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3 years ago

Jonathon saves $4,000 at the end of each quarter for 10 years. Assume 12% compounded quarterly and find the present value.

Mathematics

1 answer:

charle [14.2K]3 years ago

5 0

Answer:

Results are below.

Step-by-step explanation:

1. First, we need to calculate the Future Value:

FV= {A*[(1+i)^n-1]}/i

A= quarterly deposit

n= 10*4= 40

i= 0.12/4= 0.03

FV= {4,000*[(1.03^40) - 1]} / 0.03

FV= $301,605.04

Now, the present value:

PV= FV/(1+i)^N

PV= 301,605.04/(1.03^40)

PV= $92,459.09

2. First, we need to calculate the value of the first year of college:

FV= PV*(1+i)^n

FV= 24,000*(1.04^3)

FV= $26,996.74

Now, the quarterly payments:

FV= {A*[(1+i)^n-1]}/i

A= quarterly deposit

Isolating A:

A= (FV*i)/{[(1+i)^n]-1}

i= 0.08/4= 0.02

n= 3*4= 12

A= (26,996.74*0.02) / [(1.02^12) - 1]

A= $2,012.87

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Answer:

1) $E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

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And the variance would be given by:

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And the variance would be given by:

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And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

Step-by-step explanation:

For this case we have the following distributions given:

Probability M J

0.3 14% 22%

0.4 10% 4%

0.3 19% 12%

Part 1

The expected value is given by this formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

Part 2

$E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

Part 3

We can calculate the second moment first with the following formula:

$E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

Part 4

We can calculate the second moment first with the following formula:

$E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

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Answer:

(a) The value of fₓ (9.5) is 0.125.

(b) The value of fₓ (10.5) is 0.50.

Step-by-step explanation:

Let X denote delivery time of the mail delivered by Alice and Y denote delivery time of the mail delivered by Bob.

It i provided that:

$X\sim U(9, 11)\\Y\sim U(10, 12)$

The probability that Alice delivers the mail is, p = 1/4.

The probability that Bob delivers the mail is, q = 3/4.

The probability density function of a Uniform distribution with parameters [a, b] is:

$f(x)=\left \{ {{\frac{1}{b-a};\ a, b>0} \atop {0;\ otherwise}} \right.$

The probability density function of the delivery time of Alice is:

$f(X_{A})=\left \{ {{\frac{1}{b-a}=\frac{1}{2};\ [a, b]=[9, 11]} \atop {0;\ otherwise}} \right.$

The probability density function of the delivery time of Bob is:

$f(X_{B})=\left \{ {{\frac{1}{b-a}=\frac{1}{2};\ [a, b]=[10, 12]} \atop {0;\ otherwise}} \right.$

(a)

Compute the value of fₓ (9.5) as follows:

For delivery time 9.5, only Alice can do the delivery because Bob delivers the mail in the time interval 10 to 12.

The value of fₓ (9.5) is:

$f_{X}(9.5)=p.f(X_{A})+q.f(X_B})\\=(\frac{1}{4}\times \frac{1}{2})+(\frac{3}{4}\times0)\\=\frac{1}{8}\\=0.125$

Thus, the value of fₓ (9.5) is 0.125.

(b)

Compute the value of fₓ (10.5) as follows:

For delivery time 10.5, both Alice and Bob can do the delivery because Alice's delivery time is in the interval 9 to 11 and that of Bob's is in the time interval 10 to 12.

The value of fₓ (10.5) is:

$f_{X}(10.5)=p.f(X_{A})+q.f(X_B})\\=(\frac{1}{4}\times \frac{1}{2})+(\frac{3}{4}\times\frac{1}{2})\\=\frac{1}{8}+\frac{3}{8}\\=0.50$

Thus, the value of fₓ (10.5) is 0.50.

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