Determine the mean and the mean square value of x whose PDF is px(x) = - *°/20°1(0)

The mean square value of x is approximately 7111.11.

The given probability density function (PDF) is:

px(x) = kx(20-x), for 0 ≤ x ≤ 20

px(x) = 0, elsewhere

where k is a constant that ensures that the PDF integrates to 1 over its domain.

To find the value of k, we use the fact that the integral of the PDF over its domain equals 1:

[tex]\int_0^{20}[/tex] kx(20-x) dx = 1

Expanding and solving for k, we get:

k[tex]\int_0^{20}[/tex] (20x - x²) dx = 1

k [10x² - (1/3)x³] | from 0 to 20 = 1

k [4000 - (1/3)8000] = 1

k = 3/(8000)

Therefore, the PDF is:

px(x) = (3/(8000))x(20-x), for 0 ≤ x ≤ 20

px(x) = 0, elsewhere

To find the mean of x, we use the formula:

E[x] = [tex]\int_0^{20}[/tex]  x px(x) dx

Substituting the PDF, we get:

E[x] =[tex]\int_0^{20}[/tex]  (3/(8000))x²(20-x) dx

This integral can be evaluated using integration by parts.

Let u = x²(20-x) and dv = dx, then du/dx = 40x - 2x² and v = x.

Using the integration by parts formula, we get:

∫ u dv = uv - ∫ v du

= x³(20/3) - [tex]x^4/4[/tex] - [tex]\int_0^{20} (40x^3 - 2x^4) / 8000\: dx[/tex]

= (20/3) [tex]\int_0^{20} x^3 dx - (1/4) \int_0^{20} x^4 dx[/tex]

= (20/3)[tex](20^4/4)[/tex] - [tex](1/4)(20^5/5)[/tex]

= 2666.67

Therefore, the mean of x is approximately 2666.67.

To find the mean square value of x, we use the formula:

[tex]E[x^2][/tex]  = [tex]\int_0^{20} x^2[/tex] px(x) dx

Substituting the PDF, we get:

[tex]E[x^2][/tex] = [tex]\int_0^{20}[/tex]  (3/(8000))x³(20-x) dx

This integral can also be evaluated using integration by parts.

Let u = x³(20-x) and dv = dx, then du/dx = [tex]60x^2 - 3x^3[/tex] and v = x.

Using the integration by parts formula, we get:

∫ u dv = uv - ∫ v du

= [tex]x^4(20/4) - x^5/5 -[/tex] [tex]\int_0^{20} (60x^4 - 3x^5) / 8000\: dx[/tex]

= (20/4) [tex]\int_0^{20}x^4 dx - (1/5) \int_0^{20} x^5 dx[/tex]

= [tex](20/4)(20^5/5) - (1/5)(20^6/6)[/tex]

= 7111.11

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The mean of x is approximately 2666.67 and the mean square value of x is approximately 7111.11.

To find the mean square value of x, we use the formula:

= px(x) dx

Substituting the given PDF, we get:

= (3/(8000))x³(20-x) dx

We can use integration by parts to evaluate this integral. Let u = x³(20-x) and dv = dx, then du/dx = 60x² - 3x³ and v = x. Using the integration by parts formula, we get:

∫ u dv = uv - ∫ v du

= x⁴(20/4) - ∫ x²(60x² - 3x³) dx

= (5/2)x⁴ - 20x⁴ + x⁵/5 + C

where C is the constant of integration. Evaluating the integral between 0 and 20, we get:

= [(5/2)(20⁴) - 20(20⁴) + (20⁵/5)]/8000 - [(5/2)(0⁴) - 20(0⁴) + (0⁵/5)]/8000

= 7111.11

Therefore, the mean square value of x is approximately 7111.11.

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