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:
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