/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 In December \(2014,\) the averag... [FREE SOLUTION] | 91Ó°ÊÓ

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In December \(2014,\) the average price of regular unleaded gasoline excluding taxes in the United States was \(\$ 3.06\) per gallon, according to the Energy Information Administration. Assume that the standard deviation price per gallon is \(\$ 0.06\) per gallon to answer the following. (a) What minimum percentage of gasoline stations had prices within 3 standard deviations of the mean? (b) What minimum percentage of gasoline stations had prices within 2.5 standard deviations of the mean? What are the gasoline prices that are within 2.5 standard deviations of the mean? (c) What is the minimum percentage of gasoline stations that had prices between \(\$ 2.94\) and \(\$ 3.18 ?\)

Short Answer

Expert verified
a) 88.89%b) 84%, Prices: \(2.91 to \)3.21c) 75%

Step by step solution

01

Title - Understanding the Mean and Standard Deviation

The mean price of gasoline is given as \( \$ 3.06 \) per gallon, and the standard deviation is \( \$ 0.06 \) per gallon.
02

- Using Chebyshev's Theorem for Minimum Percentage (3 Standard Deviations)

Chebyshev's Theorem states that at least \( \left(1 - \frac{1}{k^2}\right) \times 100\% \) of the data will fall within \( k \) standard deviations of the mean. For \( k = 3 \), this gives: \[ \left(1 - \frac{1}{3^2}\right) \times 100\% = \left(1 - \frac{1}{9}\right) \times 100\% = \left(1 - 0.1111\right) \times 100\% = 88.89\% \]
03

- Using Chebyshev's Theorem for Minimum Percentage (2.5 Standard Deviations)

For \( k = 2.5 \), this gives: \[ \left(1 - \frac{1}{2.5^2}\right) \times 100\% = \left(1 - \frac{1}{6.25}\right) \times 100\% = \left(1 - 0.16\right) \times 100\% = 84\% \]
04

- Calculating Price Range within 2.5 Standard Deviations

To find the price range within 2.5 standard deviations of the mean: \( 3.06 \pm 2.5 \times 0.06 \). Calculations: Upper price: \( 3.06 + 2.5 \times 0.06 = 3.06 + 0.15 = 3.21 \) Lower price: \( 3.06 - 2.5 \times 0.06 = 3.06 - 0.15 = 2.91 \). So, the gasoline prices within 2.5 standard deviations of the mean are between \$2.91 and \$3.21.
05

- Minimum Percentage Between \(2.94 and \)3.18

Calculate the number of standard deviations from the mean for \$2.94 and \$3.18: For \( \$2.94: \frac{3.06 - 2.94}{0.06} = 2 \) For \( \$3.18: \frac{3.18 - 3.06}{0.06} = 2 \) Thus, \$2.94 and \$3.18 are both within 2 standard deviations of the mean.Using Chebyshev's Theorem: \( 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = 0.75 = 75\% \).At least 75% of gasoline stations had prices between \$2.94 and \$3.18.

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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 crucial concept in statistics that measures the extent of variation or dispersion in a set of values. In this exercise, the standard deviation of gasoline prices is given as \( \$0.06 \) per gallon. This tells us how much the price of gasoline can deviate from the mean price, on average. A smaller standard deviation means that the prices are closely packed around the mean, while a larger standard deviation indicates prices are spread out over a wider range.
To better grasp this, imagine we have a mean price of \( \$3.06 \) per gallon. If the prices at different stations deviate significantly from this value, the standard deviation tells us how typical or atypical these deviations are. The formula for calculating the standard deviation \(\sigma\) is:
\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \] where \( N \) is the number of observations, \( x_i \) represents each data point, and \( \mu \) is the mean.
Understanding standard deviation helps us determine the spread of gasoline prices around the mean price.
Mean Price
The mean price, or average price, is another fundamental concept in this exercise. The mean is calculated by summing up all individual prices and dividing it by the number of prices. In this problem, the mean price of gasoline was given as \( \$3.06 \) per gallon. This represents the central point around which the prices at different gasoline stations are distributed.
The formula for calculating the mean \( \mu \) is straightforward:
\[ \mu = \frac{1}{N} \sum_{i=1}^{N} x_i \] where \( N \) is the total number of observations, and \( x_i \) is each individual observation.
Knowing the mean price is useful as it serves as the reference point for calculating the standard deviation and for applying Chebyshev's Theorem. For instance, when we say a gas station's price is within 2.5 standard deviations of the mean, we are measuring how far that price is from the average price of \( \$3.06 \).
Thus, mean price gives us a central figure, making it easier to understand and compare distributions of different prices.
Minimum Percentage
To determine the minimum percentage of gasoline stations with prices within a certain range, we use Chebyshev's Theorem. This theorem provides a way to estimate the spread of data points in any distribution, regardless of its shape. According to Chebyshev's Theorem, at least \( \left(1 - \frac{1}{k^2}\right) \times 100\% \) of data values lie within \( k \) standard deviations from the mean.
For example, in step 2 of our solution, we calculate the minimum percentage of gasoline stations with prices within 3 standard deviations of the mean: \[ \left(1 - \frac{1}{3^2}\right) \times 100\% = 88.89\% \] This means at least 88.89% of gasoline stations had prices within 3 standard deviations of \( \$3.06 \).
Similarly, for gasoline prices between \( \$2.94 \) and \( \$3.18 \), which are within 2 standard deviations of the mean, Chebyshev's Theorem states that at least 75% of data values fall within this range: \[ \left(1 - \frac{1}{2^2}\right) \times 100\% = 75\% \] Thus, knowing how to apply Chebyshev's Theorem not only helps us understand the spread and concentration of data but also provides a useful tool for making informed decisions and predictions.

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