Q4
← Back
Basic Info
Probability
└── Homework
└── W7
└── Q4.tex
Preview
\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage[margin=1in]{geometry}
\usepackage{fancyhdr}
\usepackage{enumerate}
\usepackage[shortlabels]{enumitem}
\pagestyle{fancy}
\fancyhead[l]{Li Yifeng}
\fancyhead[c]{Homework \#7}
\fancyhead[r]{\today}
\fancyfoot[c]{\thepage}
\renewcommand{\headrulewidth}{0.2pt}
\setlength{\headheight}{15pt}
\newcommand{\bE}{\mathbb{E}}
\newcommand{\bP}{\mathbb{P}}
\begin{document}
\section*{Question 4}
\noindent An insurance company will sell a $\$90,000$ one-year term life insurance policy to an individual in a particular risk group for a premium of $\$478$.
\bigskip
\noindent Find the expected value to the company of a single policy if a person in this risk group has a $99.62\%$ chance of surviving one year.
\bigskip
\begin{enumerate}[label={},leftmargin=0in]\item
\subsection*{Solution}
Let $X$ be the random variable of the net gain of the company from a policy, then easy to get the $pdf$ of $X$
\[
p(x) =
\begin{cases}
\begin{aligned}
1 - 99.62\% = 0.38\%,&\quad x = 478 - 90\,000 = -89\,522\\
99.62\%,&\quad x = 478
\end{aligned}
\end{cases}
\]
Then we have the expectation of $X$
\[
\begin{aligned}
\bE[X] &= -89\,552p(-89\,552) + 478p(478)\\
&= -89\,522(0.38\%) + 478(99.62\%)\\
&= 136
\end{aligned}
\]
\subsection*{Answer}
\[\boxed{\bE[X] = 136}\]
\end{enumerate}
\end{document}