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

From given data, we make the following tablexx2243911121xx2Σx=16+xΣx2=134+x2But we know that, variance =Σx2n−Σxn2⇒ 134+x24−16+x42=494⇒ 134+x24−256+x2+32x16=494⇒ 3x2−32x+28016=494⇒ 280+3x2−32x=494×16⇒ 280+3x2−32x=196⇒ 3x2−32x+84=0⇒ (x−6)(3x−14)=0⇒ x=6,x=143Therefore, the values of x are 6 and 143.

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

From given data, we make the following tablexx2243911121xx2Σx=16+xΣx2=134+x2But we know that, variance =Σx2n−Σxn2⇒ 134+x24−16+x42=494⇒ 134+x24−256+x2+32x16=494⇒ 3x2−32x+28016=494⇒ 280+3x2−32x=494×16⇒ 280+3x2−32x=196⇒ 3x2−32x+84=0⇒ (x−6)(3x−14)=0⇒ x=6,x=143Therefore, the values of x are 6 and 143.

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

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

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

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

None of these

answer is A.

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Detailed Solution

From given data, we make the following table

x	x2
2	4
3	9
11	121
x	x²
$Σx = 16 + x$	${Σx}^{2} = 134 + x^{2}$

But we know that, variance $= \frac{{Σx}^{2}}{n} - {(\frac{Σx}{n})}^{2}$

$\begin{array}{lr} \Rightarrow \frac{134 + x^{2}}{4} - {(\frac{16 + x}{4})}^{2} = \frac{49}{4} \\ \Rightarrow \frac{134 + x^{2}}{4} - \frac{(256 + x^{2} + 32 x)}{16} = \frac{49}{4} \\ \Rightarrow \frac{3 x^{2} - 32 x + 280}{16} = \frac{49}{4} \\ \Rightarrow 280 + 3 x^{2} - 32 x = \frac{49}{4} \times 16 \\ \Rightarrow 280 + 3 x^{2} - 32 x = 196 \\ \Rightarrow 3 x^{2} - 32 x + 84 = 0 \\ \Rightarrow (x - 6) (3 x - 14) = 0 \\ \Rightarrow x = 6, x = \frac{14}{3} \end{array}$