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A machine that produces ball bearings has initially been set so that the mean diameter of the bearings it produces is 0.500 inches. A bearing is acceptable if its diameter is within 0.004 inches of this target value. Suppose, however, that the setting has changed during the course of production, so that the distribution of the diameters produced is now approximately normal with mean 0.499 inch and standard deviation 0.002 inch. What percentage of the bearings produced will not be acceptable?

Short Answer

Expert verified
7.3% of the bearings produced will not be acceptable.

Step by step solution

01

- Determine the Acceptable Range of Diameters

The acceptable range of diameter for the bearings is the mean ± 0.004 inches. So, the acceptable range is from 0.500 - 0.004 = 0.496 inches to 0.500 + 0.004 = 0.504 inches.
02

- Compute the Z-scores

The Z-scores for the lower and upper limits of the acceptable range can be computed using the formula \(Z = \frac{X - µ}{σ}\) where X is the value from our dataset, µ is the mean and σ is the standard deviation. For the lower limit: \(Z_1 = \frac{0.496 - 0.499}{0.002} = -1.5\). For the upper limit: \(Z_2 = \frac{0.504 - 0.499}{0.002} = 2.5\)
03

- Find the Area Under the Curve

These Z-scores correspond to areas under a standard normal distribution curve. The area under the curve between these two Z-scores is the probability that a ball bearing has a diameter within the acceptable range. Looking up these Z-scores in the standard normal distribution table, we find that the area for Z = -1.5 is 0.0668, and for Z = 2.5 is 0.9938.
04

- The unacceptable product percentage.

To find the percentage of the bearings that will not be acceptable, subtract the Z score of the lower limit from the Z score of the upper limit and then subtract from 1, i.e., (1 - (0.9938 - 0.0668)) × 100 = 7.3%

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score
Understanding the Z-score is essential when dealing with normal distributions. A Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. In the given exercise, calculating the Z-score for the acceptable range of diameters tells us where these sizes fall relative to the mean diameter.

A Z-score calculation is performed using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where \( X \) represents the data point, \( \mu \) the mean, and \( \sigma \) the standard deviation. Two Z-scores were computed for our ball bearings problem, one for the lower limit and one for the upper limit. Interpreting these Z-scores lets us determine the likelihood of a bearing being within the acceptable size range. Simply put, the Z-score transforms our specific bearing's sizes to a standard form which allows us to compare it to a known distribution—the standard normal distribution.
Standard Deviation
Standard deviation, symbolized by \( \sigma \), is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In our exercise, the standard deviation of the diameters is 0.002 inch, informing us that the diameters are fairly close to the mean (0.499 inch), with less variability. Quantifying variability is critical because it affects the probability of the bearings being within the acceptable range. The smaller the standard deviation, the more concentrated the data points are around the mean, indicating greater precision in the production of the ball bearings.
Normal Distribution Curve
The normal distribution curve, often referred to as the bell curve due to its shape, represents a distribution that has a symmetric shape around the mean, with tails that approach but never touch the horizontal axis. It's defined entirely by its mean \( \mu \) and standard deviation \( \sigma \), determining its center and width, respectively.

In our bearing scenario, the size distribution is said to be approximately normal with a known mean and standard deviation, which allows us to model the distribution and define probabilities for various diameters. The Z-scores we calculated correspond to specific points on this curve and are used to determine the proportion of sizes within the indicated range. The area under the curve between the two Z-scores gives us the probability of a bearing being within the acceptable size range.
Probability
Probability, in the context of statistics, is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

When we look for the probability that a bearing's diameter falls within the acceptable range, we are effectively seeking to find the area under the normal distribution curve between two points—our Z-scores. This area gives us the proportion of the distribution that is considered acceptable. In the ball bearings' problem, after finding the Z-scores and corresponding probabilities, we realized that 1 minus the combined area represented the proportion of bearings that wouldn't meet the acceptable criteria, equating to a 7.3% probability. Thus, understanding the probability allows us to estimate the quality of the bearing production run and expect that approximately 7.3 out of every 100 bearings produced will be outside the acceptable range.

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Most popular questions from this chapter

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1 .\) Determine each of the following probabilities: a. \(P(z < 0.10)\) b. \(P(z < -0.10)\) c. \(P(0.40 < z < 0.85)\) d. \(P(-0.85 < z < -0.40)\) e. \(P(-0.40 < z < 0.85)\) f. \(P(z > \- 1.25)\) g. \(P(z < -1.50\) or \(z > 2.50)\)

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