Question

The diameters of ball bearings are distributed normally. The mean diameter is 83 millimeters and the standard deviation is 3 millimeters. Find the probability that the diameter of a selected bearing is greater than 85 millimeters. Round your answer to four decimal places.

Answer 1

Answer:

0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.

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 question, we have that:

$\mu = 83, \sigma = 3$

Find the probability that the diameter of a selected bearing is greater than 85 millimeters.

This is 1 subtracted by the pvalue of Z when X = 85. Then

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

$Z = \frac{85 - 83}{3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

1 - 0.7486 = 0.2514

0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.