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marishachu [46]
2 years ago
14

(5) On January l, 1988, Felix inherited a perpetuity with annual payments beginning in six months. The first payment was $3,000,

and after that the payments increased by 3% each year. Find the value of this perpetuity on January 1, 1995 if the annual effective interest rate was 6% from January I, 1988 through January I, 1996 and 4% thereafter.

Mathematics
1 answer:
valkas [14]2 years ago
8 0

Answer:

Hence the value of this perpetuity on January 1, 1995 will be $55993.18404

Step-by-step explanation:

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Answer:

The 90% confidence interval would be given by (57.006;62.994)  

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X=60 represent the sample mean  

\mu population mean (variable of interest)  

\sigma=10 represent the population standard deviation  

n=30 represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma=10)

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.90=0.1,\alpha/2 =0.05 and z_\alpha/2=1.64  

Using the normal standard table, excel or a calculator we see that:  

z_{\alpha/2}=1.64

Since we have all the values we can replace:

60 - 1.64\frac{10}{\sqrt{30}}=57.006  

60 + 1.64\frac{10}{\sqrt{30}}=62.994  

So on this case the 90% confidence interval would be given by (57.006;62.994)  

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