Given that x̅ is the mean and σ2 is the variance of n observations x1, x2, …,xn . Prove that the mean and variance of the observations ax1, ax2, ax3, …., axn are ax̅ and a2σ2, respectively, (a ≠ 0).

Asked by Abhisek | 1 year ago |  164

##### Solution :-

From the question it is given that, n observations are x1, x2,…..xn

Mean of the n observation = x̅

Variance of the n observation = σ2

As we know that,

Variance = $$σ^2= \dfrac{1}{n} \displaystyle\sum_{i=1}^{n} y_i (x_i-\overline {x} )^2$$..............(i)

If each observation is multiplied by a and the new observations are yi, then

yi = axi

Thus, $$x_i =\dfrac{1}{a}y_i$$

$$\overline{y}=\dfrac{1}{n} \displaystyle\sum_{i=1}^{a} y_i$$

$$\overline{y}=\dfrac{1}{n} \displaystyle\sum_{i=1}^{a} ax_i$$

$$\overline{y}=\dfrac{a}{n} \displaystyle\sum_{i=1}^{a} x_i$$

$$\overline{y}=a\overline x$$

Therefore, mean of the observations, ax1, ax2 … axn, is a  $$\overline{x}$$ Substituting the values of xi and $$\overline{x}$$ in (1), we obtain

$$σ^2= \dfrac{1}{n} \displaystyle\sum_{i=1}^{n}(\dfrac{1}{a}y_1-\dfrac{1}{a}\overline{y})^2$$

$$a^2σ^2= \dfrac{1}{n} \displaystyle\sum_{i=1}^{n}(y_1-\overline{y})^2$$

Thus, the variance of the observations, ax1, ax2 …axn, is a2 σ2.

Answered by Pragya Singh | 1 year ago

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