what is the median of the following list of 4040 numbers? $1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2$ (a) 1974.5(b) 1975.5(c) 1976.5(d) 1977.5(e) 1978.5



Answer :

Median of the following list of 4040 numbers is 1976.5

If we solve 45² = 2025, this value is greater than 2020.

If we solve 44² that would result in 44x44 =1936, this value is less than 2020.

If we subtract 44 from 2020: 2020 - 44 = 1976

There are 1976 of the 4040 numbers that are greater than 2020.

If we add 44 into 2022: 2020 + 44 = 2064

So, there are 2064 numbers which are less than or equal to 2020.

As there are 44 duplicates, it will shift up the median's placement down 44.

List of numbers had been 1, 2, 3, ....,4040

Median of the entire given set would be:

= 1+4040/2

= 2020.5

The median of the following list of 4040 numbers is:

2020.5 - 44= 1976.5

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