/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 A machine that cuts corks for wi... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm}\). The specifications call for corks with diameters between \(2.9\) and \(3.1 \mathrm{~cm}\). 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 are defective?

Short Answer

Expert verified
Finally, computing the total value we get the proportion of defective corks produced by the machine.

Step by step solution

01

Standardize the values

To solve this exercise, first, we need to standardize the cork diameters of \(2.9 \) cm and \(3.1 \) cm, to translate them into z-scores. We use the formula \(Z = (X - μ) / σ \), where X is the value we want to standardize, μ is the mean and σ is the standard deviation. Thus, we calculate the Z score for \(2.9 \) cm and \(3.1 \) cm diameter corks.
02

Look up areas from standard normal table

Once we have the Z-scores, we look up the corresponding areas in the Standard Normal (Z) Table. The area under the curve in a normal distribution represents the probability, so we find the probability corresponding to the Z Scores for \(2.9 \) cm and \(3.1 \) cm.
03

Calculate the proportion of defective corks

Now we have two probabilities. The first represents the proportion of corks that is too small and the second represents the proportion of corks that is too large. To find out the total proportion of defective corks, we will need to add up these two probabilities.

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Key Concepts

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

Z-scores
When dealing with a normal distribution, Z-scores are a powerful tool for understanding how individual data points compare to the mean of the data set. A Z-score tells us how many standard deviations away a particular value is from the mean.

In simple terms, if you have a Z-score of 1, your value is one standard deviation above the mean. Conversely, a Z-score of -1 means the value is one standard deviation below the mean. The Z-score is calculated using the following formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \(X\) is the value you're examining, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation.

In the context of our cork exercise, we used the Z-score formula to transform cork diameters of 2.9 cm and 3.1 cm into Z-scores. This standardization allows us to determine how many standard deviations these diameters are from the average size cork the machine is supposed to produce.
standard deviation
A standard deviation is a measure of how much the values in a data set vary or spread out from the mean. A smaller standard deviation indicates that the data points are close to the mean, while a larger standard deviation shows that the points are spread out over a wider range of values.

Calculating the standard deviation involves several steps, including finding the mean (average) of the data, subtracting the mean from each data point, squaring the result, finding the average of those squared differences, and finally taking the square root. In formulas, it's expressed as:
  • \( \sigma = \sqrt{ \frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2 } \)
In our exercise, the standard deviation of 0.1 cm tells us that most of the cork diameters are close to the mean diameter, which is 3 cm. It gives insight into the typical deviation Cork diameters are expected to have under normal manufacturing conditions. Understanding the standard deviation helps in grasping how tightly or loosely the cork diameters are bound to their average size, influencing decisions about quality control and defect rates.
probability
Probability is all about measuring the likelihood of an event occurring. In the world of normal distribution, it is often represented as the area under a curve in a graph.

The probability in a standard normal distribution is obtained using Z-scores. For our cork exercise, once we found the Z-scores for the 2.9 cm and 3.1 cm corks, we turned to a standard normal table, also known as the Z-table, to find corresponding probabilities.
  • The probability that a cork is smaller than 2.9 cm corresponds to the area under the curve to the left of the Z-score for 2.9 cm.
  • Likewise, the probability that a cork is larger than 3.1 cm is associated with the area to the right of the Z-score for 3.1 cm.
By adding these two probabilities, we could ascertain the overall proportion of corks that fall outside the specified range of 2.9 cm to 3.1 cm. This is useful because it quantifies not only the defective rate but potential costs involved in discarding or reprocessing non-compliant corks, ensuring product standards and customer satisfaction.

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Most popular questions from this chapter

To assemble a piece of furniture, a wood peg must be inserted into a predrilled hole. Suppose that the diameter of a randomly selected peg is a random variable with mean \(0.25\) inch and standard deviation \(0.006\) inch and that the diameter of a randomly selected hole is a random variable with mean \(0.253\) inch and standard deviation \(0.002\) inch. Let \(x_{1}=\) peg diameter, and let \(x_{2}=\) denote hole diameter. a. Why would the random variable \(y\), defined as \(y=\) \(x_{2}-x_{1}\), be of interest to the furniture manufacturer? b. What is the mean value of the random variable \(y\) ? c. Assuming that \(x_{1}\) and \(x_{2}\) are independent, what is the standard deviation of \(y\) ? d. Is it reasonable to think that \(x_{1}\) and \(x_{2}\) are independent? Explain. e. Based on your answers to Parts (b) and (c), do you think that finding a peg that is too big to fit in the predrilled hole would be a relatively common or a relatively rare occurrence? Explain.

Of all airline flight requests received by a certain discount ticket broker, \(70 \%\) are for domestic travel (D) and \(30 \%\) are for international flights (I). Let \(x\) be the number of requests among the next three requests received that are for domestic flights. Assuming independence of successive requests, determine the probability distribution of \(x\). (Hint: One possible outcome is DID, with the probability \((.7)(.3)(.7)=.147 .\) )

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Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of 2000 of these signatures have (approximately) a binomial distribution? Explain.

Suppose a playlist on an MP3 music player consists of 100 songs, of which eight are by a particular artist. Suppose that songs are played by selecting a song at random (with replacement) from the playlist. The random variable \(x\) represents the number of songs played until a song by this artist is played. a. Explain why the probability distribution of \(x\) is not binomial. b. Find the following probabilities: i. \(p(4)\) ii. \(P(x \leq 4)\) iii. \(P(x>4)\) iv. \(P(x \geq 4)\) c. Interpret each of the probabilities in Part (b) and explain the difference between them.

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