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(C) The sum of all probabilities in a probability distribution must be equal to $1$. Therefore,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the give

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$	ext{cosech } 2$

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(C) The sum of all probabilities in a probability distribution must be equal to $1$. Therefore,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the given expression: $\sum_{x=0}^{\infty} k \frac{2^{2x+1}}{(2x+1)!} = 1$. $k \sum_{x=0}^{\infty} \frac{2^{2x+1}}{(2x+1)!} = 1$. Let $n = 2x+1$. As $x$ goes from $0$ to $\infty$,$n$ takes odd values $1, 3, 5, \ldots$. So,$k \sum_{n=1, 3, 5, \ldots}^{\infty} \frac{2^n}{n!} = 1$. The Taylor series expansion for $\sinh(z)$ is $\sum_{n=1, 3, 5, \ldots}^{\infty} \frac{z^n}{n!} = \sinh(z)$. Here,$z = 2$,so $\sum_{n=1, 3, 5, \ldots}^{\infty} \frac{2^n}{n!} = \sinh(2)$. Thus,$k \sinh(2) = 1$. $k = \frac{1}{\sinh(2)} = 	ext{cosech } 2$.

Expenses	$0$	$500$	$1000$	$1500$	$2000$	$2500$	$3000$
Probability	$0.35$	$0.25$	$0.15$	$0.10$	$0.08$	$0.05$	$0.02$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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

If a discrete random variable $X$ has the probability distribution $P(X=x) = k \frac{2^{2x+1}}{(2x+1)!}$ for $x = 0, 1, 2, \ldots, \infty$,then $k =$

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