Let X be a random variable with the following | vedclass

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(A) The sum of probabilities is $1$,so $\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1 \implies a + b = 1 - \frac{2}{5} - \frac{1}{3} = 1 - \frac

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$781$

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(A) The sum of probabilities is $1$,so $\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1 \implies a + b = 1 - \frac{2}{5} - \frac{1}{3} = 1 - \frac{11}{15} = \frac{4}{15} \dots (1)$The mean $E(X) = \sum x_i P(x_i) = 2.3 = \frac{23}{10}$.$-2(\frac{1}{5}) - 1(a) + 3(\frac{1}{3}) + 4(\frac{1}{5}) + 6(b) = \frac{23}{10}$$-\frac{2}{5} - a + 1 + \frac{4}{5} + 6b = \frac{23}{10} \implies -a + 6b + \frac{7}{5} = \frac{23}{10} \implies -a + 6b = \frac{23}{10} - \frac{14}{10} = \frac{9}{10} \dots (2)$Adding $(1)$ and $(2)$: $7b = \frac{4}{15} + \frac{9}{10} = \frac{8+27}{30} = \frac{35}{30} = \frac{7}{6} \implies b = \frac{1}{6}$.Then $a = \frac{4}{15} - \frac{1}{6} = \frac{8-5}{30} = \frac{3}{30} = \frac{1}{10}$.Variance $\sigma^{2} = E(X^{2}) - (E(X))^{2}$.$E(X^{2}) = (-2)^{2}(\frac{1}{5}) + (-1)^{2}(\frac{1}{10}) + (3)^{2}(\frac{1}{3}) + (4)^{2}(\frac{1}{5}) + (6)^{2}(\frac{1}{6})$$= \frac{4}{5} + \frac{1}{10} + 3 + \frac{16}{5} + 6 = 4 + \frac{1}{10} + 9 = 13 + 0.1 = 13.1 = \frac{131}{10}$.$\sigma^{2} = \frac{131}{10} - (2.3)^{2} = 13.1 - 5.29 = 7.81$.Therefore,$100 \sigma^{2} = 781$.

$x$	$-2$	$-1$	$3$	$4$	$6$
$P(X=x)$	$\frac{1}{5}$	$a$	$\frac{1}{3}$	$\frac{1}{5}$	$b$

Let $X$ be a random variable with the following probability distribution:

$x$ $-2$ $-1$ $3$ $4$ $6$

$P(X=x)$ $\frac{1}{5}$ $a$ $\frac{1}{3}$ $\frac{1}{5}$ $b$

If the mean of $X$ is $2.3$ and the variance of $X$ is $\sigma^{2}$,then $100 \sigma^{2}$ is equal to:

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Let $X$ be a random variable with the following probability distribution: $x$ $-2$ $-1$ $3$ $4$ $6$ $P(X=x)$ $\frac{1}{5}$ $a$ $\frac{1}{3}$ $\frac{1}{5}$ $b$ If the mean of $X$ is $2.3$ and the variance of $X$ is $\sigma^{2}$,then $100 \sigma^{2}$ is equal to: