If the mean and S.D. of the data $3,5,7, \mathrm{a}, \mathrm{b}$ are 5 and 2 respectively, then $\mathrm{a}$…
If the mean and S.D. of the data $3,5,7, \mathrm{a}, \mathrm{b}$ are 5 and 2 respectively, then $\mathrm{a}$ and $\mathrm{b}$ are the roots of the equation
$x^2-10 x+18=0$
$2 x^2-20 x+19=0$
$x^2-10 x+19=0$
$x^2-20 x+18=0$
Solution
$\text { Mean }=5$ ... [Given]
$\begin{aligned} & \therefore \quad \text { Mean }=\frac{\sum_{i=1}^{\mathrm{n}} x_i}{\mathrm{n}} \\ & \Rightarrow 5=\frac{3+5+7+\mathrm{a}+\mathrm{b}}{5} \\ & \Rightarrow \mathrm{a}+\mathrm{b}=10 ... (i)\end{aligned}$
$\begin{aligned}
& \text { S.D. }=2 ... [Given]\\
& \therefore \quad \text { S.D. }=\sqrt{\frac{\sum x_1^2}{\mathrm{n}}-(\bar{x})^2} \\
& \Rightarrow(2)^2=\frac{3^2+5^2+7^2+\mathrm{a}^2+\mathrm{b}^2}{5}-(5)^2 \\
& \Rightarrow 4=\frac{83+\mathrm{a}^2+\mathrm{b}^2}{5}-25 \\
& \Rightarrow \mathrm{a}^2+\mathrm{b}^2=62 ... (ii)
\end{aligned}$
...[Given]
Now, (i) $\Rightarrow a+b=10$
Squaring both sides, we get
$\begin{array}{ll}
& (a+b)^2=100 \\
& a^2+2 a b+b^2=100 \\
& 38=2 a b ... [From (ii)]\\
\therefore \quad & a b=19
\end{array}$
Note that the required quadratic equation is expressed as
$\begin{aligned}
& x^2-(\mathrm{a}+\mathrm{b}) x+\mathrm{ab}=0 \\
\therefore \quad & x^2-10 x+19=0
\end{aligned}$