Let X be a random variable,and let P(X=x) den | vedclass

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(D) Let the equation of the line be $P(X=x) = mx + c$.Since $\sum_{x=0}^4 P(X=x) = 1$,we have:$\sum_{x=0}^4 (mx + c) = 10m + 5c = 1 \implies 2m + c =

Vedclass · Accepted Answer

$42$

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(D) Let the equation of the line be $P(X=x) = mx + c$.Since $\sum_{x=0}^4 P(X=x) = 1$,we have:$\sum_{x=0}^4 (mx + c) = 10m + 5c = 1 \implies 2m + c = \frac{1}{5} \quad (1)$The mean $E[X] = \sum_{x=0}^4 x \cdot P(X=x) = \sum_{x=0}^4 x(mx + c) = m \sum x^2 + c \sum x = 30m + 10c = \frac{5}{2}$.Dividing by $10$,we get $3m + c = \frac{1}{4} \quad (2)$Subtracting $(1)$ from $(2)$ gives $m = \frac{1}{4} - \frac{1}{5} = \frac{1}{20}$.Substituting $m$ into $(1)$: $2(\frac{1}{20}) + c = \frac{1}{5} \implies \frac{1}{10} + c = \frac{2}{10} \implies c = \frac{1}{10}$.Now,$E[X^2] = \sum_{x=0}^4 x^2(mx + c) = m \sum x^3 + c \sum x^2 = m(100) + c(30) = 100(\frac{1}{20}) + 30(\frac{1}{10}) = 5 + 3 = 8$.Variance $\alpha = E[X^2] - (E[X])^2 = 8 - (\frac{5}{2})^2 = 8 - \frac{25}{4} = \frac{32-25}{4} = \frac{7}{4}$.Therefore,$24\alpha = 24 	imes \frac{7}{4} = 6 	imes 7 = 42$.

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

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