Which of the following data sets has the mini | vedclass

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(A) The variance measures the spread of data points around the mean. The formula for sample variance is $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$.For

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$1, 2, 3, 4, 5$

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(A) The variance measures the spread of data points around the mean. The formula for sample variance is $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$.For option $A$ $(1, 2, 3, 4, 5)$: Mean $\bar{x} = 3$. Variance $= \frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5-1} = \frac{4+1+0+1+4}{4} = \frac{10}{4} = 2.5$.For option $B$ $(1, 1, 2, 3, 6)$: Mean $\bar{x} = 2.6$. Variance $= \frac{(1-2.6)^2 + (1-2.6)^2 + (2-2.6)^2 + (3-2.6)^2 + (6-2.6)^2}{4} = \frac{2.56 + 2.56 + 0.36 + 0.16 + 11.56}{4} = \frac{17.2}{4} = 4.3$.For option $C$ $(1, 1, 2, 3, 5)$: Mean $\bar{x} = 2.4$. Variance $= \frac{(1-2.4)^2 + (1-2.4)^2 + (2-2.4)^2 + (3-2.4)^2 + (5-2.4)^2}{4} = \frac{1.96 + 1.96 + 0.16 + 0.36 + 6.76}{4} = \frac{11.2}{4} = 2.8$.For option $D$ $(1, 1, 2, 2, 5)$: Mean $\bar{x} = 2.2$. Variance $= \frac{(1-2.2)^2 + (1-2.2)^2 + (2-2.2)^2 + (2-2.2)^2 + (5-2.2)^2}{4} = \frac{1.44 + 1.44 + 0.04 + 0.04 + 7.84}{4} = \frac{10.8}{4} = 2.7$.Comparing the variances $(2.5, 4.3, 2.8, 2.7)$,the minimum variance is $2.5$ for option $A$.

$x_i$	$5$	$7$	$9$	$11$
$f_i$	$3$	$2$	$1$	$2$

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

Which of the following data sets has the minimum variance?

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