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The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?I. 3II. 7III. 12A. I onlyB. II onlyC. III onlyD. I and III onlyE. I, II, and III

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Answer:

D. I and III only

Step-by-step explanation:

Number of first fie defects are given as 9,7,10,4 and 6.

First we have to arrange the above data in ascending order which is 4,6,7,9,10

  • Now if we consider the defect in sixth car to be 3 then our data will look like:    3,4,6,7,9,10

So the mean of above data would be [tex]\frac{3+4+6+7+9+10}{6}[/tex] = 6.5

and Median of the above data would be = [tex]\frac{3^{rd} obs + 4^{th} obs}{2}[/tex] = [tex]\frac{6+7}{2}[/tex] = 6.5

Hence mean and median number of defects are same if the sixth car has 3 defects.

  • Now if we consider the defect in sixth car to be 12 then our data will look like:    4,6,7,9,10,12

So the mean of above data would be [tex]\frac{4+6+7+9+10+12}{6}[/tex] = 8

and Median of the above data would be = [tex]\frac{3^{rd} obs + 4^{th} obs}{2}[/tex] = [tex]\frac{7+9}{2}[/tex] = 8

Hence mean and median number of defects are same if the sixth car has 12 defects.

  • But Now if we consider the defect in sixth car to be 7 then our data will look like:    4,6,7,7,9,10        

So the mean of above data would be [tex]\frac{4+6+7+7+9+10}{6}[/tex] = 7.167    

and Median of the above data would be =  [tex]\frac{3^{rd} obs + 4^{th} obs}{2}[/tex] = 7

   Hence mean and median number of defects are not same if the sixth car has 7 defects.      

Therefore option D is correct.                                                                                  

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