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

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  1. Answer:

    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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    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 [tex]\mu = 25.14, \sigma = 0.08[/tex]

    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

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    [tex]Z = \frac{25.2 – 25.14}{0.08}[/tex]

    [tex]Z = 0.75[/tex]

    [tex]Z = 0.75[/tex] has a pvalue of 0.77.

    The diameter is in the 77th percentile.

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