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(C) The total frequency is $N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta$. The mean is given by $\bar{x} = \frac{\sum f_i x_i}{N} = 6$. $\frac{4(2)

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$\frac{17}{3}$

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(C) The total frequency is $N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta$. The mean is given by $\bar{x} = \frac{\sum f_i x_i}{N} = 6$. $\frac{4(2) + 4(6) + \alpha(8) + \beta(9)}{8 + \alpha + \beta} = 6$ $8 + 24 + 8\alpha + 9\beta = 48 + 6\alpha + 6\beta$ $2\alpha + 3\beta = 16 \quad \dots (i)$ The variance is $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = 6.8$. $\frac{4(4) + 4(36) + \alpha(64) + \beta(81)}{8 + \alpha + \beta} - 36 = 6.8$ $\frac{16 + 144 + 64\alpha + 81\beta}{8 + \alpha + \beta} = 42.8$ $160 + 64\alpha + 81\beta = 42.8(8 + \alpha + \beta) = 342.4 + 42.8\alpha + 42.8\beta$ $21.2\alpha + 38.2\beta = 182.4$ Multiplying by $10$: $212\alpha + 382\beta = 1824 \Rightarrow 106\alpha + 191\beta = 912 \quad \dots (ii)$ Solving $(i)$ and $(ii)$: From $(i)$,$\alpha = \frac{16 - 3\beta}{2} = 8 - 1.5\beta$. Substituting in $(ii)$: $106(8 - 1.5\beta) + 191\beta = 912$ $848 - 159\beta + 191\beta = 912 \Rightarrow 32\beta = 64 \Rightarrow \beta = 2$. Then $\alpha = 8 - 1.5(2) = 5$. Total frequency $N = 8 + 5 + 2 = 15$. New sum of observations when $x_3$ changes from $8$ to $7$: Original sum $= 6 	imes 15 = 90$. New sum $= 90 - (8 	imes 5) + (7 	imes 5) = 90 - 40 + 35 = 85$. New mean $= \frac{85}{15} = \frac{17}{3}$.

$x$	$2$	$6$	$8$	$9$
$f$	$4$	$4$	$\alpha$	$\beta$

Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively.

$x$ $2$ $6$ $8$ $9$

$f$ $4$ $4$ $\alpha$ $\beta$

If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be:

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Let the mean and variance of the frequency distribution be $6$ and $6.8$ respectively. $x$ $2$ $6$ $8$ $9$ $f$ $4$ $4$ $\alpha$ $\beta$ If $x_{3}$ is changed from $8$ to $7$,then the mean for the new data will be: