Let the mean and variance of four numbers 3,7 | vedclass

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Question

$5=\frac{3+7+x+y}{4} \Rightarrow x+y=10$

$\operatorname{Var}(x)=10=\frac{3^{2}+7^{2}+x^{2}+y^{2}}{4}-25$

$140=49+9+x^{2}+y^{2}$

$x^{2}+y^{2}=

Vedclass · Accepted Answer

$12$

Vedclass · Answer

$5=\frac{3+7+x+y}{4} \Rightarrow x+y=10$

$\operatorname{Var}(x)=10=\frac{3^{2}+7^{2}+x^{2}+y^{2}}{4}-25$

$140=49+9+x^{2}+y^{2}$

$x^{2}+y^{2}=82$

$x+y=10$

$\Rightarrow(x, y)=(9,1)$

Four numbers are $21,9,10,8$

$	ext { Mean }=\frac{48}{4}=12$

	Height	Weight
Mean	$162.6\,cm$	$52.36\,kg$
Variance	$127.69\,c{m^2}$	$23.1361\,k{g^2}$

${x_i}$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
${f_i}$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

$x_i$	$0$	$1$	$5$	$6$	$10$	$12$	$17$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

Let the mean and variance of four numbers $3,7, x$ and $y(x>y)$ be $5$ and $10$ respectively. Then the mean of four numbers $3+2 \mathrm{x}, 7+2 \mathrm{y}, \mathrm{x}+\mathrm{y}$ and $x-y$ is ..... .

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

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

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

The data is obtained in tabular form as follows.

${x_i}$ $60$ $61$ $62$ $63$ $64$ $65$ $66$ $67$ $68$

${f_i}$ $2$ $1$ $12$ $29$ $25$ $12$ $10$ $4$ $5$

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$x_i$ $0$ $1$ $5$ $6$ $10$ $12$ $17$

$f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$