MATH SOLVE

5 months ago

Q:
# At one point the average price of regular unleaded gasoline was $3.43 per gallon. Assume that the standard deviation price per gallon is $0.07 per gallon and use Chebyshev's inequality to answer the following. (a) What percentage of gasoline stations had prices within 4 standard deviations of the mean? (b) What percentage of gasoline stations had prices within 1.5 standard deviations of the mean? What are the gasoline prices that are within 1.5 standard deviations of the mean? (c) What is the minimum percentage of gasoline stations that had prices between $3.22 and $3.64?

Accepted Solution

A:

Answer:(a) The percentage of gasoline stations had prices within 4 standard deviations of the mean is 93.75%.(b) The percentage of gasoline stations had prices within 1.5 standard deviations of the mean is 55.56%. The prices for this stations goes from $3.325 to $3.535.(c) The minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.Step-by-step explanation:The Chebyshev's inequality states that at least 1-(1/K^2) of data from a sample must fall within K standard deviations from the mean, being K any positive real number greater than one.It can be expressed as[tex]P(|X-\mu| \geq k\sigma)\leq \frac{1}{k^2}[/tex]In this problem, we have, for the gasoline price, a normal distribution with mean of 3.43 and standar deviation of 0.07.(a) The percentage of gasoline stations had prices within 4 standard deviations of the mean is equal to one less the percentage of gasoline stations that had prices out of 4 standard deviations of the mean:[tex]P(|X-\mu| \leq k\sigma)=1-P(|X-\mu| \leq k\sigma) \\\\1-P(|X-\mu| \leq k\sigma) \geq 1-\frac{1}{k^2} \\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-\frac{1}{4^2}\\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-1/16\\\\1-P(|X-\mu| \leq 4\sigma) \geq 0.9375[/tex]The percentage of gasoline stations had prices within 4 standard deviations of the mean is 93.75%.(b) The percentage of gasoline stations had prices within 1.5 standard deviations of the mean is 55.56%.[tex]P(|X-\mu| \leq k\sigma)\geq 1-\frac{1}{k^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 1-\frac{1}{1.5^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 0.5556\\[/tex]The prices for this stations goes from $3.325 to $3.535.[tex]X=\mu\pm 1.5\sigma=3.43\pm 1.5*0.07=3.43 \pm 0.105\\\\X_{upper} =3.43+0.105=3.535\\X_{lower}=3.43-0.105=3.325[/tex](c) To answer, we have to calculate k for this range of prices:[tex]x=\mu\pm k\sigma\\\\k=\frac{|x-\mu|}{\sigma} =\frac{|3.64-3.43|}{0.07}=\frac{0.21}{0.07}= 3[/tex]For k=3, the Chebyshev's inequality states:[tex]P(|X-\mu| \leq 3\sigma)\geq 1-\frac{1}{3^2}\\\\P(|X-\mu| \leq 3\sigma)\geq 0.8889[/tex]So, the minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.