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3 years ago

A $250,000 home loan is used to purchase a house. The loan is for 30 years and has a 5.4% APR. Use the amortization formula to d

etermine the amount of the monthly payments.

Mathematics

1 answer:

wlad13 [49]3 years ago

8 0

$\bf ~~~~~~~~~~~~ \textit{Amortized Loan Value}
\\\\
pymt=P\left[ \cfrac{\frac{r}{n}}{1-\left( 1+ \frac{r}{n}\right)^{-nt}} \right]
\\\\\\
~~~~~~
\begin{cases}
P=
\begin{array}{llll}
\textit{original amount deposited}\\
\end{array}\to &250000\\
pymt=\textit{periodic payments}\\
r=rate\to 5.4\%\to \frac{5.4}{100}\to &0.054\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{since the payments are}\\
\textit{monthly, then twelve}
\end{array}\to &12\\
t=years\to &30
\end{cases}
$

$\bf pymt=250000\left[ \cfrac{\frac{0.054}{12}}{1-\left( 1+ \frac{0.054}{12}\right)^{-12\cdot 30}} \right]
\\\\\\
pymt=250000\left[ \cfrac{0.0045}{1-\left( 1.0045\right)^{-360}} \right]
\\\\\\
pymt\approx 250000\left[ \cfrac{0.0045}{0.80138080852472389274} \right]$

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And replacing we got:

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And replacing into the confidence interval formula we got:

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Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

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