# A process manufactures ball bearings with diameters that are normally distributed with mean 25.14 mm and standard deviation 0.08 mm. NOTE: T

A process manufactures ball bearings with diameters that are normally distributed with mean 25.14 mm and standard deviation 0.08 mm. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A particular ball bearing has a diameter of 25.2 mm. What percentile is its diameter on

### 0 thoughts on “A process manufactures ball bearings with diameters that are normally distributed with mean 25.14 mm and standard deviation 0.08 mm. NOTE: T”

The diameter is in the 77th percentile.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $$\mu$$ and standard deviation $$\sigma$$, the zscore of a measure X is given by:

$$Z = \frac{X – \mu}{\sigma}$$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

25.14 mm and standard deviation 0.08 mm.

This means that $$\mu = 25.14, \sigma = 0.08$$

A particular ball bearing has a diameter of 25.2 mm. What percentile is its diameter on?

This is the pvalue of Z when X = 25.2. So

$$Z = \frac{X – \mu}{\sigma}$$

$$Z = \frac{25.2 – 25.14}{0.08}$$

$$Z = 0.75$$

$$Z = 0.75$$ has a pvalue of 0.77.

The diameter is in the 77th percentile.