Question

Find E(X) , var (X), and std deviation (x) if a random variable X is given by its density function f(x) , such that
f(x)=0, if x≤1
f(x)=(3/8) (x^2), if 0<x≤2
f(x)=0, if x>2

Answer 1

Answer on Question #43494 – Math - Statistics and Probability

Find $E(X)$, var $(X)$, and std deviation $(x)$ if a random variable $X$ is given by its density function $f(x)$, such that

$$f(x) = 0, \text{ if } x \leq 1$$ $$f(x) = \left(\frac{3}{8}\right)(x^2), \text{ if } 0 < x \leq 2$$ $$f(x) = 0, \text{ if } x > 2$$

**Remark.**

We suppose the density function $f(x)$ is defined by

$$f(x) = \begin{cases} 0, & x \leq 0, \\ \frac{3}{8}x^2, & 0 < x \leq 2, \\ 0, & x > 2, \end{cases}$$

as then $\int_{-\infty}^{+\infty} f(x)dx = 1$.

**Solution.**

The expected value equals

$$E(X) = \int_{-\infty}^{+\infty} x f(x) dx = \int_{0}^{2} x f(x) dx = \frac{3}{8} \int_{0}^{2} x^3 dx = \frac{3x^4}{32} \Big|_{0}^{2} = 1.5.$$ $$E(X^2) = \int_{-\infty}^{+\infty} x^2 f(x) dx = \frac{3}{8} \int_{0}^{2} x^4 dx = \frac{3x^5}{40} \Big|_{0}^{2} = 2.4.$$

The variance equals

$$Var(X) = E(X^2) - E^2(X) = 2.4 - 2.25 = 0.15.$$

The standard deviation equals

$$\sigma(X) = \sqrt{Var(X)} \approx 0.387.$$

**Answer:** $E(X) = 1.5$, $Var(X) = 0.15$, $\sigma(X) = \sqrt{0.15} \approx 0.387$.

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