ここで勉強すれば数学検定１級の壁は超えられるか。

確率密度Xの確率密度関数f(x)が

[math]f\left( x\right) =\begin{cases}\dfrac {1}{5}\left( 0\leqq x\leqq 5\right) \\ 0 \left( x <0,5 <x\right) \end{cases}[/math]表せるとき、Xの分散を求めよ。

[math]E\left[ X\right] =\int ^{5}_{0}x\cdot \dfrac {1}{5}dx=\left[ \dfrac {x^{2}}{10}\right] ^{5}_{0}=\dfrac {5}{2}[/math]

[math]E\left[ X^{2}\right] =\int ^{5}_{0}x^{2}\cdot \dfrac {1}{5}dx=\left[ \dfrac {x^{3}}{15}\right] ^{5}_{0}=\dfrac {25}{3}[/math]

[math]V\left[ X\right] =E\left[ X^{2}\right] -\left\{ E\left[ X\right] \right\} ^{2}=\dfrac {25}{3}-\left( \dfrac {5}{2}\right) ^{2}=\dfrac {25}{12}[/math]

[math]\dfrac {25}{12}[/math]・・・答え