Question

Consider the first 10 positive integers. If we multiply each number by −1 and then add 1 to each number, the variance of the numbers so obtained is

Answer 1

Given numbers are first 10 positive integers i.e., 1, 2, 3, ... , 10.

Now, Variance of first n natural numbers, where n is even is

Variance \(\sigma^2=\dfrac{n^2-1}{12}\)

For n=10,

Variance\(=\dfrac{10^2-1}{12}\)

Variance \(=\dfrac{99}{12}\)

Variance = 8.25

If each observation is multiplied by a where \(a\epsilon \mathbb{R}\), then variance will also be multiplied by \(a^2\)

Here, each number is multiplied by −1

\(\therefore\) New Variance \((-1)^2\times(8.25)\)

New Variance = 8.25


If each observation is increased by 'a', where \(a\epsilon \mathbb{R}\), then the variance remains unchanged.

Hence, if 1 is added to each number 1 to 10, variance is same as of 1 to 10.

Hence, option (A) is correct.