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The extraction force required to remove a cork from a bottle of wine has a normal distribution with a mean of 310 Newtons and a standard deviation of 36 Newtons. a. The specs for this variable, given in Applied Example \(6.13,\) were "\(300\mathrm{N}+100 \mathrm{N} /-150 \mathrm{N}\)" Express these specs as an interval. b. What percentage of the corks is expected to fall within the specs? c. What percentage of the tested corks will have an extraction force of more than 250 Newtons? d. What percentage of the tested corks will have an extraction force within 50 Newtons of \(310 ?\)

Short Answer

Expert verified
a. The specs interval is \((150, 400)\) Newtons. b. The percentage of corks within the specs, and c. the percentage of corks with more than 250 Newtons, and d. the percentage within 50 Newtons of 310 will depend on the looked up proportions from the standard normal distribution table corresponding to the calculated Z-scores.

Step by step solution

01

Convert the specs into an interval

The specs \(300N + 100N/-150N\) mean we are looking at interval \((300 - 150, 300 + 100)\), which is \((150, 400)\) Newtons.
02

Calculate the percentage of corks within the specs

To calculate percentage of corks falling within specification, we need to standardize the specification limits to Z-scores. A Z-score is given by \((X - \mu) / \sigma\), where X is the value, \mu is the mean and \sigma is the standard deviation. So, standardize the limits: \[ Z_{lower} = \frac{150 - 310}{36} \text{ and } Z_{upper} = \frac{400 - 310}{36} \]. Next, we look up these Z-scores on the standard normal distribution table to find the proportion of data less than these Z-scores. The difference in these proportions gives the proportion in the specification, from which percentage can be calculated.
03

Calculate the percentage of corks with more than 250 Newtons force

To calculate the percentage of corks with more than 250 Newtons, we standardize the value to Z-score as in previous step 2's content. Next, we look up this Z-score on standard normal distribution table. Since we need to find the percentage **more than** 250, we subtract the looked up proportion from 1.
04

Calculate the percentage of corks within 50 Newtons of 310

This step is similar to step 2 where we are now looking at the interval \((310-50, 310+50)\). Standardize the limits to Z-scores and find the proportions less than these Z-scores from the standard normal distribution table. The difference gives the proportion within 50 Newtons of 310, which can be converted to percentage by multiplication by 100.

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Key Concepts

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

Standard Deviation
The standard deviation is a measure of how much variation or dispersion exists from the average (mean) value. Imagine you're at a fruit stand, and you want to have fruits of approximately the same size. If the sizes are all over the place, that's high variation, much like a high standard deviation. A small standard deviation means that the fruit sizes are more consistent, closer to the mean size. In the context of our wine corks example, a standard deviation of 36 Newtons tells us that the force required to remove most corks from their bottles deviates from the mean (average) force of 310 Newtons by up to 36 Newtons either side, more or less.

In a practical sense, a winery with a lower standard deviation in cork removal force might be perceived as more reliable or predictable, important qualities in product manufacturing and customer satisfaction.
Z-score
Think of Z-score as a way of grading your test score against the class average. If the class mean score is 80% and the standard deviation is 10%, and you scored a 90%, your Z-score would tell you how many standard deviations above the mean your score is. For the wine corks, the Z-score is calculated by taking the cork force value (X), subtracting the mean force (310 Newtons), and dividing this result by the standard deviation (36 Newtons).

The resulting Z-score helps us compare individual cork forces to the average force needed. It's like taking any force value and expressing how 'typical' or 'atypical' it is within the context of all corks tested. When you perform this standardization of the data, you're essentially placing all the corks on a common scale and can use the normal distribution to make inferences, like the percentage of corks within a certain force range.
Specification Limits
Specification limits are the boundaries of acceptability for any product characteristic, in our case, the force to remove a wine cork. They are set by design or customer requirements and often reflect the maximum permissible deviation from a target value. In our example, the specifications were given as an interval around a target: 300N plus or minus a range. So, we have an upper and lower specification limit, essentially setting a 'goalposts' scenario, where the aim is to have the cork forces fall within this spread.

To compute the percentage of cork forces within these specification limits, we first translate our interval into Z-scores, which allows us to use the normal distribution properties. By checking these Z-scores against a standard normal distribution table, we determine the percentage of corks expected to have an extraction force that falls within the ideal, consumer-friendly realm established by those limits.

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Most popular questions from this chapter

a. Find the standard \(z\)-score such that \(80 \%\) of the distribution is below (to the left of) this value. b. Find the standard \(z\)-score such that the area to the right of this value is 0.15. c. Find the two \(z\)-scores that bound the middle \(50 \%\) of a normal distribution.

Understanding the \(z\) notation, \(z(\alpha),\) requires us to know whether we have a \(z\) -score or an area. Each of the following expressions uses the \(z\) notation in a variety of ways, some typical and some not so typical. Find the value asked for in each of the following, and then with the aid of a diagram explain what your answer represents. a. \(\quad z(0.08)\) b. the area between \(z(0.98)\) and \(z(0.02)\) c. \(\quad z(1.00-0.01)\) d. \(\quad z(0.025)-z(0.975)\)

According to the Federal Highway Administration's 2006 highway statistics, the distribution of ages for licensed drivers has a mean of 47.5 years and a standard deviation of 16.6 years [www.fhwa.dot.gov]. Assuming the distribution of ages is normally distributed, what percentage of the drivers are: a. between the ages of 17 and \(22 ?\) b. younger than 25 years of age? c. older than 21 years of age? d. between the ages of 48 and \(68 ?\) e. older than 75 years of age?

Depending on where you live and on the quality of the day care, costs of day care can range from \(\$ 3000\) to \(\$ 15,000\) a year (or \(\$ 250\) to \(\$ 1250\) a month) for one child, according to the Baby Center. Day care centers in large cities such as New York and San Francisco are notoriously expensive. Suppose that day care costs are normally distributed with a mean equal to \(\$ 9000\) and a standard deviation equal to \(\$ 1800 .\) a. What percentage of day care centers cost between \(\$ 7200\) and \(\$ 10,800 ?\) b. What percentage of day care centers cost between \(\$ 5400\) and \(\$ 12,600 ?\) c. What percentage of day care centers cost between \(\$ 3600\) and \(\$ 14,400 ?\) d. Compare the results in a through c with the empirical rule. Explain the relationship.

a. Find the area under the normal curve for \(z\) between \(z(0.95)\) and \(z(0.025).\) b. \(\quad\) Find \(z(0.025)-z(0.95).\)

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