Pt= 4x-6 and TQ =3x+4
1 answer:
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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 = 6.5
and Median of the above data would be = =
= 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 = 8
and Median of the above data would be = =
= 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 = 7.167
and Median of the above data would be = = 7
Hence mean and median number of defects are not same if the sixth car has 7 defects.
Therefore option D is correct.