Problem 50 - An Introduction to Mathematical Statistics and Its Applications

Question

Econo-Tire is planning an advertising campaign for its newest product, an inexpensive radial. Preliminary road tests conducted by the firm's quality- control department have suggested that the lifetimes of these tires will be normally distributed with an average of thirty thousand miles and a standard deviation of five thousand miles. The marketing division would like to run a commercial that makes the claim that at least nine out of ten drivers will get at least twenty-five thousand miles on a set of EconoTires. Based on the road test data, is the company justified in making that assertion?

Step-by-Step Solution

Answer

The claim made by Econo-Tire is not justified based on the road test data, as only approximately 84.13% of tires are expected to last at least 25,000 miles, which is less than the asserted 90%.

Step 1: Identify the given values

From the data given, we have the average (mean) lifetime of these tires, \( \mu = 30,000 \) miles, and the standard deviation, \( \sigma = 5,000 \) miles. We must also consider the claim - that 90% of the tires will last at least 25,000 miles.

Step 2: Calculate the Z-Score

We have to calculate the Z-Score for the value of 25,000 miles. The Z-Score is calculated using the formula \( Z = \frac{X - \mu}{\sigma} \), where X is the value we are converting to a standard score. Therefore, \( Z = \frac{25,000 - 30,000}{5,000} = -1 \). -1 is the Z-Score for 25,000 miles.

Step 3: Use the standard normal table

We need to find the proportion of data below this Z-Score, using a standard normal (Z) table or calculator. The value associated with Z = -1 is 0.1587, which means that roughly 15.87% of tires will not meet the 25,000 mileage mark.

Step 4: Calculate the Percentage

To find the percentage of drivers who will get at least 25,000 miles from their tires, subtract this value from 1 (or 100%). Therefore, \( 1 - 0.1587 = 0.8413 \) or 84.13%, which is less than the 90% in the claim.

Key Concepts

Z-Score Calculation

In statistics, the Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. It indicates how many standard deviations an element is from the mean.
To calculate the Z-score, use the formula: \[Z = \frac{X - \mu}{\sigma}\]In this formula, \(X\) represents the value for which the Z-Score is being calculated, \(\mu\) is the mean of the population, and \(\sigma\) is the standard deviation.
For example, if the lifetime of a tire is 25,000 miles, with a mean lifetime of 30,000 miles and a standard deviation of 5,000 miles, the Z-score is \(-1\), as calculated by substituting these values into the formula.
This negative Z-score indicates that 25,000 miles is one standard deviation below the mean, helping us understand the position of this score in the context of the overall dataset.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation suggests that values are close to the mean, whereas a high standard deviation indicates greater variability.
To determine the standard deviation, you can calculate it with the following steps:

• Find the mean (average) of all data values.
• Subtract the mean from each data value and square the result.
• Calculate the mean of these squared differences.
• Take the square root of this mean to get the standard deviation.

In our tire example, the standard deviation of 5,000 miles means that on average, the tire lifetime deviates from the mean lifetime of 30,000 miles by 5,000 miles.
This gives us insight into the spread of tire lifetimes and helps in making judgments about individual tire lifetimes compared to the average.

Statistical Claims Validation

Validating statistical claims involves using statistical methods to determine the validity of a claim based on data analysis. With normal distribution and Z-scores, you can assess whether claims hold up under scrutiny.
For instance, Econo-Tire's claim that "at least nine out of ten drivers will achieve 25,000 miles" can be examined by calculating probabilities related to the normal distribution.
We discovered that the Z-score for 25,000 miles is \(-1\), which corresponds to approximately 15.87% of tires falling below this mileage.

• Subtracting this from 100% reveals that only 84.13% of tires meet the mileage mark.
• Since this percentage is less than 90%, the claim does not hold.

Understanding this concept allows businesses and individuals to make informed decisions based on statistical evidence, rather than assumptions, thus ensuring accuracy in their promotional messages.