Mathematics, 02.03.2020 21:21 dimontsho4717

At one point the average price of regular unleaded gasoline was $3.52 per gallon. Assume that the standard deviation price per gallon is $0.06 per gallon and use Chebyshev�s inequality to answer the following.

a. At least % of gasoline stations had prices within 4 standard deviations of the mean. (round to the nearest hundredth as needed)
b. At least % of gasoline stations had prices within 1.5 standard deviations of the mean.
c. The gasoline prices that are within 1.5 standard deviations of the mean are $ to $ (Use ascending order)
d. % is the minimum percentage of gasoline stations that had prices between $3.40 and $3.64. (round to the nearest hundredth as needed)

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Answer from: rue61

a) $1-\frac{1}{4^2} = 1- \frac{1}{16} = \frac{15}{16} =0.9375 = 93.75\%$

At least 93.75% of gasoline stations had prices within 4 standard deviations of the mean. (round to the nearest hundredth as needed)

b) $1-\frac{1}{1.5^2} =0.5556= 55.56\%$

At least 55.56% of gasoline stations had prices within 1.5 standard deviations of the mean. (round to the nearest hundredth as needed)

c) Lower = 3.52 -1.5*0.06 = 3.43

Upper = 3.52 +1.5*0.06 = 3.61

d) $k = \frac{3.52-3.40}{0.06} = 2$

And since we have the value of k we have:

$1- \frac{1}{2^2}= 1-\frac{1}{4} = \frac{3}{4} = 75\%$

75% is the minimum percentage of gasoline stations that had prices between $3.40 and $3.64. (round to the nearest hundredth as needed)

Step-by-step explanation:

Data given

$\bar X =3.52$ reprsent the sample mean

s=0.06 represent the sample standard deviation

The Chebyshev's Theorem states that for any dataset

•We have at least 75% of all the data within two deviations from the mean.

•We have at least 88.9% of all the data within three deviations from the mean.

•We have at least 93.8% of all the data within four deviations from the mean.

Or in general words "For any set of data (either population or sample) and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least: $1-\frac{1}{k^2}"$

Part a

$1-\frac{1}{4^2} = 1- \frac{1}{16} = \frac{15}{16} =0.9375 = 93.75\%$

At least 93.75% of gasoline stations had prices within 4 standard deviations of the mean. (round to the nearest hundredth as needed)

Part b

$1-\frac{1}{1.5^2} =0.5556= 55.56\%$

At least 55.56% of gasoline stations had prices within 1.5 standard deviations of the mean. (round to the nearest hundredth as needed)

Part c

For this case we have:

Lower = 3.52 -1.5*0.06 = 3.43

Upper = 3.52 +1.5*0.06 = 3.61

Part d

For this case we can find the number of k like this:

$k = \frac{3.52-3.40}{0.06} = 2$

And since we have the value of k we have:

$1- \frac{1}{2^2}= 1-\frac{1}{4} = \frac{3}{4} = 75\%$

75% is the minimum percentage of gasoline stations that had prices between $3.40 and $3.64. (round to the nearest hundredth as needed)

Answer from: Quest

incomplete question. how do you expect anyone to you when you do not post the question in its entirety?

Answer from: Quest

$w\geq 14$

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

$2l+2w\geq 40$

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

$2\times 6+2w\geq 40$

now, we can solve for w

$12+2w\geq 40$

subtract both sides by 12

$12+2w-12\geq 40-12$

$2w\geq 28$

divide both sides by 2

and we get

$w\geq 14$

so,

the smallest possible measurement of the width is 14 feet

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At one point the average price of regular unleaded gasoline was $3.52 per gallon. Assume that the st...