Respuesta :
Answer:
The value of the constant is [tex]c=\frac{3}{4}[/tex].
The cumulative distribution function of X is [tex]F(x)=\left \{ {{\frac{1}{4}\left(-x^3+3x+2\right)\quad{-1\:<\:x\:<\:1}} \atop {0}\quad {\mathrm{elsewhere}}} \right.[/tex].
The expected value E(X) is 0.
Step-by-step explanation:
1. The function f(x) is a probability density function (pdf) for the continuous random variable X, defined over the set of real numbers, if
- [tex]f(x)\geq 0,[/tex] for all x∈R.
- [tex]\int\limits^{\infty}_{-\infty} {f(x)} \, dx=1[/tex]
- [tex]P(a<X<b)=\int\limits^b_a {f(x)} \, dx[/tex]
We have the following probability density function
[tex]f(x)=\left \{ {{c(1-x^2)\quad{-1\:<\:x\:<\:1}} \atop {0\quad {\mathrm{Otherwise}}}} \right.[/tex]
We know that for f(x) to be a probability distribution [tex]\int\limits^{\infty}_{-\infty} {f(x)} \, dx=1[/tex]. We integrate f(x) with respect to x, set the result equal to 1 and solve for c.
[tex]\int\limits^{1}_{-1} {c(1-x^2)} \, dx=1 \\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\c\cdot \int _{-1}^11-x^2dx=1\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\c\left(\int _{-1}^11dx-\int _{-1}^1x^2dx\right)=1\\\\c\left(2-\frac{2}{3}\right)=1\\\\\frac{4}{3}c=1\\\\c=\frac{3}{4}[/tex]
2. The cumulative distribution function F(x) of a continuous random variable X with density function f(x) is
[tex]F(x)=P(X\leq x)=\int\limits^{x}_{-\infty} {f(t)} \, dt, \mathrm{\:for} -\infty<x<\infty[/tex]
To find the cumulative distribution function F(x) integrate f(x) from the lower bound of the domain on which f(x) ≠0 to x.
[tex]F(x)=\int\limits^{x}_{-\infty} {f(t)} \, dt=\int\limits^x_{-1} {c(1-x^2)} \, dx =\int\limits^x_{-1} {\frac{3}{4} (1-x^2)} \, dx[/tex]
[tex]\int\limits^x_{-1} {\frac{3}{4} (1-x^2)} \, dx \\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\\frac{3}{4}\cdot \int _{-1}^x1-x^2dx\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\frac{3}{4}\left(\int _{-1}^x1dx-\int _{-1}^xx^2dx\right)\\\\\frac{1}{4}\left(-x^3+3x+2\right)[/tex]
[tex]F(x)=\left \{ {{\frac{1}{4}\left(-x^3+3x+2\right)\quad{-1\:<\:x\:<\:1}} \atop {0}\quad {\mathrm{elsewhere}}} \right.[/tex]
3. Let X be a random variable with probability distribution f (x). The mean, or expected value, of X is
[tex]\mu=E(X)=\int\limits^{\infty}_{-\infty} {xf(x)} \, dx[/tex]
[tex]E(X)=\int\limits^1_{-1} {x(c(1-x^2))} \, dx =\int\limits^1_{-1} {x(\frac{3}{4}(1-x^2))} \, dx[/tex]
[tex]E(X)=\int _{-1}^1x\frac{3}{4}\left(1-x^2\right)dx=0[/tex]
