For the random variable X with the probability function below, answer the following questions.
f(x)= { 3x^2 , 0<x<1
{ 0, otherwise
a. Find the cdf of X.
b. Find the probability that X is less than 0.5.
c. Find E(X).
The probability function "f(x)=\\begin{cases}\n3x^2,\\ \\ \\ 0<x<1\n\\\\0, \\ \\ \\text{otherwise}\n\\end{cases}"
a. The CDF of X: "F(x)=P(X\\leq x)=\\int\\limits _{-\\infin}^x f(t)dt"
If "x\\leq 0:" "F(x)= \\int\\limits _{-\\infin}^x 0dt =0"
If "0<x<1" : "F(x)= \\int\\limits _{0}^x 3t^2dt =t^3\\big|^x_0=x^3"
If "1\\leq x" : "F(x)= \\int\\limits _{0}^1 3t^2dt =t^3\\big|^1_0=1"
"F(x)=\\begin{cases}\n0,\\ \\ x\\leq 0\n\\\\\nx^3,\\ \\ 0<x<1\n\\\\\n1,\\ \\ 1\\leq x\n\\end{cases}"
b. "P(X<0.5)= F(0.5)=(0.5)^3=0.125"
c. "E(x)=\\int\\limits _{-\\infin}^{+\\infin}xf(x)dx=\\int\\limits _0^1 3x^3dx=\\frac{3}{4}x^4\\big|^1_0=\\frac{3}{4}"
Answer:
a. "F(x)=\\begin{cases}\n0,\\ \\ x\\leq 0\n\\\\\nx^3,\\ \\ 0<x<1\n\\\\\n1,\\ \\ 1\\leq x\n\\end{cases}"
b. "P(X<0.5)=0.125"
c. "E(x)=\\frac{3}{4}"
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