If the mean and variance of the frequency distribution x _ i 2 4 6...

Given,

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,

Now mean is given by,

Mean $= \dfrac{8 + 16 + 120 + 80 + 56 + 80 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$

$⇒ 360 + 9\alpha + 9\beta = 360 + 6\alpha + 12\beta$

$⇒ 3\alpha - 3\beta = 0$

$⇒ \alpha = \beta . . . . . . \left(1\right)$

Now variance is given by,

$15 . 08 = \dfrac{16 + 64 + 36\alpha + 960 + 800 + 144\alpha + 784 + 1280}{40 + 2\alpha} - {9}^{2}$

$⇒ 96 . 08 = \dfrac{3904 + 180\alpha}{40 + 2\alpha}$

$⇒ \left(40 + 2\alpha\right)\left(96 . 08\right) = 3904 + 180\alpha$

$⇒ 3843 . 20 + ( 192 . 16 ) \alpha = 3904 + 180\alpha$

$⇒ ( 12 . 16 ) \alpha = 60 . 80$

$⇒ \alpha = 5 = \beta$

Hence, ${\alpha}^{2} + {\beta}^{2} - \alpha\beta = 25 + 25 - 25 = 25$