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lys-0071 [83]
3 years ago
12

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
Mathematics
1 answer:
Misha Larkins [42]3 years ago
6 0

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, we will get:

P(-0.1875

P(-0.1875  

P(-0.1875

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

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