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In establishing warranties on HDTV sets, the manufacturer wants to set the limits so that few will need repair at manufacturer expense. On the other hand, the warranty period must be long enough to make the purchase attractive to the buyer. For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer's expense?

Short Answer

Expert verified
Set the warranty limit to approximately 41.12 months.

Step by step solution

01

Understand the Problem Context

We need to determine the warranty limit for HDTVs such that only 10% of them need repairs before the warranty period expires. This implies setting a warranty period corresponding to the 90th percentile of the normal distribution described.
02

Identify Key Values and Distribution

We have the mean (\( \mu = 36.84 \) months) and the standard deviation (\( \sigma = 3.34 \) months). We are dealing with a normally distributed variable, representing the time until repairs are needed.
03

Use the Standard Normal Distribution

Convert our problem into a standard normal distribution problem to find the corresponding z-score for the 90th percentile. In a standard normal distribution, the z-score for the 90th percentile (0.90 percentile) is approximately 1.28.
04

Calculate the Warranty Period Using Z-Score Formula

Use the formula\[ x = \mu + z \times \sigma \]to calculate the required warranty period, where:\[ x \] is the warranty period,\( \mu \) is the mean (36.84 months),\( z \) is the z-score (1.28), and\( \sigma \) is the standard deviation (3.34 months).So:\[ x = 36.84 + 1.28 \times 3.34 \approx 41.12 \text{ months}\]
05

Finalize the Warranty Limit

The warranty period should be set to approximately 41.12 months. This ensures that only 10% of the HDTVs are repaired at the manufacturer's expense, aligning with their initial warranty coverage idea.

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Key Concepts

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

Normal Distribution
Understanding the normal distribution is crucial when dealing with warranty limits. The normal distribution, often known as the bell curve due to its shape, represents how the values of a data set are typically distributed. In statistics, many naturally occurring phenomena tend to have a normal distribution. This type of distribution is symmetric around its mean, indicating that most values cluster around a central region and the probabilities of outcomes further from the mean taper off equally in both directions.
For example, the life span of electronic devices, like HDTVs, can often be modeled with a normal distribution. In our exercise, we have a mean of 36.84 months and a standard deviation of 3.34 months for the HDTV's lifespan until repair is needed. These parameters guide us in setting a probability threshold for the warranty limit.
When a manufacturer sets a warranty limit, they often seek to encompass a high percentage of their products without requiring repairs. Here, the goal is for only 10% of the devices to fail before the warranty period, or in other words, cover 90% under warranty without issue. This forms the essence of our task, utilizing the normal distribution to make informed decisions.
Z-Score Calculation
Calculating the z-score is a fundamental step in transforming our warranty limit challenge into a standard format. A z-score, or standard score, is a method of describing how far a point is from the mean of a data set in terms of standard deviations.
This calculation is essential when you want to understand a value's position within a normally distributed dataset. The formula for a z-score is:
  • The z-score = \( \frac{x - \mu}{\sigma} \)
  • Here, \( x \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In our scenario, we are interested in finding the 90th percentile to set our warranty limit. The 90th percentile corresponds to a standard z-score of approximately 1.28. This score implies that the warranty period should account for 1.28 times the standard deviation added to the mean, thus covering a vast majority (90%) of the devices without premature failures.
By using this z-score, the calculation of the warranty timeframe becomes straightforward. It aids in locating where our warranty should sit on the bell curve, ensuring most HDTVs (90%) won't need repair during the warranty period.
Percentile
Percentiles are a way to determine the value below which a given percentage of observations fall. In manufacturing, percentiles are used to set benchmarks, like warranty periods, ensuring a certain proportion of products meet a specific lifespan
In our exercise, determining the 90th percentile of a normally distributed dataset allows the manufacturer to decide the upper threshold of their warranty limit.
When you set a warranty at the 90th percentile, you are ensuring that 90% of the products last until or beyond this time without needing repairs. Thus, only 10% might require repair services during the warranty period. This implies setting a warranty period to ensure customer satisfaction while minimizing repair costs.
Calculating percentiles involves:
  • Understanding the dataset's distribution (here, normal distribution)
  • Identifying the corresponding z-score (here, 1.28 for the 90th percentile)
  • Applying the z-score in conjunction with the mean and standard deviation to find the target value within the dataset.
In this example, the 90th percentile value of the time until needed repairs helps establish a precise and balanced warranty period. This insights-driven approach lets manufacturers optimize their warranties, aligning customer satisfaction with business practicalities.

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Most popular questions from this chapter

A uniform distribution is defined over the interval from 2 to \(5 .\) a. What are the values for \(a\) and \(b\) ? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00 . e. Find the probability of a value more than 2.6 . f. Find the probability of a value between 2.9 and 3.7 .

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