Question

If x is the mean of 10 natural numbers x1, x2, x3, …………x10, then sum of deviation from its mean is

Answer 1

The mean of the above given natural numbers

=  X̄ = x₁ + x₂ + x₃ + .....x₁₀ / 10 -- (1)

(x₁ -  X̄) + (x₂ -  X̄) + ... + (x₁₀ -  X̄)

= x₁ + x₂ + x₁₀ - 10 X̄

= x₁ + x₂ + x₁₀ - 10

(x₁ + x₂ + x₁₀/10)

(x₁ + x₂ + x₁₀) - (x₁ + x₂ + x₁₀)

= 0 = RHS

Answer 2

Answer:

Sum of deviations from the mean is 0.

Step-by-step explanation:

$\text{Given , 10 natural numbers are : }x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}\\\\\text{Now, mean of the above natural numbers = x} =(\frac{x_1+x_2......+x_{10}}{10}) ..............(1)\\\\\text{Therefore, Sum of deviations from mean is given by : }\\(x_1-x)+(x_2-x)+........+(x_{10}-x)\\=x_1+x_2......+x_{10}-10\cdot x\\=x_1+x_2......+x_{10}-10\cdot (\frac{x_1+x_2......+x_{10}}{10})\text{ , By using (1)}\\\\=(x_1+x_2......+x_{10})-(x_1+x_2......+x_{10})\\=0$

Hence, the sum of deviation from its mean = 0