3. A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else
a. Find the probability that X is between 2 and 4
b. Find the probability that the value of X exceeds 3
c. Find the expected value of X
d. Find the Variance of X

Respuesta :

The solutions to the questions are

  • The probability that X is between 2 and 4 is 0.314
  • The probability that X exceeds 3 is 0.199
  • The expected value of X is 2
  • The variance of X is 2

Find the probability that X is between 2 and 4

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

[tex]P(x) = \int\limits^a_b {f(x) \, dx[/tex]

So, we have:

[tex]P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx[/tex]

Using an integral calculator, we have:

[tex]P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2[/tex]

Expand the expression

[tex]P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}[/tex]

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

Find the probability that the value of X exceeds 3

This is represented as:

[tex]P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx[/tex]

Using an integral calculator, we have:

[tex]P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3[/tex]

Expand the expression

[tex]P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}[/tex]

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

Find the expected value of X

This is calculated as:

[tex]E(x) = \int\limits^a_b {x * f(x) \, dx[/tex]

So, we have:

[tex]E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx[/tex]

This gives

[tex]E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx[/tex]

Using an integral calculator, we have:

[tex]E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0[/tex]

Expand the expression

[tex]E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}[/tex]

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

Find the Variance of X

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

[tex]E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx[/tex]

This gives

[tex]E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx[/tex]

Using an integral calculator, we have:

[tex]E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0[/tex]

Expand the expression

[tex]E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}[/tex]

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

https://brainly.com/question/15318348

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