Question

Ball-Bearings, Inc. produces ball bearings on a Kronar BBX machine. For one specific ball bearing produced, the mean diameter is set at 20.00 mm (millimeters). The standard deviation over a long period of time was computed to be 0.150 mm. Production is normally distributed. What percent of the ball bearings will have diameters of 20.27 mm or more

Answer 1

Answer:

3.59% of the ball bearings will have diameters of 20.27 mm or more.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 20, \sigma = 0.15$

What percent of the ball bearings will have diameters of 20.27 mm or more?

This is 1 subtracted by the pvalue of Z when X = 20.27. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.27 - 20}{0.15}$

$Z = 1.8$

$Z = 1.8$ has a pvalue of 0.9641.

1-0.9641 = 0.0359

3.59% of the ball bearings will have diameters of 20.27 mm or more.