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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

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Answer:

The diameter is in the 77th percentile.

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 = (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 = (X - \mu)/(\sigma)


Z = (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.

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