Question

If the mean and the standard deviation of the data $3,5,7,a,b$ are $5$ and $2$ respectively, then $a$ and $b$ are the roots of the equation

• A. $x^2-20x+18=0$
• B. $x^2-10x+19=0$
• C. $2x^2-20x+19=0$
• D. $x^2-10x+18=0$

Answer 1

The correct option is B

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

Explanation of the correct option:

Find the required equation

Given: mean$=5$, S.D.$=2$

We know that, mean$=\frac{{\displaystyle \sum _{i=1}^n}{x}_i}{n}$

$$\begin{gathered} \Rightarrow 5=\frac{3+5+7+a+b}{5} \\ \Rightarrow 25=15+a+b \\ \Rightarrow 10=a+b..................\left(1\right) \end{gathered}$$

We know that, $S.D.=\sqrt{\frac{{\displaystyle \sum _{}}{x_i}^2}{n}-{\left(\overline{x}\right)}^2}$

$$\begin{gathered} \Rightarrow {\left(2\right)}^2=\frac{3^2+5^2+7^2+a^2+b^2}{5}-{\left(5\right)}^2 \\ \Rightarrow 4=\frac{83+a^2+b^2}{5}-25 \\ \Rightarrow 20=83+a^2+b^2-125 \\ \Rightarrow 62=a^2+b^2.............\left(2\right) \end{gathered}$$

From equation $\left(1\right)$,

$10=\left(a+b\right)$

Squaring both the sides,

$$\begin{gathered} \Rightarrow {\left(10\right)}^2={\left(a+b\right)}^2 \\ \Rightarrow 100=a^2+b^2+2ab \\ \Rightarrow 100=62+2ab\mathbf{}\left[\because a^2+b^2=62\right] \\ \Rightarrow 38=2ab \\ \Rightarrow ab=19..................\left(3\right) \end{gathered}$$

We know that general quadratic equation is expressed as $x^2+(a+b)x+ab=0$, where $(a+b)$ is the sum of roots and $ab$ is the product.

So, from equation $\left(1\right)$ and $\left(3\right)$,

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

Hence, option $\left(B\right)$ is the correct answer.