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Nuetrik [128]
3 years ago
13

There is a 0.9989 probability that a randomly selected 31​-year-old male lives through the year. A life insurance company charge

s ​$156 for insuring that the male will live through the year. If the male does not survive the​ year, the policy pays out ​$120 comma 000 as a death benefit. Complete parts​ (a) through​ (c) below. a. From the perspective of the 31​-year-old ​male, what are the monetary values corresponding to the two events of surviving the year and not​ surviving?
Mathematics
1 answer:
Anna35 [415]3 years ago
7 0

Answer:

Monetary value of surviving: -$156.

Monetary value of not surviving: $120,000.

Step-by-step explanation:

If he survives, he is going to have to pay $156 to the company.

If he dies, his family is paid $120,000.

So,

Monetary value of surviving: -$156.

Monetary value of not surviving: $120,000.

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\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12
\end{cases}
\\\\\\
\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\


\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
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\end{array}\to &2
\end{cases}
\\\\\\
\left(1+\frac{0.040798}{2}\right)^{2}-1\\\\
-------------------------------\\\\


\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
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n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{daily, thus 365}
\end{array}\to &365
\end{cases}
\\\\\\
\left(1+\frac{0.040730}{365}\right)^{365}-1
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