The variance of the numbers 2, 3,11 and x is494.find the value x .

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

1. A

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

2. B

$6,\frac{14}{5}$

3. C

$6,\frac{16}{3}$

4. D

None of these

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### Solution:

From given data, we make the following table

But we know that, variance $=\frac{{\mathrm{\Sigma x}}^{2}}{\mathrm{n}}-{\left(\frac{\mathrm{\Sigma x}}{\mathrm{n}}\right)}^{2}$

$\begin{array}{l}⇒ \frac{134+{\mathrm{x}}^{2}}{4}-{\left(\frac{16+\mathrm{x}}{4}\right)}^{2}=\frac{49}{4}\\ ⇒ \frac{134+{\mathrm{x}}^{2}}{4}-\frac{\left(256+{\mathrm{x}}^{2}+32\mathrm{x}\right)}{16}=\frac{49}{4}\\ ⇒ \frac{3{\mathrm{x}}^{2}-32\mathrm{x}+280}{16}=\frac{49}{4}\\ ⇒ 280+3{\mathrm{x}}^{2}-32\mathrm{x}=\frac{49}{4}×16\\ ⇒ 280+3{\mathrm{x}}^{2}-32\mathrm{x}=196\\ ⇒ 3{\mathrm{x}}^{2}-32\mathrm{x}+84=0\\ ⇒ \left(\mathrm{x}-6\right)\left(3\mathrm{x}-14\right)=0\\ ⇒ \mathrm{x}=6,\mathrm{x}=\frac{14}{3}\end{array}$

Therefore, the values of x are 6 and $\frac{14}{3}$.

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