SOLUTION: check the data for outliers 4, 6, 12, 14, 18, 24, 30, 60 A. No outliers exist B.The value 6 can be considered an outlier C.The value 60 can be considered an outlier D.The v

To find any outliers:

1.  Find the first and third quartiles:

a.  The first quartile:
There are 8 data values. 25% of 8 is 2.  Since that is a whole
number, the first quartile is the mean of the 2nd and 3rd data values.
So we average 6 and 12:  
 =  = 9

b. The third quartile:
There are 8 data values. 75% of 8 is 6.  Since that is a whole
number, the third quartile is the mean of the 6th and 7th data values.
So we average 24 and 30:  
 =  = 27

2.  Find the IQR or interquartile range by subtracting them:

    27 - 9 = 18 = IQR

3. Calculate the "fences" 

   a. Multiply the IQR by 1.5

      18×1.5 = 27

   b. Subtract 27 from the first quartile:
      
      9 - 27 = -18 = lower fence.

   c. Add 27 to the third quartile:

      27 + 27 = 54 = upper fence.

4. Outliers are any values that are below the lower fence or above
   the upper fence.

There are no values below the lower fence.  There is one value
above the upper fence, which is 60.

Therefore 60 is the only outlier.  

Answer:  C

Edwin