Question

1. Let n be an odd positive integer. Consider a list of n consecutive integers.

Show that the average is the middle number (that is the number in the

middle of the list when they are arranged in an increasing order). What

is the average when n is an even positive integer instead?

2.

Let x1,x2,...,xn be a list of numbers, and let ¯ x be the average of the list.Which of the following

statements must be true? There might be more than one such statement, or one, or none;

a) At least half of the numbers on the list must be bigger than ¯ x.

b) Half of the numbers on the list must be bigger than ¯ x.

c) Some of the numbers on the list must be bigger than ¯ x.

d) Not all of the numbers on the list can be bigger than ¯ x.

Answer 1

1) Let n be an odd positive integer

Consider a list of n consecutive integers.

let n=3, then 3 consecutive numbers (increasing order) will be a-1, a, a+1

similarly if n=5, then 5 consecutive numbers (increasing order) will be a-2, a-1, a, a+1, a+2

and so on...

so we can see that when n=3,

Average= (a-1+a+a+1)/3= a i,e the middle number

Similarly for n= 5, Average= (a-2+a-1+a+a+1+a+2)/5= a i,e the middle number

So, when n is odd then the average of the n consecutive integers is the middle number of the list i.e, (n+1)/2^th number of the list.

When n is an even positive integer instead,

Then the average of the n consecutive integers arranged in increasing order will be the mean of the (n/2)th and ((n/2)+1)th number of the list.

B) The correct statements are

c) Some of the numbers on the list must be bigger than ¯ x.

d) Not all of the numbers on the list can be bigger than ¯ x.