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If variance of $x_1, x_2 \ldots \ldots, x_{\mathrm{n}}$ is $\sigma_x^2$, then the variance of $\lambda x_1, \lambda x_2, \ldots \ldots, \lambda x_n(\lambda \neq 0)$ is
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The correct answer is:
$\lambda^2 \cdot \sigma_x^2$
When each item of a data is multiplied by $\lambda$, variance is multiplied by $\lambda^2$.
$\therefore \quad$ New variance $=\lambda^2 \cdot \sigma_x^2$
$\therefore \quad$ New variance $=\lambda^2 \cdot \sigma_x^2$
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