Question

Suppose f(x) = 0.125x for 0 < x < 4. determine the mean and variance of x. round your answers to 3 decimal places.

Answer 1

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 $0 \leq f(x) \leq 1$ for any values of x in the domain of f, then f is a probability density function (PDF). 

Note that

$0 \ \textless \ x \ \textless \ 4 \\ \Leftrightarrow 0.125(0) \ \textless \ 0.125x \ \textless \ 0.125(4) \\ \Leftrightarrow 0 \ \textless \ 0.125x \ \textless \ 0.5 \\ \Leftrightarrow 0 \ \textless \ f(x) \ \textless \ 0.5 \\ \Rightarrow 0 \ \textless \ f(x) \ \textless \ 1 (\text{Because }0 \ \textless \ f(x) \ \textless \ 0.5 \ \textless \ 1)$

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:

$\boxed{\text{mean of }x = E[x] = \int_{-\infty}^{\infty}{xf(x)}dx}$

$\boxed{\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2}}$

To compute for the mean of x,

$\text{mean of }x = E[x] \\ = \int_{-\infty}^{\infty}{xf(x)}dx} \\ = \int_{-\infty}^{\infty}{x(0.125x)}dx} \\ \boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx}}$

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,

$\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}$

the bounds of the integral for the mean can be changed from $-\infty \ \textless \ x \ \textless \ \infty$ to $0 \ \textless \ x \ \textless \ 4$ so that 

$\boxed{\text{mean of }x = \int_{-\infty}^{\infty}{0.125x^2}dx = \int_{0}^{4}{0.125x^2}dx}$

Hence, the mean is computed as 

$\text{mean of }x = \int_{0}^{4}{0.125x^2}dx \\ = \left[ \frac{0.125x^3}{3} \right]_{0}^{4} \\ \\ = \left[ \frac{0.125(4)^3}{3} \right] - \left[ \frac{0.125(0)^3}{3} \right] \\ \\ \boxed{\text{mean of }x = \frac{8}{3} \approx 2.667}$

Since the formula for variance is computed as 

$\text{Variance of }x = \text{Var}[x] = E[x^2] - (E[x])^2$

we must first compute for $E[x^2]$ for which

$E[x^2] = \int_{-\infty}^{\infty}{x^2 f(x)}dx \\ \\ = \int_{-\infty}^{\infty}{x^2(0.125x)}dx \\ \\ \boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx}$

Similar to the computation of integral of the mean, we take note that 

$\text{domain of }f = \left \{x : 0 \ \textless \ x \ \textless \ 4 \right \}$

so that we can change the bounds of the integral, that is,

$\boxed{E[x^2] = \int_{-\infty}^{\infty}{0.125x^3}dx = \int_{0}^{4}{0.125x^3}dx}$

Hence,

$E[x^2] = \int_{0}^{4}{0.125x^3}dx \\ \\ = \int_{0}^{4}{0.125x^3}dx \\ \\= \left[ \frac{0.125x^4}{4} \right]_{0}^{4} \\ \\ = \left[ \frac{0.125(4)^4}{4} \right] - \left[ \frac{0.125(0)^4}{4} \right] \\ \\ \boxed{E[x^2] = 8}$

Because $E[x] = \frac{8}{3}$,

$\text{Variance of }x \\ \\ = E[x^2] - (E[x])^2 \\ \\ = 8 - \left( \frac{8}{3} \right)^2 \\ \\ \boxed{\text{Variance of }x = \frac{8}{9} \approx 0.889}$