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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)

Respuesta :

Answer:

a)  [tex] 1-\frac{1}{4^2} = 1- \frac{1}{16} = \frac{15}{16} =0.9375 = 93.75\% [/tex]

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

b)  [tex] 1-\frac{1}{1.5^2} =0.5556= 55.56\% [/tex]

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) [tex] Lower = 3.52 -1.5*0.06 = 3.43[/tex]

[tex] Upper = 3.52 +1.5*0.06 = 3.61[/tex]

d) [tex] k = \frac{3.52-3.40}{0.06} = 2[/tex]

And since we have the value of k we have:

[tex] 1- \frac{1}{2^2}= 1-\frac{1}{4} = \frac{3}{4} = 75\%[/tex]

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  

[tex]\bar X =3.52[/tex] reprsent the sample mean

[tex]s=0.06[/tex] 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: [tex] 1-\frac{1}{k^2}"[/tex]

Part a

[tex] 1-\frac{1}{4^2} = 1- \frac{1}{16} = \frac{15}{16} =0.9375 = 93.75\% [/tex]

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

[tex] 1-\frac{1}{1.5^2} =0.5556= 55.56\% [/tex]

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:

[tex] Lower = 3.52 -1.5*0.06 = 3.43[/tex]

[tex] Upper = 3.52 +1.5*0.06 = 3.61[/tex]

Part d

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

[tex] k = \frac{3.52-3.40}{0.06} = 2[/tex]

And since we have the value of k we have:

[tex] 1- \frac{1}{2^2}= 1-\frac{1}{4} = \frac{3}{4} = 75\%[/tex]

75% 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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