Question

The variance of the numbers $2, 3, 11$ and $x$ is $\frac{49}{4}$. Find the value of $x$.

• A. $6, \frac{14}{3}$
• B. $6, \frac{14}{5}$
• C. $6, \frac{16}{3}$
• D. $4, \frac{13}{5}$

Answer 1

Answer: (A)

$6, \frac{14}{3}$

Answer 2

From given data, we make the following table. But we know that, variance =$\frac{\Sigma x^{2}}{n} -\left(\frac{\Sigma x}{n}\right)^{2}$ $\Rightarrow \frac{134+x^{2}}{4} -\left(\frac{16+x}{4}\right)^{2} = \frac{49}{4}$ (given) $\Rightarrow \frac{134+x^{2}}{4} - \frac{\left(256+x^{2}+32x\right)}{16} = \frac{49}{4}$ $\Rightarrow \frac{3x^{2}-32x+280}{16} = \frac{49}{4}$ $\Rightarrow 3x^{2}-32x+280 = 196$ $\Rightarrow 3x^{2}-32x+84 = 0$ $\Rightarrow \left(x-6\right)\left(3x-14\right)=0$ $\Rightarrow x=6, x= \frac{14}{3}$ Therefore, the values of $x$ are $6$ and $\frac{14}{3}$.