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One of the cars sold by Walt's car dealership is a very popular subcompact car called Rhino. The final sale price of the basic model of this car varies from customer to customer depending on the negotiating skills and persistence of the customer. Assume that these sale prices of this car are normally distributed with a mean of \(\$ 19,800\) and a standard deviation of \(\$ 350\). a. Dolores paid \(\$ 19,445\) for her Rhino. What percentage of Walt's customers paid less than Dolores for a Rhino? b. Cuthbert paid \(\$ 20,300\) for a Rhino. What percentage of Walt's customers paid more than Cuthbert for a Rhino?

Short Answer

Expert verified
The percentages of customers who paid less than Dolores and more than Cuthbert, respectively, can be found by calculating their z-scores and looking up the corresponding percentiles in a standard normal distribution table. The necessary calculations are shown in the steps above

Step by step solution

01

Find the z-score for Dolores

Firstly, we need to calculate the z-score for Dolores. This can be done with the z-score formula: \( z = \frac{x - \mu}{\sigma} \), where \(x\) is the value from the sample, \(\mu\) is the mean and \(\sigma\) is the standard deviation. In this case, \(x = \$ 19,445\), \(\mu = \$ 19,800\) and \(\sigma = \$ 350\). The z-value will tell us how many standard deviations below or above the mean Dolores' payment is.
02

Find the percentile for Dolores

Having calculated the z-score, the next step is to find the corresponding percentile value from a standard normal distribution table. This will give us the proportion of the distribution that lies below this z-value, which represents the percentage of customers who paid less than Dolores.
03

Repeat the process for Cuthbert

We can then follow the same steps for Cuthbert. This time though, we will find the proportion of the distribution that lies above Cuthbert's z-value as the problem asks what percentage of customers paid more than him.
04

Calculation for Cuthbert

We would start by calculating the z-score as like Dolores. Then subtract the normal distribution value derived from the z-score from 1, since we want to find the percentage of people who paid more. This will return the percentage of people who paid more than Cuthbert

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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 numerical measurement that describes a value's relationship to the mean of a group of values. It's a critical concept in statistics, especially when dealing with a normal distribution. The formula for calculating the z-score is \[ z = \frac{x - \mu}{\sigma} \]where:
  • \( x \) is the value from the sample set,
  • \( \mu \) is the mean of the data set,
  • \( \sigma \) is the standard deviation.
A z-score tells us how many standard deviations an element is from the mean. A z-score can be positive or negative, indicating that the value is either above or below the mean, respectively. For example, Dolores's z-score calculation in the exercise shows us whether the price she paid is more or less than what is considered "average" for the Rhino car.
Demystifying Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It gives insight into how much individual measurements typically deviate from the mean. In essence, it provides a summary of how data points spread out over a distribution and is calculated using the formula:\[ \sigma = \sqrt{\frac{1}{n} \sum (x_i - \mu)^2} \]where:
  • \( x_i \) is each individual value,
  • \( \mu \) is the mean of all values,
  • \( n \) is the total number of values.
In the context of the exercise, a standard deviation of \( \\(350 \) for the price of Rhino cars shows how much customers typically pay above or below the mean price of \( \\)19,800 \). A smaller standard deviation would mean prices are closely clustered around the mean, whereas a larger standard deviation indicates prices vary more widely.
Exploring Percentiles
Percentiles are statistical measures that represent the rank of a particular value within a given data set. In the context of a normal distribution, percentiles help us understand what proportion of the data falls below a certain point. When you calculate the z-score, you can use it to find the percentile from a standard normal distribution table, which helps to interpret the overall standing of a specific value. For Dolores, once her z-score is determined, the corresponding percentile indicates what percentage of customers paid less than she did for their car. Thus, if her score falls at the 25th percentile, it means 25% of other customers paid less than she did. For Cuthbert, calculating percentiles enables us to determine what percentage of customers paid more by subtracting his percentile from 100%. This process provides insight into how their payments compare within the population of all customers.
The Significance of Mean in Normal Distribution
The mean is a fundamental statistical measure that acts as a "central" point in a data set. For data that fits a normal distribution, the mean is particularly informative since it represents the most common value, around which all other data points are symmetrically distributed. It is calculated using the following formula:\[ \mu = \frac{1}{n} \sum_{i=1}^{n} x_i \]Where:
  • \( x_i \) are each of the values in the set,
  • \( n \) is the number of values.
In the car dealership example, the mean price of \( \$19,800 \) indicates the typical selling price of the Rhino car model. The mean acts as the benchmark against which individual sales are compared, as seen through the calculation of z-scores. Understanding the concept of mean helps contextualize how individual sales, like those of Dolores and Cuthbert, fit into the larger picture of overall sales trends.

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Most popular questions from this chapter

According to a May 27, 2009 Minneapolis Star-Tribune article (Source: http://www.startribune.com politics \(/ 45797562 . \mathrm{htm} 1\) ), \(78 \%\) of U.S. households have at least one credit card. Find the probability that in a random sample of 500 U.S. households, 375 to 385 households have at least one credit card

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