If the mean of the observations x₁, x₂, x₃, ……, x_n is x̄, then the mean of x₁ − a, x₂ − a, x₃ − a, ……, x_n − a is :

1. $\bar x$

2. $\bar x$ + a

3. $\bar x$ − a

4. $\Big(\dfrac{n \bar x - a}{n}\Big)$

The original mean =

$\bar{x} = \dfrac{x$1 + x2 + x3 + \dots + xn}{n}

The new mean is the sum of the new observations divided by :

$\text{New Mean} = \dfrac{(x$1 - a) + (x2 - a) + (x3 - a) + \dots + (xn - a)}{n} \\[1em] = \dfrac{(x1 + x2 + \dots + x_n) - (a + a + \dots + a)}{n}

Since there are n terms of a, the sum of the a is na:

$\text{New Mean} = \dfrac{(x$1 + x2 + \dots + xn) - na}{n}\\[1em] = \dfrac{x1 + x2 + \dots + xn}{n} - \dfrac{na}{n} \\[1em] = \bar{x} - a.

Hence, option 3 is the correct option.