a) We know that for "f(x)" to be a probability distribution
We integrate "f(x)" with respect to "x," set the result equal to 1 and solve for "c."
"=c\\big({1^3 \\over 3}-{(-1)^3 \\over 3}\\big)={2 \\over 3}c=1"
"c={3 \\over 2}"
b) Cumulative distribution function (cdf):
"\\displaystyle\\int_{-1}^x {3 \\over 2}x^2 dx ={3 \\over 2}\\big[{x^3 \\over 3}\\big]\\begin{array}{c}\n x \\\\\n -1\n\\end{array}={x^3 \\over 2}+{1 \\over 2}"
"F(x) = \\begin{cases}\n 0 &\\text{if } x<-1 \\\\\n {x^3 \\over 2}+{1 \\over 2} &\\text{if }-1\\leq x<1 \\\\\n 1 &\\text{if } x\\geq 1\n\\end{cases}"
c)
"=1-F({1 \\over 2})=1-({({1 \\over 2})^3 \\over 2}+{1 \\over 2})={7 \\over 16}"
d)
"={3 \\over 2}\\big[{x^4 \\over 4}\\big]\\begin{array}{c}\n 1 \\\\\n -1\n\\end{array}=0"
"Var(X)=\\sigma^2=E[(X-\\mu)^2]=\\displaystyle\\int_{-\\infin}^\\infin (x-0)^2f(x)dx="
"=\\displaystyle\\int_{-1}^{1} {3 \\over 2}x^4dx={3 \\over 2}\\big[{x^5 \\over 5}\\big]\\begin{array}{c}\n 1 \\\\\n -1\n\\end{array}={3 \\over 5}"
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