Question

Fuel consumption is commonly measured in miles per gallon​ (mi/gal). An agency designed new fuel consumption tests to be used starting with 2008 car models. Listed below are randomly selected amounts by which the measured MPG ratings decreased because of the new 2008 standards. Find the​ range, variance, and standard deviation for the sample data. Is the decrease of 4​ mi/gal unusual? Why or why​ not?

22    

11    

33    

22    

44    

11    

33    

22    

22    

22    

22    

22    

11    

22    

22    

22    

11    

22    

22    

22

  

The range of the sample data is

nothing

​mi/gal. ​(Type an integer or a​ decimal.)The variance of the sample data is

nothing.

​(Round to one decimal place as​ needed.)The standard deviation of the sample data is

nothing

​mi/gal.

​(Round to one decimal place as​ needed.)

Is the largest​ decrease, 4​ mi/gal,

unusual​?

Why or why​ not?

A.

The decrease of 4​ mi/gal is unusual because the smallest value in a data set is usually an outlier.  

B.

The decrease of 4​ mi/gal is not unusual because the sample is a simple random​ sample, in which no values are considered unusual.

C.The decrease of 4​ mi/gal is

unusualunusual

because it is

more thanmore than

two standard deviations

fromfrom

the mean.

D.The decrease of 4​ mi/gal is

not unusualnot unusual

because it is

withinwithin

two standard deviations

ofof

the mean.

Answer 1

range of data = maximum - minimum = 44 - 11 = 33
------------------------------------------------------------------------------
Mean = Sum of observations/ Count of observations
Mean = (22 + 11 + 33 + 22 + 44 + 11 + 33 + 22 + 22 + 22 + 22 + 22 + 11 + 22 + 22 + 22 + 11 + 22 + 22 + / 20) = 22
------------------------------------------------------------------------------
Variance
Step 1: Add them up
22 + 11 + 33 + 22 + 44 + 11 + 33 + 22 + 22 + 22 + 22 + 22 + 11 + 22 + 22 + 22 + 11 + 22 + 22 + = 440
Step 2: Square your answer
440*440 =193600
…and divide by the number of items. We have 20 items , 193600/20 = 9680
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
22^2 + 11^2 + 33^2 + 22^2 + 44^2 + 11^2 + 33^2 + 22^2 + 22^2 + 22^2 + 22^2 + 22^2 + 11^2 + 22^2 + 22^2 + 22^2 + 11^2 + 22^2 + 22^2 + = 10890
Step 4: Subtract the amount in Step 2 from the amount in Step 3
10890 - 9680 = 1210
Step 5: Subtract 1 from the number of items in your data set, 20 - 1 = 19
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
1210 / 19 = 63.6842
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
7.9802
------------------------------------------------------------------------------
The decrease of 4​ mi/gal is not unusual because the sample is a simple random​ sample, in which no values are considered unusual.

[ANSWERS]

(a) the range of the sample data is 33 mi/gal
(b) the variance of the sample data is 63.6842 ​mi/gal
(c) the standard deviation of the sample data is 7.9802 ​mi/gal.
(d) b.the decrease of 4​ mi/gal is not unusual because the sample is a simple random​ sample,
in which no values are considered unusual.