If the standard deviation of the numbers 2, 3 | vedclass

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Question

(D) The formula for standard deviation $(SD)$ is $SD = \sqrt{\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2}$.Given $SD = 3.5 = \frac{7}{2}$

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$3a^2 - 32a + 84 = 0$

Vedclass · Answer

(D) The formula for standard deviation $(SD)$ is $SD = \sqrt{\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}ight)^2}$.Given $SD = 3.5 = \frac{7}{2}$,so $SD^2 = \frac{49}{4}$.The numbers are $2, 3, a, 11$,so $n = 4$.$\sum x_i = 2 + 3 + a + 11 = 16 + a$.$\sum x_i^2 = 2^2 + 3^2 + a^2 + 11^2 = 4 + 9 + a^2 + 121 = 134 + a^2$.Substituting into the variance formula:$\frac{49}{4} = \frac{134 + a^2}{4} - \left(\frac{16 + a}{4}ight)^2$.Multiply by $16$ to clear the denominators:$49 	imes 4 = 4(134 + a^2) - (16 + a)^2$.$196 = 536 + 4a^2 - (256 + 32a + a^2)$.$196 = 536 + 4a^2 - 256 - 32a - a^2$.$196 = 280 + 3a^2 - 32a$.$3a^2 - 32a + 84 = 0$.

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$k^2-1$	$k^2-1$	$k^2-1$	$k-3$

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

If the standard deviation of the numbers $2, 3, a$,and $11$ is $3.5$,then which of the following is true?

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