Задача 40807 Теория вероятности...

leronmarkevich

2019-10-21 16:23:28

Теория вероятности

sova

2019-10-21 16:38:57

★

[m]M(X)=\int_{-\infty }^{+\infty }xf(x)dx=\int_{0}^{4}\frac{x}{4\sqrt{x}}dx=\frac{1}{4}\int_{0}^{4}\sqrt{x}dx=[/m]

[m]=\frac{1}{4}\cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}|^{4}_{0}=\frac{1}{12}\cdot 4^{\frac{3}{2}}=\frac{2}{12}\cdot \sqrt{4^{3}}=\frac{16}{12}=\frac{4}{3}[/m]

[m]D(X)=\int_{-\infty }^{+\infty }(x-M(X))^{2}\cdot f(x)dx=\int_{0}^{4}\frac{( x-\frac{4}{3})^{2}}{4\sqrt{x}}dx=[/m]

[m]=\frac{1}{4}\int_{0}^{4}\frac{x^{2}-2\cdot \frac{4}{3}x+\frac{16}{9}}{x^{\frac{1}{2}}}dx=\frac{1}{4}\int_{0}^{4}x^{\frac{3}{2}}dx-\frac{2}{3}\int_{0}^{4}x^{\frac{1}{2}}dx+\frac{4}{9}\int_{0}^{4}x^{-\frac{1}{2}}dx=[/m]

[m]=\frac{1}{4}\cdot \frac{x^{\frac{5}{2}}}{\frac{5}{2}}|_{0}^{4}-\frac{2}{3}\cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}|_{0}^{4}+\frac{4}{9}\cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}}|_{0}^{4}=\frac{1}{10}\cdot4^{\frac{5}{2}}-\frac{4}{9} \cdot4^{\frac{3}{2}}+\frac{8}{9}\cdot4^{\frac{1}{2}}=[/m]

[m]=3,2-\frac{32}{9}+\frac{16}{9}=\frac{64}{45}[/m]

проще считать так:

[m]D(X)=\int_{0 }^{4 }x^{2}\cdot f(x)dx-(\frac{4}{3})^2=\frac{1}{4}\int_{0}^{4}\frac{ x^{2}}{\sqrt{x}}dx-\frac{16}{9}=\frac{1}{4}\frac{ x^{\frac{5}{2}}}{\frac{5}{2}}|_{0}^{4}-\frac{16}{9}=[/m]

[m]=\frac{1}{10}\cdot 4^{\frac{5}{2}}-\frac{16}{9}=3,2-\frac{16}{9}=\frac{64}{45}[/m]