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Nina [5.8K]
3 years ago
12

2. A depositor puts $25,000 in a saving account that pays 5% interest, compounded semiannually. Equal annual withdrawals are to

be made from the account, beginning one year from now and continuing forever. What is the maximum annual withdrawal.
Business
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

The correct answer is $1265.60.

Explanation:

According to the scenario, the given data are as follows:

Present Value (PV) = $25,000

Rate of interest = 5%

Rate of interest ( semi annual) (r) = 2.5%

Time period (semi annual) = 2

So, First we calculate the effective annual interest rate,

Effective annual interest rate =  ( 1 + r)^n  = (1.025)^2 -1

=5.0625%

So, Annual Withdrawal = PV × Effective annual interest rate

by putting the value, we get

Annual withdrawal = $25,000 × 5.0625%

= $1265.60

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a) $y=0.00991 x+1.042$b) $r^2=0.7503^2=0.563$\\C) $r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503$

x: 500, 700, 750, 590 , 540, 650, 480

y: 7.00, 7.50 , 9.00, 6.5, 7.50 , 7.0, 4.50

We want to create a linear model like this :

$$y=m x+b$$

Where

$$m=\frac{S_{x y}}{S_{x x}}$$

And:  

$$\begin{aligned}& S_{x y}=\sum_{i=1}^n x_i y_i-\frac{\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right)}{n} \\& S_{x x}=\sum_{i=1}^n x_i^2-\frac{\left(\sum_{i=1}^n x_i\right)^2}{n}\end{aligned}$$

With these we can find the sums:  

$$\begin{aligned}& S_{x x}=\sum_{i=1}^n x_i^2-\frac{\left(\sum_{i=1}^n x_i\right)^2}{n}=2595100-\frac{4210^2}{7}=63085.714 \\& S_{x y}=\sum_{i=1}^n x_i y_i-\frac{\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right) n}{=} 30095-\frac{4210 * 49}{7}=625\end{aligned}$$

And the slope would be:  

m=\frac{625}{63085.714}=0.00991

Now we can find the means for x and y like this:

$$\begin{aligned}& \bar{x}=\frac{\sum x_i}{n}=\frac{4210}{7}=601.429 \\& \bar{y}=\frac{\sum y_i}{n}=\frac{49}{7}=7\end{aligned}$$

And we can find the intercept using this:

$$b=\bar{y}-m \bar{x}=7-(0.00991 * 601.429)=1.042$$

And the line would be:

$$y=0.00991 x+1.042$$

Part b

The correlation coefficient is given by:

r=\frac{n\left(\sum x y\right)-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left[n \sum x^2-\left(\sum x\right)^2\right]\left[n \sum y^2-\left(\sum y\right)^2\right]}}

For our case we have this:

$$\begin{aligned}& \mathrm{n}=7 \sum x=4210, \sum y=49, \sum x y=30095, \sum x^2=2595100, \sum y^2=354 \\& r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503\end{aligned}$$

The determination coefficient is given by:

$$r^2=0.7503^2=0.563$$

Part c

r=\frac{7(30095)-(4210)(49)}{\sqrt{\left[7(2595100)-(4210)^2\right]\left[7(354)-(49)^2\right]}}=0.7503

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