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Chapter 7: Problem 30

A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean 3 centimeters and standard deviation 0.1 centimeters. The specifications call for corks with diameters between 2.9 and 3.1 centimeters. A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine is defective?

Short Answer

Expert verified
The proportion of corks produced by the machine that are defective is 0.3174 or 31.74%.

Step by step solution

01

Identify the Given Parameters

In this scenario, the mean (\( \mu \)) is 3 cm, and the standard deviation (\( \sigma \)) is 0.1 cm. The required diameter of the corks lies within the range 2.9 - 3.1 cm.
02

Compute the Z-scores

The z-score is calculated using the formula - \( Z = \frac{(X - \mu)}{\sigma} \) where X is the observation. We calculate two z-scores: \( Z1 = \frac{(2.9 - 3.0)}{0.1} = -1 \) and \( Z2 = \frac{(3.1 - 3.0)}{0.1} = 1 \). These are the scores for the lower limit (2.9 cm) and upper limit (3.1 cm) of the diameter respectively.
03

Determine the Proportion of Corks within Specifications

To find the proportion of diameters that lie between these two limits, consult a standard z-table or use the cumulative distribution function. The values corresponding to -1 and 1 are 0.1587 and 0.8413 respectively. Therefore, the proportion of corks within the required specifications is \( 0.8413 - 0.1587 = 0.6826 \).
04

Determine the Proportion of Defective Corks

The defective corks are those which do not lie within the 2.9 cm to 3.1 cm diameter range. Thus, the proportion of defective corks will be the total proportion minus the proportion that meet the specifications: \( 1 - 0.6826 = 0.3174 \).

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Key Concepts

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

Understanding Z-score Calculation
Z-score calculation is a fundamental concept in statistics, especially when dealing with the normal distribution. The z-score is a measurement that describes a value's position relative to the mean of a group of values. In simpler terms, it tells us how many standard deviations an element is from the mean.

The formula for calculating the z-score is: \[Z = \frac{(X - \mu)}{\sigma}\]where:
  • \(X\) is the value or observation
  • \(\mu\) is the mean of the distribution
  • \(\sigma\) is the standard deviation
If a z-score is 0, it indicates that the observation is exactly at the mean. Positive z-scores indicate values above the mean, while negative z-scores represent values below it.

In the exercise, two z-scores were calculated for cork diameters of 2.9 cm and 3.1 cm, resulting in z-scores of -1 and 1, respectively. These scores help determine how unlikely it is for a cork鈥檚 diameter to fall outside the specified range.
The Role of Standard Deviation
Standard deviation is a key concept in statistics and vital in understanding any data set's spread or variation. It quantifies how much the observations deviate from the mean, giving insight into the distribution's variability.

A small standard deviation indicates the data points are close to the mean, while a large standard deviation shows they are spread out over a wider range. In a normal distribution like the one in the cork cutting machine scenario, approximately 68% of the data falls within one standard deviation (\(\mu \pm \sigma\)) of the mean.

In this exercise, the machine鈥檚 cork diameter distribution has a standard deviation of 0.1 cm. This value not only helps calculate z-scores but also aids in assessing the variation in cork sizes, directly impacting the proportion of defective corks.
Defective Products Analysis
Analyzing defective products is crucial for maintaining quality control in manufacturing processes. In the given exercise, corks are considered defective if they don't meet the specified diameter range of 2.9 to 3.1 cm. Such defects can lead to functional problems, like leaking or improper sealing of wine bottles.

To determine the number of defective corks, we use the normal distribution model to find the proportion of corks that exceed the acceptable range. First, we calculate the z-scores for the boundary values (2.9 cm and 3.1 cm) and then use a z-table or cumulative distribution function to find the proportion of corks within these limits, which in this case is 68.26%.

The percentage of defective corks is simply the complement of this value, indicating that 31.74% of the corks are either too small or too large. Such analysis helps in identifying process improvements and minimizing waste.

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Most popular questions from this chapter

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