yezz wrote:
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. 3
II. 7
III. 12
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
Intuitively each answer choice except for 7 , together with the given forms the union of 2 AP that has the same difference (d =3) and same number of terms.
3,4,6,7,9,10 = {3,6,9} U {4,7,10}
4,6,7,9,10,12 = { 4,7,10} U {6,9,12}
mean and median of such union is equal (symmetric distribution) and is equivalent to the average of both sets median (means)
Another intuitive way to see that mean will be equal to median is to imagine them on a number line. Both sets (with 3 and with 12) are symmetrical about the centre and hence mean = median.
------3-4--6-7--9-10-----
The centre is between 6 and 7 and the elements are symmetrical about it.
-------4--6-7--9-10--12-------
The centre is between 7 and 9 and the elements are symmetrical about it.
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