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(C) The formula for the combined variance $\sigma^{2}$ of two sets is given by:$\sigma^{2} = \frac{n_{1}\sigma_{1}^{2} + n_{2}\sigma_{2}^{2}}{n_{1} +

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$5$

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(C) The formula for the combined variance $\sigma^{2}$ of two sets is given by:$\sigma^{2} = \frac{n_{1}\sigma_{1}^{2} + n_{2}\sigma_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1}n_{2}}{(n_{1} + n_{2})^{2}}(\bar{x}_{1} - \bar{x}_{2})^{2}$Given values:$n_{1} = 10, n_{2} = n, \sigma_{1}^{2} = 2, \sigma_{2}^{2} = 1$$\bar{x}_{1} = 2, \bar{x}_{2} = 3, \sigma^{2} = \frac{17}{9}$Substituting these values into the formula:$\frac{17}{9} = \frac{10(2) + n(1)}{10 + n} + \frac{10n}{(10 + n)^{2}}(2 - 3)^{2}$$\frac{17}{9} = \frac{20 + n}{10 + n} + \frac{10n}{(10 + n)^{2}}$Multiply by $9(10 + n)^{2}$ to clear denominators:$17(10 + n)^{2} = 9[(20 + n)(10 + n) + 10n]$$17(100 + 20n + n^{2}) = 9[200 + 30n + n^{2} + 10n]$$1700 + 340n + 17n^{2} = 1800 + 360n + 9n^{2}$$8n^{2} - 20n - 100 = 0$$2n^{2} - 5n - 25 = 0$$(2n + 5)(n - 5) = 0$Since $n$ must be positive,$n = 5$.

Set	Size	Mean	Variance
Observation $I$	$10$	$2$	$2$
Observation $II$	$n$	$3$	$1$

$x_i$	$1$	$2$	$3$	$4$	$5$
$f_i$	$5$	$4$	$f$	$2$	$3$

Consider the statistics of two sets of observations as follows:

Set Size Mean Variance

Observation $I$ $10$ $2$ $2$

Observation $II$ $n$ $3$ $1$

If the variance of the combined set of these two observations is $\frac{17}{9}$,then the value of $n$ is equal to:

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Consider the statistics of two sets of observations as follows: Set Size Mean Variance Observation $I$ $10$ $2$ $2$ Observation $II$ $n$ $3$ $1$ If the variance of the combined set of these two observations is $\frac{17}{9}$,then the value of $n$ is equal to: