If the mean and standard deviation (S.D.) of | vedclass

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(C) Given,$	ext{Mean} = 5$ and $	ext{S.D.} = 2$. For the data $3, 5, 7, a, b$ with $n=5$: $	ext{Mean} = \frac{3+5+7+a+b}{5} = 5$ $\Rightarrow 15+a+

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$x^2-10x+19=0$

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(C) Given,$	ext{Mean} = 5$ and $	ext{S.D.} = 2$. For the data $3, 5, 7, a, b$ with $n=5$: $	ext{Mean} = \frac{3+5+7+a+b}{5} = 5$ $\Rightarrow 15+a+b = 25$ $\Rightarrow a+b = 10$ ... $(i)$ Using the formula for variance,$	ext{Var} = 	ext{S.D.}^2 = 2^2 = 4$: $	ext{Var} = \frac{\sum x_i^2}{n} - (	ext{Mean})^2$ $4 = \frac{3^2+5^2+7^2+a^2+b^2}{5} - 5^2$ $4 = \frac{9+25+49+a^2+b^2}{5} - 25$ $29 = \frac{83+a^2+b^2}{5}$ $145 = 83 + a^2 + b^2$ $a^2 + b^2 = 62$ ... $(ii)$ We know that $(a+b)^2 = a^2 + b^2 + 2ab$. $10^2 = 62 + 2ab$ $100 - 62 = 2ab$ $38 = 2ab \Rightarrow ab = 19$. The quadratic equation with roots $a$ and $b$ is given by $x^2 - (a+b)x + ab = 0$. Substituting the values,we get $x^2 - 10x + 19 = 0$.

If the mean and standard deviation ($S$.$D$.) of the data $3, 5, 7, a, b$ are $5$ and $2$ respectively,then $a$ and $b$ are the roots of the equation:

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