The possible value of x if SD of the numbers 2, 3, 2x and 11 is 3.5 is equal to
The possible value of x if SD of the numbers 2, 3, 2x and 11 is 3.5 is equal to
- $4, \frac{5}{3}$
- $5, \frac{7}{3}$
- $3, \frac{7}{3}$
- $2, \frac{5}{3}$
Solution
Given Standard deviation of 2, 3, 2x and 11 is 3.5.
$\begin{aligned} & \because \text { Mean }(\bar{x})=\frac{2+3+2 x+11}{4}=\frac{8+x}{2} \\ & \text { Now, } \sum_{i=1}^4\left(x_i-\bar{x}\right)^2\left\{2-\left(\frac{8+x}{2}\right)\right\}^2+\left\{3-\left(\frac{8+x}{2}\right)\right\}^2 \\ &+\left\{2 x-\left(\frac{8+x}{2}\right)\right\}^2+\left\{11-\left(\frac{8+x}{2}\right)\right\}^2 \\ &=\left(\frac{4+x}{2}\right)^2+\left(\frac{2+x}{2}\right)^2+\left(\frac{3 x-8}{2}\right)^2+\left(\frac{14-x}{2}\right)^2\end{aligned}$
$\begin{aligned} & \because \mathrm{SD}=\sqrt{\left[\begin{array}{l}{\left[\frac{4+x}{4}\right)^2+\left(\frac{2+x}{2}\right)^2+\left(\frac{3 x-8}{2}\right)^2} \\ +\left(\frac{14-x}{2}\right)^2\end{array}\right]} \\ & (3 \cdot 5)^2 \times 4=\frac{1}{4}\end{aligned}$
$\begin{aligned} & {\left[16+x^2+8 x+4+x^2+4 x+9 x^2+64-48 x\right.} \\ & \left.\quad+196+x^2-28 x\right] \\ & \\ & \Rightarrow 12 x^2-64 x+84=0 \\ & \Rightarrow \quad x=3,7 / 3\end{aligned}$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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