If x_1, x_2, x_3, ldots, x_n are n observatio | vedclass

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(A) Given that $\sum(x_i+2)^2 = 28n$ $\Rightarrow \sum x_i^2 + 4\sum x_i + 4n = 28n$ $\Rightarrow \sum x_i^2 + 4\sum x_i = 24n$ $... (i)$ Similarly,$\

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$12$

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(A) Given that $\sum(x_i+2)^2 = 28n$ $\Rightarrow \sum x_i^2 + 4\sum x_i + 4n = 28n$ $\Rightarrow \sum x_i^2 + 4\sum x_i = 24n$ $... (i)$ Similarly,$\sum(x_i-2)^2 = 12n$ $\Rightarrow \sum x_i^2 - 4\sum x_i + 4n = 12n$ $\Rightarrow \sum x_i^2 - 4\sum x_i = 8n$ $... (ii)$ Adding $(i)$ and $(ii)$: $2\sum x_i^2 = 32n \Rightarrow \sum x_i^2 = 16n$ Subtracting $(ii)$ from $(i)$: $8\sum x_i = 16n \Rightarrow \sum x_i = 2n$ The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$ $\sigma^2 = \frac{16n}{n} - \left(\frac{2n}{n}\right)^2 = 16 - 4 = 12$

Measure	Height	Weight
Mean	$162.6 \ cm$	$52.36 \ kg$
Variance	$127.69 \ cm^2$	$23.1361 \ kg^2$

If $x_1, x_2, x_3, \ldots, x_n$ are $n$ observations such that $\sum(x_i+2)^2 = 28n$ and $\sum(x_i-2)^2 = 12n$,then the variance is:

Similar Questions

The standard deviation of $a, a+d, a+2 d, \ldots, a+2 n d$ is

The following values are calculated in respect of heights and weights of the students of a section of Class $XI$:

Measure Height Weight

Mean $162.6 \ cm$ $52.36 \ kg$

Variance $127.69 \ cm^2$ $23.1361 \ kg^2$

Can we say that the weights show greater variation than the heights?

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