What Percent of Ball Bearings Have a Diameter of 20.27 mm or More?

E

z = (x-u)/sigma = 20.27 - 20/0.15 = 1.8. from the z-table, z = 1.8 is 0.4641. So 1.0 - 0.9641 = 0.0359.

To solve this problem, we will use the concept of the normal distribution and the z-score.

The given information tells us that the mean diameter of the ball bearings is 20mm and the standard deviation is 0.150mm. We want to find the percentage of ball bearings that have a diameter of 20.27mm or more.

First, we need to calculate the z-score for the diameter of 20.27mm. The z-score is a measure of how many standard deviations a particular value is away from the mean. It is calculated using the formula:

z = (x - μ) / σ

where:

• z is the z-score,
• x is the value we are interested in (20.27mm in this case),
• μ is the mean (20mm in this case), and
• σ is the standard deviation (0.150mm in this case).

Plugging in the values, we get:

z = (20.27 - 20) / 0.150 = 1.8

Next, we need to find the corresponding area under the normal distribution curve for this z-score. This area represents the percentage of values that are less than or equal to 20.27mm.

We can use a standard normal distribution table (also known as a z-table) to find this area. The table provides the cumulative probability up to a given z-score. However, since the table usually gives probabilities for z-scores that are less than a given value, we need to find the complement of the desired area.

The complement of the area represents the percentage of values that are greater than the given value. In this case, we want to find the percentage of ball bearings that have a diameter of 20.27mm or more, so we need to find the complement of the area.

From the standard normal distribution table, we find that the area to the left of 1.8 is approximately 0.9641. To find the complement, we subtract this value from 1:

Complement = 1 - 0.9641 = 0.0359

Therefore, approximately 3.59% of the ball bearings will have a diameter of 20.27mm or more.

The correct answer is E. 3.59%.