The diameters of ball bearings are distributed normally. The mean diameter is 126 millimeters and the standard deviation is 3 millimeters. Find the probability that the diameter of a selected bearing is greater than 123 millimeters. Round your answer to four decimal places.

0.8413 = 84.13% probability that the diameter of a selected bearing is greater than 123 millimeters.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

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:

Find the probability that the diameter of a selected bearing is greater than 123 millimeters.

This is 1 subtracted by the pvalue of Z when X = 123. So

has a pvalue of 0.1587

1 - 0.1587 = 0.8413

0.8413 = 84.13% probability that the diameter of a selected bearing is greater than 123 millimeters.

42

step-by-step explanation:

256/6=42