Question

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​?

Answer 1

a)
`1-(1)/(k^2) =1- (1)/(2^2)= 1-0.25 = 0.75`

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

b)
`1-(1)/(k^2) =1- (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 =(x-\mu)/(\sigma)`

And using the lower limit we got:

`z = (3.05-3.41)/(0.09)=-4`

And with the upper limit we got:

`z = (3.77-3.41)/(0.09)=4`

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

`1-(1)/(k^2) =1- (1)/(4^2)= 1-0.0625 = 0.9375`

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-(1)/(k^2)`

Part a

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

`1-(1)/(k^2) =1- (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-(1)/(k^2) =1- (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 =(x-\mu)/(\sigma)`

And using the lower limit we got:

`z = (3.05-3.41)/(0.09)=-4`

And with the upper limit we got:

`z = (3.77-3.41)/(0.09)=4`

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

`1-(1)/(k^2) =1- (1)/(4^2)= 1-0.0625 = 0.9375`