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Suppose ten distinct, positive integers have a median of \(10\). ("Distinct integers" means that no two integers are the same.)

What is the smallest the average of those ten integers could be?

 

 Apr 7, 2019
 #1
avatar+6045 
+2

As the numbers are distinct they can be ordered.

The median will be the average of the 5th and 6th elements so they must be 9 and 11

Then the 4 numbers less than 9 should be 1 2 3 4

and the 4 numbers greater than 11 should be 12 13 14 15

The average of all the numbers is

 

(10+20+54)/10 = 8.4

 Apr 7, 2019
 #2
avatar+142 
+2

But how did you get 9 and 11?? Never mind I just realized thanks

 Apr 7, 2019
edited by KeyLimePi  Apr 7, 2019
 #3
avatar+105180 
0

Note, KeyLime

The median of  a set of an even number n of ordered data values  is  the  average of the (n/2)th data value  and the (n + 2)/2 th data value

 

So...here....we want the  average of  the  (10/2)th data value  = 5th data value  and the (10+2)/2 = 6th data value

 

Since we want the average to be as small as possible....the 5th and 6th data values need to be the integers on either side of 10 that are closest to 10.....i.e.,  9 and 11   [ just as Rom said ]

 

Note that any other two values will produce an average  > 8.4

 

To see this....let's suppose that the 5th and 6th data values are  8  and 12

The median is  [ 8 + 12] / 2  = 10

 

And the elements < 8   need to be as small as possible...i.e, 1,2, 3 , 4

An d the elements  > 12 need to be as small as possible......i.e., 13, 14, 15, 16

 

Note that the average becomes   [ 1 + 2 + 3 + 4 + 8 + 12 + 13 + 14 + 15 + 16 ]  / 10   =  8.8

 

Which is > 8.4

 

Try this with the 5th and 6th data values  being 7 and 13

So....the elements need to be  1,2,3,4, 7, 13, 14, 15, 16, 17

Convince yourself that this average will also be > 8.4

 

 

cool cool cool

 Apr 7, 2019

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