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Andru [333]
3 years ago
11

Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa

st returns. He learns that over the entire 20th century, the real (that is, adjusted for inflation) annual returns on U.S. common stocks had mean 8.7% and standard deviation 20.2%. The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal. (a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 32 years will exceed 14%
Mathematics
1 answer:
erica [24]3 years ago
4 0

Answer: 0.3974

Step-by-step explanation:

Given : The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal.

Real annual returns on U.S. common stocks had mean : \mu=0.087

Standard deviation : \sigma=0.202

We assume that the past pattern of variation continues.

Let x be the random variable that represents the annual returns on common stocks over the next 32 years .

The formula for z-score : z=\dfrac{x-\mu}{\sigma}

For x= 0.14, z=\dfrac{0.14-0.087}{0.202}\approx0.26

By using the standard normal distribution table , we have

The  probability that the mean annual return on common stocks over the next 32 years will exceed 14% :-

P(x>0.14)=P(z>0.26)=1-P(z\leq0.26)\\\\=1-0.6025681=0.3974319\approx0.3974

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Let's move like a crab, backwards some.

after 2 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
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t=years\to &2
\end{cases}
\\\\\\
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after 3 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
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is that enough to pay the $1100?


now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

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1000r² + 2000r + 1000   <---- standard form.
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