follows:

`z=(x-\mu)/(\sigma)=(18.12-18)/(0.10)=(0.12)/(0.10)=1.20`

Thus, the z score for bolt of diameter 18.12 mm is 1.20.