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laila [671]
3 years ago
13

$1689 invested for 4 years at 3% compounded annually ?

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

The amount gotten after $1689 invested for 4 years at 3% compounded annually is $1901

The amount of money gained after an investment is compounded is given by:

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

Where P is principal, A is the final amount, r is the rate, n is the number of times compounded per period and t is the time

Given that P = $1689, t = 4, r = 3% = 0.03, n = 1, hence:

A=P(1+\frac{r}{n} )^{nt}\\\\A=1689(1+\frac{0.03}{1} )^{4*1}\\\\A=\$1901

The amount gotten after $1689 invested for 4 years at 3% compounded annually is $1901

Find out more at: brainly.com/question/14295570

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Tpy6a [65]

Answer:

a) 0.048 = 4.8% probability of exactly seven successes.

b) 100% probability of at least twenty-two successes.

Step-by-step explanation:

The first question we use the binomial distribution, while for the second we use the approximation to the normal.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

Question a:

Here, we have n = 10, p = 0.41.

This probability is P(X = 7). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.048

0.048 = 4.8% probability of exactly seven successes.

b) With 150 trials and a 39 chance of success, the probability of at least twenty-two successes.

Here we have n = 150, p = 0.39

Mean and standard deviation:

\mu = 150*0.39 = 58.5

\sigma = \sqrt{150*0.39*0.61} = 5.9737

This probability is, using continuity correction, P(X \geq 22 - 0.5) = P(X \geq 21.5), which is 1 subtracted by the p-value of Z when X = 21.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21.5 - 58.5}{5.9737}

Z = -6.19

Z = -6.19 has a p-value of 0

1 - 0 = 1

100% probability of at least twenty-two successes.

5 0
3 years ago
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