Value of an Integral Type Expected

Expected value is one of an important concept in probability. In probability, expected value of a given real-values are chance the variables as present a compute of the center of the distribution of the variable. In online, few websites are providing math tutoring. Tutor, will give step by step explanation for the expected value problems. Expected problems are deals with, probability, geometry distribution, etc. in this article we shall discuss for value of an integral type expected.


Sample Problem for Value of an Integral Type Expected:

Value of an integral type expected problem 1:

Evaluate the expected value from the given continuous random variable using uniform distribution. Value is expected from the interval value is 3 < x < 7.

Solution:

Given:

A given interval value for the uniform distribution is a = 3 and b = 7

Formula for finding the expected value for uniform distribution is

E(X) = `int_a^bx f(x)dx` --------------- (1)

Here, the value of f(x) is `1/(b - a)` , for 3 < x < 7

= `1/(7 - 3)`

= `1/4`

f(x) = `1/4`

In the next step we put the f(x) value in the above equation, we get

= `int_3^7x (1/4) dx`

= `1/4int_3^7x dx`

= `1/4[ x^2/2]_3^7`

= `1/4[(7)^2/2-(3)^2/2 ]`

= `1/4 ` [1/2 (7)2 - (3)2]

= `1/8` [49 - 9]

= `1/8` (40)

= `40/8`

= 5

We get expected value E(x) is 5.

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Value of an Integral Type Expected Problem 2:

Evaluate the expected value from the given probability density function using exponential distribution `e^-2x` with the interval of [0, `oo` ].

Solution:

Given:

A given probability density function f(x) is `e^-2x` .

Formula for finding the expected value for uniform distribution is

E(X) =` int_0^ooxf(x)dx` --------------- (1)

In the first we find out the expected value of the exponential distribution function


=` int_0^oox(e^(-2x))dx`

=` int_0^ooxe^(-2x)dx`

Here we use `int udv = uv - int vdu`

u = x       dv = `e^(-2x)`

u' = 1      v = `(e^(-2x))/-2 `

u'' = 0      v' = `(e^-2x)/4`

in the next step we substitute the above values, we get

= ` [(xe^(-2x))/-2- (e^(-2x))/4]_0^oo`

= ` [((ooe^(-2oo))/-2-(e^(-2oo))/4 )- ((0 - e^(0))/4)]`

Here we use `e^-oo` = 0, e0 = 1/4

= 0.25

We get the expected value E(X) is 0.25.