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2 years ago

13

suppose you could make a single "lump sum" deposit of $3679, in an investment that provides an annual percentage rate(apr) of 4%

compounded daily. determine the future value(fv) of the investment after 11 years.

1 answer:

yaroslaw [1]2 years ago

3 0

Answer:

$\$5,698.30$

Step-by-step explanation:

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value (future value)

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=11\ years\\ P=\$3,679\\ r=4\%=4/100=0.04\\n=365$

substitute in the formula above

$A=3,670(1+\frac{0.04}{365})^{365*11}$

$A=\$5,698.30$

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1 year ago

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr

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Answer:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

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And using a calculator, excel ir the normal standard table we have that:

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And we can calculate the probability like this: $P(-1.5 \leq Z \leq 1.5) = P(z$ Step-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(50,12)$

Where $\mu=50$ and $\sigma=12$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

We can find the probability required like this:

$z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5$

$z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5$

And using a calculator, excel ir the normal standard table we have that:

$P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)$

And we can calculate the probability like this:

$P(-1.5 \leq Z \leq 1.5) = P(z$

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Answer:

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Answer:

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Step-by-step explanation:

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