Answer to Question #159406 in Statistics and Probability for TasRak

Question #159406

The hospitalization period, in days, for patients following treatment for a certain

type of virus X, where X has the density function f(x) = 4/{π(x2 +1)}

, 0<x <58. Find

the expected value of X. that a person is hospitalized following treatment for this

disorder.


1
Expert's answer
2021-02-01T07:10:29-0500
f(x)dx=0584π(x2+1)dx\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{58}\dfrac{4}{\pi(x^2+1)}dx

=4π[arctan(x)]580=4arctan(58)π1=\dfrac{4}{\pi}[\arctan(x)]\begin{matrix} 58 \\ 0 \end{matrix}=\dfrac{4\arctan(58)}{\pi}\not=1

Therefore


f(x)={1arctan(58)(x2+1)if 0<x<580otherwise f(x) = \begin{cases} \dfrac{1}{\arctan(58)\cdot(x^2+1)} &\text{if } 0<x<58 \\ 0 &\text{otherwise } \end{cases}

E(X)=xf(x)dxE(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx


=058x(1arctan(58)(x2+1))dx=\displaystyle\int_{0}^{58}x(\dfrac{1}{\arctan(58)(x^2+1)})dx

=12arctan(58)[ln(x2+1)]580==\dfrac{1}{2\arctan(58)}[\ln(x^2+1)]\begin{matrix} 58 \\ 0 \end{matrix}=

=ln(3365)2arctan(58)2.613739=\dfrac{\ln(3365)}{2\arctan(58)}\approx2.613739



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