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The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is \(12,200 .\) The distribution of pages printed per cartridge closely follows the normal probability distribution \(\begin{array}{lllll}\text { and the standard deviation is } 820 & \text { pages. The }\end{array}\) manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last. How many pages should the manufacturer advertise for each cartridge if it wants to be correct 99 percent of the time?

Short Answer

Expert verified
The manufacturer should advertise each cartridge lasts about 14,111 pages.

Step by step solution

01

Understanding the Problem

We need to determine the number of pages a cartridge should last to ensure that it meets the expectation 99% of the time, given that the mean and standard deviation of printing pages are known and follow a normal distribution.
02

Identify the Given Values

The problem provides us with a mean (\(\mu = 12,200\) pages) and a standard deviation (\(\sigma = 820\) pages). We need to find the page count value (\(x\)) that falls below 99% of the distribution.
03

Find the Z-value Corresponding to 99%

To achieve a 99% reliability, we need to find the Z-value that corresponds to a cumulative probability of 0.99 in the standard normal distribution table. From a standard normal distribution table, the Z-value for 0.99 is approximately 2.33.
04

Calculate the Required Page Count

Using the formula for conversion from a Z-score to a value \(x = \mu + Z \cdot \sigma\),we substitute the known values:\(x = 12,200 + 2.33 \times 820\).
05

Perform the Calculation

Compute the expression:\(x = 12,200 + 2.33 \times 820\)\,\(x = 12,200 + 1,910.6\),so \(x \approx 14,110.6\).Thus, the manufacturer should advertise that each cartridge lasts approximately 14,111 pages.

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Key Concepts

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

Standard Deviation
Standard deviation is a key concept in statistics that measures the amount of variation or dispersion of a set of values. In simpler terms, it tells us how much the individual data points differ from the average (mean), and hence from each other.
If we have a set of data, like the number of pages a printer cartridge can print, the standard deviation helps us understand how spread out these values are around their average or mean value.
  • If the standard deviation is large, the data points are more spread out from the mean.
  • If it is small, the points are closer to the mean, indicating less variability.
In the given context of the printer cartridges, the standard deviation is 820 pages. This means that for most cartridges, the number of printed pages will vary by about 820 pages above or below the average of 12,200 pages.
Z-score
A Z-score is another fundamental statistical concept that measures how many standard deviations a data point is from the mean. To understand it fully, think of the Z-score as a way to standardize different data points so they can be easily compared. When dealing with normal distribution, the Z-score helps translate the particular data point into a number that indicates how far and in what direction it deviates from the mean. In our example, if we want to find out how many pages a cartridge should last 99% of the time, we need a Z-score. Here’s how we use it:
  • The Z-score is found using standard Z-tables, which tell us how much of the data lies below a particular Z-value.
  • For 99% reliability or probability, the Z-score is approximately 2.33, meaning the data point is 2.33 standard deviations above the mean.
This Z-score is crucial to determining how many pages should be advertised to cover 99% of cases. By multiplying the Z-score by the standard deviation and adding it to the mean, we can find the desired page count.
Cumulative Probability
Cumulative probability refers to the likelihood that a random variable will be less than or equal to a certain value. When dealing with a normal distribution, this concept is often visualized as the area under the curve. In the printer cartridge example, cumulative probability helps determine the number of printed pages that can be expected most of the time. When the problem states a 99% reliability goal, it means the printer cartridges should meet or exceed a certain number of pages in 99% of the cases.
  • To find this, cumulative probability tables (or Z-tables) are used.
  • The desired cumulative probability of 0.99 corresponds to a Z-score that assists in finding the cutoff value for the page count.
By comprehending cumulative probability, manufacturers can set realistic expectations for their products, ensuring that customers are satisfied with predictable performance outcomes.

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Most popular questions from this chapter

A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone's privacy. a. How much toothpaste would you expect to be remaining in the tube? b. What is the standard deviation of the amount remaining in the tube? c. What is the likelihood there is less than 3.0 ounces remaining in the tube? d. What is the probability there is more than 1.5 ounces remaining in the tube?

The goal at U.S. airports handling international flights is to clear these flights within 45 minutes. Let's interpret this to mean that 95 percent of the flights are cleared in 45 minutes, so 5 percent of the flights take longer to clear. Let's also assume that the distribution is approximately normal. a. If the standard deviation of the time to clear an international flight is 5 minutes, what is the mean time to clear a flight? b. Suppose the standard deviation is 10 minutes, not the 5 minutes suggested in part (a). What is the new mean? c. A customer has 30 minutes from the time her flight lands to catch her limousine. Assuming a standard deviation of 10 minutes, what is the likelihood that she will be cleared in time?

A normal distribution has a mean of 80 and a standard deviation of 14. Determine the value above which 80 percent of the values will occur.

According to the South Dakota Department of Health the mean number of hours of TV viewing per week is higher among adult women than men. A recent study showed women spent an average of 34 hours per week watching TV and men 29 hours per week. Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 4.5 hours and is 5.1 hours for the men. a. What percent of the women watch TV less than 40 hours per week? b. What percent of the men watch TV more than 25 hours per week? c. How many hours of TV do the one percent of women who watch the most TV per week watch? Find the comparable value for the men.

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was \(\$ 20.50,\) with a standard deviation of \(\$ 3.50 .\) Assume the distribution of hourly wages follows the normal probability distribution. If we select a crew member at random, what is the probability the crew member earns: a. Between \(\$ 20.50\) and \(\$ 24.00\) per hour? b. More than \(\$ 24.00\) per hour? c. Less than \(\$ 19.00\) per hour?

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