Respuesta :
Answers:
- mean: 2.667
- variance: 0.889
Explanation:
To get the mean and variance of x, we need to verify first whether...
- x is discrete or continuous random variable
- f is probability mass or probability density function
because if we cannot verify the 2 statements above, we can't compute the mean and the variance.
Since 0 < x < 4, x is a continuous random variable because x can be any positive number less than, which includes a non-integer.
Note that if the random variable is continuous and [tex]0 \leq f(x) \leq 1[/tex] for any values of x in the domain of f, then f is a probability density function (PDF).
Note that
[tex]0 \ \textless \ x \ \textless \ 4 \\ \Leftrightarrow 0.125(0) \ \textless \ 0.125x \ \textless \ 0.125(4) \\ \Leftrightarrow 0 \ \textless \ 0.125x \ \textless \ 0.5 \\ \Leftrightarrow 0 \ \textless \ f(x) \ \textless \ 0.5 \\ \Rightarrow 0 \ \textless \ f(x) \ \textless \ 1 (\text{Because }0 \ \textless \ f(x) \ \textless \ 0.5 \ \textless \ 1)[/tex]
Hence, for any x in the domain of f, 0 < f(x) < 1. Moreover, since x is a continuous random variable, thus f is a PDF.
First, we use the following notations for mean and variance:
E[x] = mean of x
Var[x] = variance of x
Since f is a probability density function, we can use the following formulas for the mean and the variance of x:
[tex]\boxed{\text{mean of }x = E[x] = \int_{-\infty}^{\infty}{xf(x)}dx}[/tex]
[tex]\boxed{\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2}} [/tex]
To compute for the mean of x,
[tex]\text{mean of }x = E[x] \\ = \int_{-\infty}^{\infty}{xf(x)}dx} \\ = \int_{-\infty}^{\infty}{x(0.125x)}dx} \\ \boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx}}[/tex]
The integral seems complicated because of the infinity sign. But because the domain of f is the set of positive numbers less than 4, that is,
[tex]\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}[/tex]
the bounds of the integral for the mean can be changed from [tex]-\infty \ \textless \ x \ \textless \ \infty[/tex] to [tex]0 \ \textless \ x \ \textless \ 4[/tex] so that
[tex]\boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx = \int_{0}^{4}{0.125x^2}dx}[/tex]
Hence, the mean is computed as
[tex]\text{mean of }x = \int_{0}^{4}{0.125x^2}dx \\ = \left[ \frac{0.125x^3}{3} \right]_{0}^{4} \\ \\ = \left[ \frac{0.125(4)^3}{3} \right] - \left[ \frac{0.125(0)^3}{3} \right] \\ \\ \boxed{\text{mean of }x = \frac{8}{3} \approx 2.667} [/tex]
Since the formula for variance is computed as
[tex]\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2[/tex]
we must first compute for [tex]E[x^2][/tex] for which
[tex]E[x^2] = \int_{-\infty}^{\infty}{x^2 f(x)}dx \\ \\ = \int_{-\infty}^{\infty}{x^2(0.125x)}dx \\ \\ \boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx}[/tex]
Similar to the computation of integral of the mean, we take note that
[tex]\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}[/tex]
so that we can change the bounds of the integral, that is,
[tex]\boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx = \int_{0}^{4}{0.125x^3}dx}[/tex]
Hence,
[tex]E[x^2] = \int_{0}^{4}{0.125x^3}dx \\ \\ = \int_{0}^{4}{0.125x^3}dx \\ \\= \left[ \frac{0.125x^4}{4} \right]_{0}^{4} \\ \\ = \left[ \frac{0.125(4)^4}{4} \right] - \left[ \frac{0.125(0)^4}{4} \right] \\ \\ \boxed{E[x^2] = 8}[/tex]
Because [tex]E[x] = \frac{8}{3} [/tex],
[tex]\text{Variance of }x \\ \\ = E[x^2] - (E[x])^2 \\ \\ = 8 - \left( \frac{8}{3} \right)^2 \\ \\ \boxed{\text{Variance of }x = \frac{8}{9} \approx 0.889}[/tex]
- mean: 2.667
- variance: 0.889
Explanation:
To get the mean and variance of x, we need to verify first whether...
- x is discrete or continuous random variable
- f is probability mass or probability density function
because if we cannot verify the 2 statements above, we can't compute the mean and the variance.
Since 0 < x < 4, x is a continuous random variable because x can be any positive number less than, which includes a non-integer.
Note that if the random variable is continuous and [tex]0 \leq f(x) \leq 1[/tex] for any values of x in the domain of f, then f is a probability density function (PDF).
Note that
[tex]0 \ \textless \ x \ \textless \ 4 \\ \Leftrightarrow 0.125(0) \ \textless \ 0.125x \ \textless \ 0.125(4) \\ \Leftrightarrow 0 \ \textless \ 0.125x \ \textless \ 0.5 \\ \Leftrightarrow 0 \ \textless \ f(x) \ \textless \ 0.5 \\ \Rightarrow 0 \ \textless \ f(x) \ \textless \ 1 (\text{Because }0 \ \textless \ f(x) \ \textless \ 0.5 \ \textless \ 1)[/tex]
Hence, for any x in the domain of f, 0 < f(x) < 1. Moreover, since x is a continuous random variable, thus f is a PDF.
First, we use the following notations for mean and variance:
E[x] = mean of x
Var[x] = variance of x
Since f is a probability density function, we can use the following formulas for the mean and the variance of x:
[tex]\boxed{\text{mean of }x = E[x] = \int_{-\infty}^{\infty}{xf(x)}dx}[/tex]
[tex]\boxed{\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2}} [/tex]
To compute for the mean of x,
[tex]\text{mean of }x = E[x] \\ = \int_{-\infty}^{\infty}{xf(x)}dx} \\ = \int_{-\infty}^{\infty}{x(0.125x)}dx} \\ \boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx}}[/tex]
The integral seems complicated because of the infinity sign. But because the domain of f is the set of positive numbers less than 4, that is,
[tex]\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}[/tex]
the bounds of the integral for the mean can be changed from [tex]-\infty \ \textless \ x \ \textless \ \infty[/tex] to [tex]0 \ \textless \ x \ \textless \ 4[/tex] so that
[tex]\boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx = \int_{0}^{4}{0.125x^2}dx}[/tex]
Hence, the mean is computed as
[tex]\text{mean of }x = \int_{0}^{4}{0.125x^2}dx \\ = \left[ \frac{0.125x^3}{3} \right]_{0}^{4} \\ \\ = \left[ \frac{0.125(4)^3}{3} \right] - \left[ \frac{0.125(0)^3}{3} \right] \\ \\ \boxed{\text{mean of }x = \frac{8}{3} \approx 2.667} [/tex]
Since the formula for variance is computed as
[tex]\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2[/tex]
we must first compute for [tex]E[x^2][/tex] for which
[tex]E[x^2] = \int_{-\infty}^{\infty}{x^2 f(x)}dx \\ \\ = \int_{-\infty}^{\infty}{x^2(0.125x)}dx \\ \\ \boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx}[/tex]
Similar to the computation of integral of the mean, we take note that
[tex]\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}[/tex]
so that we can change the bounds of the integral, that is,
[tex]\boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx = \int_{0}^{4}{0.125x^3}dx}[/tex]
Hence,
[tex]E[x^2] = \int_{0}^{4}{0.125x^3}dx \\ \\ = \int_{0}^{4}{0.125x^3}dx \\ \\= \left[ \frac{0.125x^4}{4} \right]_{0}^{4} \\ \\ = \left[ \frac{0.125(4)^4}{4} \right] - \left[ \frac{0.125(0)^4}{4} \right] \\ \\ \boxed{E[x^2] = 8}[/tex]
Because [tex]E[x] = \frac{8}{3} [/tex],
[tex]\text{Variance of }x \\ \\ = E[x^2] - (E[x])^2 \\ \\ = 8 - \left( \frac{8}{3} \right)^2 \\ \\ \boxed{\text{Variance of }x = \frac{8}{9} \approx 0.889}[/tex]
