A random variable X takes values 0, 1, 2, 3 w | vedclass

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(C) The sum of probabilities must be $1$: $\frac{2a + 1}{30} + \frac{8a - 1}{30} + \frac{4a + 1}{30} + b = 1$ $\frac{14a + 1}{30} + b = 1 \Rightarrow

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$60$

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(C) The sum of probabilities must be $1$: $\frac{2a + 1}{30} + \frac{8a - 1}{30} + \frac{4a + 1}{30} + b = 1$ $\frac{14a + 1}{30} + b = 1 \Rightarrow 14a + 30b = 29 \dots (1)$ We are given $\sigma^{2} + \mu^{2} = 2$. Since $\sigma^{2} = E[X^{2}] - \mu^{2}$, we have $E[X^{2}] = 2$. $E[X^{2}] = \sum x_{i}^{2} p(x_{i}) = 0^{2} \cdot \frac{2a+1}{30} + 1^{2} \cdot \frac{8a-1}{30} + 2^{2} \cdot \frac{4a+1}{30} + 3^{2} \cdot b = 2$ $\frac{8a - 1 + 16a + 4}{30} + 9b = 2$ $\frac{24a + 3}{30} + 9b = 2$ $\Rightarrow 24a + 3 + 270b = 60$ $\Rightarrow 24a + 270b = 57$ Dividing by $3$: $8a + 90b = 19 \dots (2)$ From $(1)$, $30b = 29 - 14a$. Substitute into $(2)$: $8a + 3(29 - 14a) = 19$ $8a + 87 - 42a = 19$ $-34a = -68 \Rightarrow a = 2$ Substituting $a = 2$ into $(1)$: $14(2) + 30b = 29$ $\Rightarrow 28 + 30b = 29$ $\Rightarrow 30b = 1$ $\Rightarrow b = \frac{1}{30}$ Therefore, $\frac{a}{b} = \frac{2}{1/30} = 60$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

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