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

This article aims to find the mean and variance of $ x$ given $ f(x) $ and the range of $x$. The article uses the concept of mean and variance.

The formula for mean and variance is given as:

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

\[Variance\: of\: x = Var(x) = E[x^{2}] – (E[x])^{2}\]

Expert Answer

To get the mean and variance of $ x $, we first need to verify that…

– $x$ is a discrete or continuous random variable

– $f$ is the probability weight or probability density function

because if we can’t verify the above $2$ statements, then we can’t calculate the mean and variance.

Since $0 < x < 4$, $x$ is a continuous random variable because $x$ can be any positive number less than that 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 $f$, then $f$ is a probability density function $(PDF)$.

Note that:

\[0<x<4\]

\[\Leftrightarrow 0.125(0) < 0.125x < 0.125(4) \]

\[\Leftrightarrow 0 < 0.125x < 0.5 \]

\[\Leftrightarrow 0 < f(x) < 0.5 \]

\[\Rightarrow 0<f(x) <1 \]

Thus, for any $x$ in the domain $f$, $0 < f(x) < 1$. Furthermore, since $x$ is a continuous random variable, $f$ is a $PDF$.

First, we use the following notation for mean and variance:

\[E(x) = mean \: of \: x\]

\[Var(x) = variance\: of \: x\]

Since $f$ represents probability density function, we can use the following formulas for the mean and variance of $x$:

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

\[Variance\: of\: x = Var(x) = E[x^{2}] – (E[x])^{2}\]

To find the mean of $ x$:

\[mean\: of \: x = E[x] \]

\[= \int_{-\infty}^{\infty} xf(x)dx\]

\[mean\: of \: x= \int_{-\infty}^{\infty} 0.125x^{2}dx \]

The integral seems complicated because of the infinity sign, but since the domain of $f$ is the set of positive numbers smaller than $4$, i.e.

\[domain\: of \: f = {x: 0<x<4}\]

The bounds of the integral for mean value can be changed from $-\infty <x < \infty$ to $0<x<4$ such that:

\[mean\: of \: x = \int_{-\infty}^{\infty} 0.125x^{2}dx = \int_{0}^{4} 0.125 x^{2} dx\]

Hence, the mean is computed as:

\[= |\dfrac{0.125 x^{3}}{3}|_{0}^{4} = \dfrac{8}{3}\]

\[mean \: of \: x = 2.667\]

The formula for the variance of the $ x$ is

\[Variance\: of\: x = Var(x) = E[x^{2}] – (E[x])^{2}\]

We need to compute $E[x^{2}]$

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

\[=\int_{-\infty}^{\infty} x^{2} (0.125x) dx \]

\[=\int_{-\infty}^{\infty} 0.125x^{3} dx \]

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

\[= |\dfrac {0.125x^{4}}{4}|_{0}^{4}\]

\[E[x^{2}] = 8\]

\[Variance\: of\: x = Var(x) = E[x^{2}] – (E[x])^{2}\]

\[variance \: of \: x = 8- (\dfrac{8}{3})^{2} \]

\[variance \: of \: x = 0.889\]

Numerical Result

The mean of $x$ is $2.667$.

The variance of $x$ is $0.889$.

Example

Suppose $f(x) = 0.125x$ for $0 < x < 2$. Determine the mean and variance of $x$.

Solution

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

\[Variance\: of\: x = Var(x) = E[x^{2}] – (E[x])^{2}\]

Hence, the mean is computed as:

\[mean \: of \: x = 0.33\]

The formula for the variance of the $ x$ is:

\[variance \: of \: x = 0.3911\]

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