Question #350150

Suppose f(x) = 1. 5 𝑥 for -1 < x <1. Determine the mean and variance of X


Expert's answer

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∫−∞∞f(x)dx=∫−111.5x2dx\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{-1}^{1}1.5x^2dx

=[0.5x3]1−1=1,True=[0.5x^3]\begin{matrix} 1 \\ -1 \end{matrix}=1, True


μ=E(X)=∫−∞∞xf(x)dx=∫−111.5x3dx\mu=E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{-1}^{1}1.5x^3dx

=[1.54x4]1−1=0=[\dfrac{1.5}{4}x^4]\begin{matrix} 1 \\ -1 \end{matrix}=0

E(X2)=∫−∞∞x2f(x)dx=∫−111.5x4dxE(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{-1}^{1}1.5x^4dx

=[0.3x5]1−1=0.6=[0.3x^5]\begin{matrix} 1 \\ -1 \end{matrix}=0.6

Var(X)=E(X2)−(E(X))2Var(X)=E(X^2)-(E(X))^2

=0.6−(0)2=0.6=0.6-(0)^2=0.6


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