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

Answer:

0.1507 or 15.07%.

Step-by-step 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.

$z=\frac{x-\mu}{\sigma}$, where,

z = z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

$z=\frac{21.97-22}{0.016}$

$z=\frac{-0.03}{0.016}$

$z=-0.1875$

Let us find z-score of data point 22.03.

$z=\frac{22.03-22}{0.016}$

$z=\frac{0.03}{0.016}$

$z=0.1875$

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%.