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(A) The sum of probabilities in a distribution is $1$:$\alpha + \alpha + \alpha + \beta + \beta + 0.3 = 1 \implies 3\alpha + 2\beta = 0.7$ (Equation $

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$20.4$

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(A) The sum of probabilities in a distribution is $1$:$\alpha + \alpha + \alpha + \beta + \beta + 0.3 = 1 \implies 3\alpha + 2\beta = 0.7$ (Equation $1$).The mean $\mu$ is given by $\sum x_i P(x_i) = 4.2$:$1(\alpha) + 2(\alpha) + 3(\alpha) + 4(\beta) + 5(\beta) + 6(0.3) = 4.2$$6\alpha + 9\beta + 1.8 = 4.2 \implies 6\alpha + 9\beta = 2.4 \implies 2\alpha + 3\beta = 0.8$ (Equation $2$).Solving Equations $1$ and $2$:Multiply Eq $1$ by $2$: $6\alpha + 4\beta = 1.4$.Multiply Eq $2$ by $3$: $6\alpha + 9\beta = 2.4$.Subtracting: $5\beta = 1.0 \implies \beta = 0.2$.Substituting $\beta = 0.2$ into Eq $1$: $3\alpha + 2(0.2) = 0.7 \implies 3\alpha = 0.3 \implies \alpha = 0.1$.We need to find $\sigma^2 + \mu^2$. Since $\sigma^2 = E(X^2) - \mu^2$,then $\sigma^2 + \mu^2 = E(X^2)$.$E(X^2) = \sum x_i^2 P(x_i) = 1^2(0.1) + 2^2(0.1) + 3^2(0.1) + 4^2(0.2) + 5^2(0.2) + 6^2(0.3)$$E(X^2) = 0.1 + 0.4 + 0.9 + 3.2 + 5.0 + 10.8 = 20.4$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x_i)$	$\alpha$	$\alpha$	$\alpha$	$\beta$	$\beta$	$0.3$

$X = x$	$-1$	$0$	$1$	$2$
$F(X = x)$	$0.3$	$0.7$	$0.8$	$1$

The probability distribution of a random variable $X$ is given below:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x_i)$ $\alpha$ $\alpha$ $\alpha$ $\beta$ $\beta$ $0.3$

If $\mu$ and $\sigma^2$ represent the mean and variance of $X$ and $\mu=4.2$,then $\sigma^2+\mu^2=$

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The probability distribution of a random variable $X$ is given below:$X$$1$$2$$3$$4$$5$$6$$P(X=x_i)$$\alpha$$\alpha$$\alpha$$\beta$$\beta$$0.3$If $\mu$ and $\sigma^2$ represent the mean and variance of $X$ and $\mu=4.2$,then $\sigma^2+\mu^2=$