Find the variance of the distribution: | $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | |---|---|---|---|---|---|---| | $P(x)$ | $\dfrac{1}{6}$ | $\dfrac{5}{18}$ | $\dfrac{2}{9}$ | $\dfrac{1}{6}$ | $\dfrac{1}{9}$ | $\dfrac{1}{18}$ |

> [!TLDR] > Using $\operatorname{Var}(X)=E(X^{2})-[E(X)]^{2}$ with $E(X)=\tfrac{35}{18}$ and $E(X^{2})=\tfrac{35}{6}$ gives $\operatorname{Var}(X)=\tfrac{665}{324}\approx 2.05$. ## Put the probabilities on a common denominator …

Find the variance of the distribution: | x | 0 | 1

NCERT Exemplar · Q37

Find the variance of the distribution:

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$\dfrac{1}{6}$	$\dfrac{5}{18}$	$\dfrac{2}{9}$	$\dfrac{1}{6}$	$\dfrac{1}{9}$	$\dfrac{1}{18}$

CBSEShort· 5mImportance★★★★★est

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Variance $=E(X^{2})-[E(X)]^{2}$ . Write every probability over $18$ : the weights for $x=0,\dots,5$ are $3,5,4,3,2,1$ (out of $18$ ). …

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