Solving Expected Net Present Value

Introduction to solving expected net present value:

Expected value is one of an essential concept in probability. In online, few websites are providing math tutoring. Tutor, will give step by step explanation for the expected value problems. In probability, expected value of a given real-values are ability to the variables as present to compute of the center of the distribution of the variable. Expected problems are dealwith, probability, geometry distribution, etc. In this article we shall discuss for solving expected net present value. Having problem with Expected Value of Binomial Distribution keep reading my upcoming posts, i will try to help you.

Sample Problem for Solving Expected Net Present Value:

solving expected net present value problem 1:

In company makes a profit of 2830 dollars with the probability 0.46 or it can have a loss of 470 dollars with the probability 0.37. Calculate the expected profit of the company?

Solution:

Let the discrete random variable ‘x’ is

x1 = 2830 dollars (profit of the company)

x2 = 470 dollars (loss of the company)

With probabilities P1 = 0.46 and P2 = 0.37 correspondingly of the company profit and loss

Then expected profit is specified by,

E(x) = `sum_(i=0)^3 (x_i)p(x_i)`

= (2830)*(0.46) - (470)*(0.37)

= 1301.8 – 173.9

= 1127.9

solving expected net present value problem 2:

Estimate the expected value from given value using the geometric distribution method where the number of possibility (p) is 3.7.

Solution:

Given:

A number of possibilities (p) are 3.7.

Formula for calculate the expected value in a given value is `mu = 1/p`

Here, p is number of possibility in the event

`mu = 1/p`

= 1/3.7

Expected value = 0.2702

An expected value for given event is 0.2702.

solving expected net present value problem 3:

Evaluate the expected value from the given discrete chance variable (`1/6` ). Where the x value is start from 2 to 5.

Solution:

Expected value is predictable for the discrete chance variable by employ the formula,

E(x) = `sum_(i=0)^4 (x_i)p(x_i)`

E(x) =` 2xx(1/6) + 3xx(1/6) + 4xx(1/6) + 5xx(1/6)`

E(x) = 0 .333 + 0.5 + 0.666 + 0.833

E(x) = 2.333

We get an expected value as 2.333


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Practice Problem for Solving Expected Net Present Value:

Evaluate the expected value from the given discrete chance variable `1/14` . Where the ‘x’ is start from 0 to 3.

Answer:  0.4285

Evaluate the expected value from the given discrete chance variable `-1/16` from 0 to 2.

Answer:  - 0.1875