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A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weights \(\bar{x}\). If test results over a long period of time show that \(5 \%\) of the \(\bar{x}\) values are over 2.1 pounds and \(5 \%\) are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process?

Short Answer

Expert verified
Mean is 1.9 pounds, standard deviation is approximately 0.6667 pounds.

Step by step solution

01

Understand the Problem

We are given a situation where the sample mean weights \(\bar{x}\) follow a normal distribution and that the distribution has 5% of the values above 2.1 pounds and 5% below 1.9 pounds. We need to find the mean \(\mu\) and standard deviation \(\sigma\) of the population.
02

Identify the Z-Scores

Since the sample mean follows a normal distribution and uses samples of size 30, we use the fact that 5% on either side of the normal distribution corresponds to Z-scores of approximately \(-1.645\) and \(1.645\). This is based on the standard normal distribution properties.
03

Set Up the Equations for Z-Scores

We have two equations based on the Z-score formula, \((\bar{x} - \mu) / (\sigma/\sqrt{n})\). For the upper limit: \(1.645 = (2.1 - \mu) / (\sigma/\sqrt{30})\), and for the lower limit: \(-1.645 = (1.9 - \mu) / (\sigma/\sqrt{30})\).
04

Solve Equation for Upper Limit

Substituting in the upper limit equation: \((2.1 - \mu) = 1.645(\sigma/\sqrt{30})\). From here, we deduce: \(2.1 - \mu = 1.645 \times \frac{\sigma}{\sqrt{30}} = 0.30045\sigma\) (approximately).
05

Solve Equation for Lower Limit

For the lower limit: \(1.9 - \mu = -1.645(\sigma/\sqrt{30})\). Rearranging gives: \((1.9 - \mu) = -0.30045\sigma\).
06

Solve for Mean and Standard Deviation

We have two equations: \(2.1 - \mu = 0.30045\sigma\) and \(1.9 - \mu = -0.30045\sigma\). Solving these simultaneously, we add them to eliminate \(\mu\), giving \(0.4 = 2(0.30045\sigma)\), so \(\sigma = \frac{0.4}{0.6009} \approx 0.6667\). Now solve for \(\mu\) using either equation: \(2.1 - \mu = 0.30045(0.6667)\), thus \(\mu = 2.1 - 0.2002 \approx 1.9\).
07

Conclusion

The population mean \(\mu\) is 1.9 and the standard deviation \(\sigma\) is approximately 0.6667.

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Key Concepts

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

Normal Distribution
Normal distribution is one of the most important statistical concepts that describes how data is spread out over a range of values. It is also known as the Gaussian distribution. This distribution is symmetric around a central value, often referred to as the mean. The curve of the normal distribution is bell-shaped, where most of the data points gather around the mean, and the probability of deviations decreases as you move away from the center.

The normal distribution is significant because many variables in nature, such as heights, test scores, and manufacturing measurements, tend to follow this pattern. In the context of our exercise, the sample mean weights from the production process are assumed to follow a normal distribution. This understanding helps in predicting probabilities and making quality control assessments in manufacturing.
Z-Scores
Z-scores are a statistical measurement that describe a value's relation to the mean of a group of values. They are expressed in terms of standard deviations from the mean. A Z-score tells us how many standard deviations an element is from the mean.
  • A Z-score of 0 indicates that the value is exactly at the mean.
  • Positive Z-scores indicate values above the mean.
  • Negative Z-scores signify values below the mean.
In our exercise, we identify Z-scores based on the percentage distribution of sample means that fall 5% above and below certain weights. These correspond to Z-scores of approximately 1.645 and -1.645, respectively. This allows us to create equations to find the mean and standard deviation of the population using these boundary points.
Sample Mean
The sample mean \( \bar{x} \) is the average weight found by the quality control inspector for each sample of 30 products. It is calculated by adding all the product weights in a sample and then dividing by the number of products. This value provides a snapshot of each sample's central tendency, allowing inspectors to monitor and maintain production quality.

In statistical terms, the sample mean is a point estimate of the population mean \( \mu \). When sample means are collected over time, their distribution tends to be normal, which is a key assumption in applying statistical process control. This property of the sample mean helps in making predictions and deductions about the overall production process.
Population Mean
The population mean \( \mu \) is the average weight of all the products in the entire production. Unlike the sample mean, which is calculated from a subset, the population mean reflects the central value of the entire dataset or production output. In our problem, we determine the population mean by using characteristics of the normal distribution and Z-scores to set up equations for bounds provided by the sample mean values.

Through solving these equations, we find that the population mean \( \mu \) is 1.9 pounds. This value serves as a benchmark or target quality control goal that helps keep production consistent and within specified limits.
Standard Deviation
Standard deviation \( \sigma \) is a measure that tells us how much variation or dispersion exists from the average. In the context of a normal distribution, it shows how tightly data points are clustered around the mean. A small standard deviation indicates that data points tend to be close to the mean, while a large standard deviation implies wider dispersion.

In our problem, through calculated Z-scores and solving equations, we find the standard deviation to be approximately 0.6667 pounds. This suggests that the product weights in the production process tend to vary by this amount around the mean weight. Understanding standard deviation is crucial as it provides insights into the consistency and reliability of the manufacturing process.

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Most popular questions from this chapter

The American Association of Individual Investors (AAII) polls its subscribers on a weekly basis to determine the number who are bullish, bearish, or neutral on the short-term prospects for the stock market. Their findings for the week ending March \(2,2006,\) are consistent with the following sample results (AAII website, March 7, 2006). Bullish \(409 \quad\) Neutral \(299 \quad\) Bearish \(291\) Develop a point estimate of the following population parameters. a. The proportion of all AAII subscribers who are bullish on the stock market. b. The proportion of all AAII subscribers who are neutral on the stock market. c. The proportion of all AAII subscribers who are bearish on the stock market.

Three firms carry inventories that differ in size. Firm A's inventory contains 2000 items, firm B's inventory contains 5000 items, and firm C's inventory contains 10,000 items. The population standard deviation for the cost of the items in each firm's inventory is \(\sigma=144\) A statistical consultant recommends that each firm take a sample of 50 items from its inventory to provide statistically valid estimates of the average cost per item. Managers of the small firm state that because it has the smallest population, it should be able to make the estimate from a much smaller sample than that required by the larger firms. However, the consultant states that to obtain the same standard error and thus the same precision in the sample results, all firms should use the same sample size regardless of population size. a. Using the finite population correction factor, compute the standard error for each of the three firms given a sample of size 50 b. What is the probability that for each firm the sample mean \(\bar{x}\) will be within ±25 of the population mean \(\mu ?\)

People end up tossing \(12 \%\) of what they buy at the grocery store (Reader 's Digest, March, 2009 ). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior. a. Show the sampling distribution of \(\bar{p},\) the proportion of groceries thrown out by your sample respondents. b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion? c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion?

Barron 's reported that the average number of weeks an individual is unemployed is 17.5 weeks (Barron \(s\), February 18,2008 ). Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study. a. Show the sampling distribution of \(\bar{x}\), the sample mean average for a sample of 50 unemployed individuals. b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean? c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within \(1 / 2\) week of the population mean?

The average price of a gallon of unleaded regular gasoline was reported to be \(\$ 2.34\) in northern Kentucky (The Cincinnati Enquirer, January 21,2006 ). Use this price as the population mean, and assume the population standard deviation is \(\$ .20\). a. What is the probability that the mean price for a sample of 30 service stations is within \(\$ .03\) of the population mean? b. What is the probability that the mean price for a sample of 50 service stations is within \(\$ .03\) of the population mean? c. What is the probability that the mean price for a sample of 100 service stations is within \(\$ .03\) of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within \(\$ .03\) of the population mean?

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