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Question

$N=\sum f_i=40+\alpha+\beta$

$\sum f_i x_i=360+6 \alpha+12 \beta$

$\sum f _{ i } x _{ i }^2=3904+36 \alpha+144 \beta$

$\operatorname{Mean}(\o

Vedclass · Accepted Answer

$25$

Vedclass · Answer

$N=\sum f_i=40+\alpha+\beta$

$\sum f_i x_i=360+6 \alpha+12 \beta$

$\sum f _{ i } x _{ i }^2=3904+36 \alpha+144 \beta$

$\operatorname{Mean}(\overline{ x })=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}=9$

$\Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta)$

$3 \alpha=3 \beta \Rightarrow \alpha=\beta$

$\sigma^2=\frac{\sum f _{ i } x _1^2}{\sum f _{ i }}-\left(\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}\right)^2$

$\Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(\overline{ x })^2=15.08$

$\Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}-(9)^2=15.08$

$\Rightarrow \alpha=5$

Now, $\alpha^2+\beta^2-\alpha \beta=\alpha^2=25$

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

$\mathrm{x}$	$\mathrm{x}_{1}=2$	$\mathrm{x}_{2}=6$	$\mathrm{x}_{3}=8$	$\mathrm{x}_{4}=9$
$\mathrm{f}$	$4$	$4$	$\alpha$	$\beta$

If the mean and variance of the frequency distribution

$x_i$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ $16$

$f_i$ $4$ $4$ $\alpha$ $15$ $8$ $\beta$ $4$ $5$

are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.

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