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Tests of a new tire developed by a tire manufacturer led to an estimated mean tread life of 67,350 miles and standard deviation of 1,120 miles. The manufacturer will advertise the lifetime of the tire (for example, a " 50,000 mile tire") using the largest value for which it is expected that \(98 \%\) of the tires will last at least that long. Assuming tire life is normally distributed, find that advertised value.

Short Answer

Expert verified
Advertise the tire as having a lifespan of 65,054 miles.

Step by step solution

01

Understand the Problem

We are given a mean tread life of 67,350 miles and a standard deviation of 1,120 miles. We're asked to find an advertised tire lifetime such that 98% of the tires last at least that long.
02

Identify the Z-Score for 98%

The tire lifetime follows a normal distribution. To find the advertised tire lifetime, we need the Z-score corresponding to the lower 2% tail, as the tires should last longer for 98% of the cases. Using a standard normal distribution table or calculator, find the Z-score where 98% of the area lies to the right.
03

Calculate the Z-Score Value

For the lower 2% of the Z distribution curve, the Z-score value is approximately -2.05. This means that the value of concern lies about 2.05 standard deviations below the mean.
04

Apply the Z-score Formula

Using the Z-score formula \( X = \mu + Z \cdot \sigma \), calculate the advertised value. Here, \( \mu = 67,350 \), \( Z = -2.05 \), and \( \sigma = 1,120 \). Substitute these values into the formula to find \( X \).
05

Perform the Calculation

Substitute the values into the formula: \[ X = 67,350 + (-2.05) \times 1,120 \] Compute: \[ X = 67,350 - 2,296 = 65,054 \]
06

Conclusion

The largest value for which it is expected that 98% of the tires will last at least that long is approximately 65,054 miles.

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Key Concepts

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

Understanding the Z-Score
The Z-score is a statistical measurement that matters a lot when we deal with normal distributions. It tells you how many standard deviations a particular value is from the mean. Think of it as a way to understand where a specific piece of data falls within a standardized group of data. For instance, for the tire manufacturer who wants 98% of tires to last beyond a certain mileage, the Z-score helps identify precisely which mileage value corresponds to the boundary where only 2% of tires last below it. This means 98% of the tires last longer than this mileage.

To find this out, we look up this 2% in the Z-score table, which tells us we need a Z-score of around -2.05. This Z-score value gives us a clear way to translate this threshold into an actual mileage number.
Standard Deviation Demystified
Standard deviation is a key concept that measures how much variation exists in a dataset. Here, we're talking about the lifespan of tires, which has a standard deviation of 1,120 miles. This number gives us insight into how much the mileage of different tires varies from the average of 67,350 miles.

In simpler terms, the standard deviation tells us how spread out the tire lifetimes are. A small standard deviation would mean most tires have a similar lifespan to the average, while a larger one indicates a greater spread or variability in tire life.
  • Smaller deviations show less variety among the values.
  • Larger deviations indicate more diversity among the values.
Understanding this helps the manufacturer determine the reliability and predictability of how long their tires will last.
Calculating the Mean of a Dataset
The mean is simply the average value of a dataset. It is found by adding all individual values and dividing by the count of those values. Here, the mean we are dealing with is the average tread life of 67,350 miles. This value represents the mid-point of all the tire mileages we expect.

The mean provides the central value around which the normal distribution is centered. Knowing the mean allows us to anchor our Z-score calculations because the Z-score uses the mean as a baseline for determining how unusual or typical a particular data point is. In the tire example, the mean helps determine the average performance of all tires.
Probability Calculation Made Easy
Probability calculations in a normal distribution context are about determining the likelihood of a certain event. When using probability in this tire problem, we calculate to find an advertised tire life such that 98% of tires meet or exceed that lifespan.

Here's how we do it:
  • First, know the desired probability (here, 98%) that you want your tires to last at least a certain distance.
  • Find the Z-score from a standard normal distribution table that matches the leftover probability (here, 2% in the tail).
  • Use this Z-score to calculate the specific mileage using the formula: \( X = \mu + Z \cdot \sigma \).
This process allows you to translate percentage sales pitching ("98% of tires last this long!") into a quantifiable number (like 65,054 miles). It involves a few straightforward steps and uses the inherent properties of the normal distribution to make predictions.

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Most popular questions from this chapter

The useful life of a particular make and type of automotive tire is normally distributed with mean 57,500 miles and standard deviation 950 miles. a. Find the probability that such a tire will have a useful life of between 57,000 and 58,000 miles. b. Hamlet buys four such tires. Assuming that their lifetimes are independent, find the probability that all four will last between 57,000 and 58,000 miles. (If so, the best tire will have no more than 1,000 miles left on it when the first tire fails.)

A continuous random variable \(X\) has a normal distribution with mean \(169 .\) The probability that \(X\) takes a value greater than 180 is 0.17 . Use this information and the symmetry of the density function to find the probability that \(X\) takes a value less than 158 . Sketch the density curve with relevant regions shaded to illustrate the computation.

A continuous random variable \(X\) has a normal distribution with mean 100 and standard deviation \(10 .\) Sketch a qualitatively accurate graph of its density function.

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The systolic blood pressure \(X\) of adults in a region is normally distributed with mean \(112 \mathrm{~mm} \mathrm{Hg}\) and standard deviation \(15 \mathrm{~mm}\) Hg. A person is considered "prehypertensive" if his systolic blood pressure is between 120 and \(130 \mathrm{~mm} \mathrm{Hg}\). Find the probability that the blood pressure of a randomly selected person is prehypertensive.

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