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

To find the probability that the diameter of a selected bearing is greater than 85 millimeters, calculate the z-score and use the standard normal distribution table. The resulting probability is approximately 0.7486.

Step-by-step explanation:

To find the probability that the diameter of a selected bearing is greater than 85 millimeters, we need to calculate the z-score and then use the standard normal distribution table. The z-score is found by using the formula:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. Substituting the given values into the formula, we get:

z = (85 - 83) / 3 = 0.67

Using the standard normal distribution table, we find the probability corresponding to a z-score of 0.67 is approximately 0.7486. Therefore, the probability that the diameter of a selected bearing is greater than 85 millimeters is 0.7486.

Answer 2

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

`Z = (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.