document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1167857>Question 1167857</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and $10600 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business has a Poisson distribution with mean 1.8.<BR>\n" );
document.write( "Find the expected amount paid to the company under this policy during a one-year period.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852087 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852087\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $N$ be the number of major snowstorms per year that shut down business.<BR>\n" );
document.write( "$N$ follows a Poisson distribution with mean $\lambda = 1.8$.<BR>\n" );
document.write( "The probability mass function (PMF) is $P(N=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ for $k = 0, 1, 2, \dots$.\r<BR>\n" );
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document.write( "Let $X$ be the amount paid to the company under the policy.<BR>\n" );
document.write( "The policy pays nothing for the first snowstorm.<BR>\n" );
document.write( "The policy pays $\$10600$ for each one thereafter.\r<BR>\n" );
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document.write( "We can define $X$ as follows:<BR>\n" );
document.write( "* If $N=0$ (no snowstorms), $X=0$.<BR>\n" );
document.write( "* If $N=1$ (one snowstorm), $X=0$ (the first one is free).<BR>\n" );
document.write( "* If $N=k$ for $k \ge 2$, the number of payable snowstorms is $k-1$. So, $X = 10600(k-1)$.\r<BR>\n" );
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document.write( "We want to find the expected amount paid, $E[X]$.<BR>\n" );
document.write( "$E[X] = \sum_{k=0}^{\infty} X(k) P(N=k)$<BR>\n" );
document.write( "$E[X] = 0 \cdot P(N=0) + 0 \cdot P(N=1) + \sum_{k=2}^{\infty} 10600(k-1) P(N=k)$<BR>\n" );
document.write( "$E[X] = 10600 \sum_{k=2}^{\infty} (k-1) \frac{e^{-\lambda} \lambda^k}{k!}$<BR>\n" );
document.write( "We can factor out $e^{-\lambda}$:<BR>\n" );
document.write( "$E[X] = 10600 e^{-\lambda} \sum_{k=2}^{\infty} (k-1) \frac{\lambda^k}{k!}$\r<BR>\n" );
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document.write( "Let's evaluate the sum $S = \sum_{k=2}^{\infty} (k-1) \frac{\lambda^k}{k!}$:<BR>\n" );
document.write( "$S = (2-1)\frac{\lambda^2}{2!} + (3-1)\frac{\lambda^3}{3!} + (4-1)\frac{\lambda^4}{4!} + \dots$<BR>\n" );
document.write( "$S = \frac{\lambda^2}{2!} + \frac{2\lambda^3}{3!} + \frac{3\lambda^4}{4!} + \dots$\r<BR>\n" );
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document.write( "We can use the property that $\frac{k-1}{k!} = \frac{k}{k!} - \frac{1}{k!} = \frac{1}{(k-1)!} - \frac{1}{k!}$.<BR>\n" );
document.write( "So, $S = \sum_{k=2}^{\infty} \left( \frac{1}{(k-1)!} - \frac{1}{k!} \right) \lambda^k$<BR>\n" );
document.write( "$S = \sum_{k=2}^{\infty} \frac{\lambda^k}{(k-1)!} - \sum_{k=2}^{\infty} \frac{\lambda^k}{k!}$\r<BR>\n" );
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document.write( "For the first part of the sum: $\sum_{k=2}^{\infty} \frac{\lambda^k}{(k-1)!} = \lambda \sum_{k=2}^{\infty} \frac{\lambda^{k-1}}{(k-1)!}$<BR>\n" );
document.write( "Let $j=k-1$. When $k=2$, $j=1$.<BR>\n" );
document.write( "$= \lambda \sum_{j=1}^{\infty} \frac{\lambda^j}{j!} = \lambda \left( \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \dots \right)$<BR>\n" );
document.write( "We know the Taylor series for $e^\lambda$ is $\sum_{j=0}^{\infty} \frac{\lambda^j}{j!} = 1 + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!} + \dots = e^\lambda$.<BR>\n" );
document.write( "So, $\sum_{j=1}^{\infty} \frac{\lambda^j}{j!} = e^\lambda - 1$.<BR>\n" );
document.write( "Thus, the first part is $\lambda(e^\lambda - 1)$.\r<BR>\n" );
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document.write( "For the second part of the sum: $\sum_{k=2}^{\infty} \frac{\lambda^k}{k!}$<BR>\n" );
document.write( "This is the Taylor series for $e^\lambda$ minus the first two terms ($k=0$ and $k=1$):<BR>\n" );
document.write( "$\sum_{k=2}^{\infty} \frac{\lambda^k}{k!} = \left( \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} \right) - \frac{\lambda^0}{0!} - \frac{\lambda^1}{1!} = e^\lambda - 1 - \lambda$.\r<BR>\n" );
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document.write( "Now substitute these back into $S$:<BR>\n" );
document.write( "$S = \lambda(e^\lambda - 1) - (e^\lambda - 1 - \lambda)$<BR>\n" );
document.write( "$S = \lambda e^\lambda - \lambda - e^\lambda + 1 + \lambda$<BR>\n" );
document.write( "$S = \lambda e^\lambda - e^\lambda + 1$\r<BR>\n" );
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document.write( "Finally, multiply by $e^{-\lambda}$ to get $E[X]$:<BR>\n" );
document.write( "$E[X] = 10600 e^{-\lambda} (\lambda e^\lambda - e^\lambda + 1)$<BR>\n" );
document.write( "$E[X] = 10600 (\lambda - 1 + e^{-\lambda})$\r<BR>\n" );
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document.write( "Now, substitute the given value $\lambda = 1.8$:<BR>\n" );
document.write( "$E[X] = 10600 (1.8 - 1 + e^{-1.8})$<BR>\n" );
document.write( "$E[X] = 10600 (0.8 + e^{-1.8})$\r<BR>\n" );
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document.write( "Using a calculator, $e^{-1.8} \approx 0.1652988$<BR>\n" );
document.write( "$E[X] = 10600 (0.8 + 0.1652988)$<BR>\n" );
document.write( "$E[X] = 10600 (0.9652988)$<BR>\n" );
document.write( "$E[X] \approx 10232.167288$\r<BR>\n" );
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document.write( "Rounding to two decimal places for currency:<BR>\n" );
document.write( "$E[X] \approx \$10232.17$\r<BR>\n" );
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document.write( "The final answer is $\boxed{10232.17}$. </SPAN>\n" );
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