Given that x is the mean and σ² is the variation of n observations x1, x2, .... , xn Prove that the mean and variance of the observations ax1, ax2, .... , axn recpectively. (a ≠ 0)
Solution:
The given n observations are x1, x2, .... , xn
Mean = x
Variance = σ2
Therefore,
σ² = 1/n ∑ni = 1yi (xi - x)² ....(1)
If each observation is multiplied by a and the new observations are yi, then
yi = axi i.e., xi = 1/n yi
Hence,
y = 1/n ∑ni = 1yi
= 1/n ∑ni = 1axi
= a/n ∑ni = 1xi
= a x [∵ x = 1/n ∑ni = 1xi]
Therefore, mean of the observations, x1, x2, .... , xn is a x
Substituting the values of x and xi in (1) , we obtain
σ² = 1/n ∑ni = 1 (1/a yi - 1/a y)²
a²σ² = 1/n ∑ni = 1 (yi - y)²
Thus, the variance of the observations ax1, ax2, .... , axn is a a²σ²
NCERT Solutions Class 11 Maths Chapter 15 Exercise ME Question 4
Given that x is the mean and σ² is the variation of n observations x1, x2, .... , xn Prove that the mean and variance of the observations ax1, ax2, .... , axn recpectively. (a ≠ 0)
Summary:
Therefore, mean of the observations, x1, x2, .... , xn is a x and thus the variance of the observations ax1, ax2, .... , axn is a a2σ2
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