If the variance of the numbers 2, 3, 11 and x | vedclass

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(A) The mean of the given numbers is $\overline{x} = \frac{2+3+11+x}{4} = \frac{16+x}{4}$.Variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \ov

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The mean of the given numbers is $\overline{x} = \frac{2+3+11+x}{4} = \frac{16+x}{4}$.Variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \overline{x})^2$.$\frac{49}{4} = \frac{1}{4} [(\frac{16+x}{4} - 2)^2 + (\frac{16+x}{4} - 3)^2 + (\frac{16+x}{4} - 11)^2 + (\frac{16+x}{4} - x)^2]$.$49 = (\frac{8+x}{4})^2 + (\frac{4+x}{4})^2 + (\frac{x-28}{4})^2 + (\frac{16-3x}{4})^2$.$49 	imes 16 = (64 + x^2 + 16x) + (16 + x^2 + 8x) + (x^2 - 56x + 784) + (256 - 96x + 9x^2)$.$784 = 12x^2 - 128x + 1120$.$12x^2 - 128x + 336 = 0$.Dividing by $4$,we get $3x^2 - 32x + 84 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{32 \pm \sqrt{1024 - 1008}}{6} = \frac{32 \pm 4}{6}$.Thus,$x = \frac{36}{6} = 6$ or $x = \frac{28}{6} = \frac{14}{3}$.

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
$f_i$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

If the variance of the numbers $2, 3, 11$ and $x$ is $\frac{49}{4}$,then the values of $x$ are

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