Question

The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters. a. What is the probability that a randomly selected ball bearing will be acceptable

Answer 1

0.1507 or 15.07%.

Explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

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

z = z-score,

x = Sample score,

`\mu` = Mean,

`\sigma` = Standard deviation.

`z=(21.97-22)/(0.016)`

`z=(-0.03)/(0.016)`

`z=-0.1875`

Let us find z-score of data point 22.03.

`z=(22.03-22)/(0.016)`

`z=(0.03)/(0.016)`

`z=0.1875`

Using probability formula
`P(a<z<b)=P(z<b)-P(z<a)`, we will get:

`P(-0.1875<z<0.1875)=P(z<0.1875)-P(z<-0.1875)`

`P(-0.1875<z<0.1875)=0.57535-0.42465`

`P(-0.1875<z<0.1875)=0.1507`

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.