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Vikentia [17]
3 years ago
13

At one point the average price of regular unleaded gasoline was ​$3.41 per gallon. Assume that the standard deviation price per

gallon is ​$0.09 per gallon and use​ Chebyshev's inequality to answer the following. ​
(a) What percentage of gasoline stations had prices within 2 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 ​$3.05 and ​$3.77​?
Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

a)  1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75

So we expected about 75% within two deviations from the mean

b) 1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

Lower = 3.41 -1.5*0.09 = 3.275

Upper = 3.41 +1.5*0.09 = 3.545

c) We can calculate how many deviations we are within the mean with the limits with this formula:

z =\frac{x-\mu}{\sigma}

And using the lower limit we got:

z = \frac{3.05-3.41}{0.09}=-4

And with the upper limit we got:

z = \frac{3.77-3.41}{0.09}=4

So then the value of k =4 and the percentage is given by:

1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375

Step-by-step explanation:

Previous concepts and Data given  

\mu =3.41 reprsent the population mean

\sigma=0.09 represent the population 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

For this case we can find the percentage required replaincg k =2 and we got:

1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75

So we expected about 75% within two deviations from the mean

Part b

For this case we can find the percentage required replaincg k =2 and we got:

1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

Lower = 3.41 -1.5*0.09 = 3.275

Upper = 3.41 +1.5*0.09 = 3.545

Part c

We can calculate how many deviations we are within the mean with the limits with this formula:

z =\frac{x-\mu}{\sigma}

And using the lower limit we got:

z = \frac{3.05-3.41}{0.09}=-4

And with the upper limit we got:

z = \frac{3.77-3.41}{0.09}=4

So then the value of k =4 and the percentage is given by:

1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375

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