Mathematics, 05.10.2019 04:20 bullockarwen

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?

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Answer from: coollid876

(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.

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

$P(|X-\mu| \geq k\sigma)\leq \frac{1}{k^2}$

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:

$P(|X-\mu| \leq k\sigma)=1-P(|X-\mu| \leq k\sigma) \\\\1-P(|X-\mu| \leq k\sigma) \geq 1-\frac{1}{k^2} \\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-\frac{1}{4^2}\\\\1-P(|X-\mu| \leq 4\sigma) \geq 1-1/16\\\\1-P(|X-\mu| \leq 4\sigma) \geq 0.9375$

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%.

$P(|X-\mu| \leq k\sigma)\geq 1-\frac{1}{k^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 1-\frac{1}{1.5^2}\\\\P(|X-\mu| \leq 1.5\sigma)\geq 0.5556\\$

The prices for this stations goes from $3.325 to $3.535.

$X=\mu\pm 1.5\sigma=3.43\pm 1.5*0.07=3.43 \pm 0.105\\\\X_{upper} =3.43+0.105=3.535\\X_{lower}=3.43-0.105=3.325$

$x=\mu\pm k\sigma\\\\k=\frac{|x-\mu|}{\sigma} =\frac{|3.64-3.43|}{0.07}=\frac{0.21}{0.07}= 3$

For k=3, the Chebyshev's inequality states:

$P(|X-\mu| \leq 3\sigma)\geq 1-\frac{1}{3^2}\\\\P(|X-\mu| \leq 3\sigma)\geq 0.8889$

So, the minimum percentage of stations that had prices between $3.22 and $3.64 is 88.88%.

Answer from: Quest

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true

Bwill make the inequalities y < 2x + 1 and y> -x- 2 true. o a. region b o b. region c o c. r

Answer from: Quest

its kinda hard to answer this when we dont have all the information that we need.

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At one point the average price of regular unleaded gasoline was $3.43 per gallon. assume that the st...