Question 1167857
Let $N$ be the number of major snowstorms per year that shut down business.
$N$ follows a Poisson distribution with mean $\lambda = 1.8$.
The probability mass function (PMF) is $P(N=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ for $k = 0, 1, 2, \dots$.

Let $X$ be the amount paid to the company under the policy.
The policy pays nothing for the first snowstorm.
The policy pays $\$10600$ for each one thereafter.

We can define $X$ as follows:
* If $N=0$ (no snowstorms), $X=0$.
* If $N=1$ (one snowstorm), $X=0$ (the first one is free).
* If $N=k$ for $k \ge 2$, the number of payable snowstorms is $k-1$. So, $X = 10600(k-1)$.

We want to find the expected amount paid, $E[X]$.
$E[X] = \sum_{k=0}^{\infty} X(k) P(N=k)$
$E[X] = 0 \cdot P(N=0) + 0 \cdot P(N=1) + \sum_{k=2}^{\infty} 10600(k-1) P(N=k)$
$E[X] = 10600 \sum_{k=2}^{\infty} (k-1) \frac{e^{-\lambda} \lambda^k}{k!}$
We can factor out $e^{-\lambda}$:
$E[X] = 10600 e^{-\lambda} \sum_{k=2}^{\infty} (k-1) \frac{\lambda^k}{k!}$

Let's evaluate the sum $S = \sum_{k=2}^{\infty} (k-1) \frac{\lambda^k}{k!}$:
$S = (2-1)\frac{\lambda^2}{2!} + (3-1)\frac{\lambda^3}{3!} + (4-1)\frac{\lambda^4}{4!} + \dots$
$S = \frac{\lambda^2}{2!} + \frac{2\lambda^3}{3!} + \frac{3\lambda^4}{4!} + \dots$

We can use the property that $\frac{k-1}{k!} = \frac{k}{k!} - \frac{1}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$.
So, $S = \sum_{k=2}^{\infty} \left( \frac{1}{(k-1)!} - \frac{1}{k!} \right) \lambda^k$
$S = \sum_{k=2}^{\infty} \frac{\lambda^k}{(k-1)!} - \sum_{k=2}^{\infty} \frac{\lambda^k}{k!}$

For the first part of the sum: $\sum_{k=2}^{\infty} \frac{\lambda^k}{(k-1)!} = \lambda \sum_{k=2}^{\infty} \frac{\lambda^{k-1}}{(k-1)!}$
Let $j=k-1$. When $k=2$, $j=1$.
$= \lambda \sum_{j=1}^{\infty} \frac{\lambda^j}{j!} = \lambda \left( \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \dots \right)$
We know the Taylor series for $e^\lambda$ is $\sum_{j=0}^{\infty} \frac{\lambda^j}{j!} = 1 + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \dots = e^\lambda$.
So, $\sum_{j=1}^{\infty} \frac{\lambda^j}{j!} = e^\lambda - 1$.
Thus, the first part is $\lambda(e^\lambda - 1)$.

For the second part of the sum: $\sum_{k=2}^{\infty} \frac{\lambda^k}{k!}$
This is the Taylor series for $e^\lambda$ minus the first two terms ($k=0$ and $k=1$):
$\sum_{k=2}^{\infty} \frac{\lambda^k}{k!} = \left( \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \right) - \frac{\lambda^0}{0!} - \frac{\lambda^1}{1!} = e^\lambda - 1 - \lambda$.

Now substitute these back into $S$:
$S = \lambda(e^\lambda - 1) - (e^\lambda - 1 - \lambda)$
$S = \lambda e^\lambda - \lambda - e^\lambda + 1 + \lambda$
$S = \lambda e^\lambda - e^\lambda + 1$

Finally, multiply by $e^{-\lambda}$ to get $E[X]$:
$E[X] = 10600 e^{-\lambda} (\lambda e^\lambda - e^\lambda + 1)$
$E[X] = 10600 (\lambda - 1 + e^{-\lambda})$

Now, substitute the given value $\lambda = 1.8$:
$E[X] = 10600 (1.8 - 1 + e^{-1.8})$
$E[X] = 10600 (0.8 + e^{-1.8})$

Using a calculator, $e^{-1.8} \approx 0.1652988$
$E[X] = 10600 (0.8 + 0.1652988)$
$E[X] = 10600 (0.9652988)$
$E[X] \approx 10232.167288$

Rounding to two decimal places for currency:
$E[X] \approx \$10232.17$

The final answer is $\boxed{10232.17}$.