2019　横浜市立大学　0108解答MathJax

【５解答】(1)　${\displaystyle {\int }_{-1}^3}f(x)\mathit{dx}=1\text{，}$かつ${\displaystyle {\int }_{-1}^3}xf(x)\mathit{dx}={\displaystyle \frac{2}{3}}$より，

${\displaystyle {\int }_{-1}^3}f(x)\mathit{dx}={\displaystyle {\int }_{-1}^0}a(x+1)\mathit{dx}+{\displaystyle {\int }_0^3}(bx+a)\mathit{dx}$$=a{[{\displaystyle \frac{x^2}{2}}+x]}_{-1}^0+{[{\displaystyle \frac{b}{2}}{x}^2+ax]}_0^3$$=-a({\displaystyle \frac{1}{2}}-1)+({\displaystyle \frac{9}{2}}b+3a)$$={\displaystyle \frac{7}{2}}a+{\displaystyle \frac{9}{2}}b=1$$\cdots \text{①}$

${\displaystyle {\int }_{-1}^3}xf(x)\mathit{dx}$$={\displaystyle {\int }_{-1}^0}a(x^2+x)\mathit{dx}+{\displaystyle {\int }_0^3}(b{x}^2+ax)\mathit{dx}$$=a{[{\displaystyle \frac{x^3}{3}}+{\displaystyle \frac{x^2}{2}}]}_{-1}^0+{[{\displaystyle \frac{b}{3}}{x}^3+{\displaystyle \frac{a}{2}}{x}^2]}_0^3$$=-a(-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{2}})+(9b+{\displaystyle \frac{9}{2}}a)$$={\displaystyle \frac{26}{6}}a+9b={\displaystyle \frac{2}{3}}$$\cdots \text{②}$

$\text{①}\text{，}$$\text{②}$を連立して，$\mathit{a}\mathbf{=}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\text{，}$$\mathit{b}\mathbf{=}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{\mathbf{6}}}$

(2)

$E(X^2)={\displaystyle {\int }_{-1}^3}{x}^{2}f(x)\mathit{dx}$$={\displaystyle {\int }_{-1}^0}\{{\displaystyle \frac{1}{2}}(x^3+x^2)\}\mathit{dx}+{\displaystyle {\int }_0^3}(-{\displaystyle \frac{1}{6}}{x}^3+{\displaystyle \frac{1}{2}}{x}^2)\mathit{dx}$$={\displaystyle {[{\displaystyle \frac{x^4}{8}}+{\displaystyle \frac{x^3}{6}}]}_{-1}^0}+{[-{\displaystyle \frac{1}{24}}{x}^4+{\displaystyle \frac{1}{6}}{x}^3]}_0^3$$=-({\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{6}})+(-{\displaystyle \frac{27}{8}}+{\displaystyle \frac{9}{2}})={\displaystyle \frac{7}{6}}$

$V(X)=E(X^2)-{\{E(X)\}}^2$$={\displaystyle \frac{7}{6}}-{\left({\displaystyle \frac{2}{3}}\right)}^2={\displaystyle \frac{\mathbf{13}}{\mathbf{18}}}$

※$E(X)={\displaystyle \frac{2}{3}}$より，$V(X)={\displaystyle {\int }_{-1}^3}{(x-{\displaystyle \frac{2}{3}})}^{2}f(x)\mathit{dx}$から計算してもよい．

(3)　(ア)

$E(Y)=E(18X+5)$$=18E(X)+5$$=18\cdot {\displaystyle \frac{2}{3}}+5=17$

$V(Y)=V(18X+5)$$={18}^2\cdot V(X)$$={18}^2\cdot {\displaystyle \frac{13}{18}}=234$

これらを母平均，母分散とする母集団から大きさ$117$の標本を無作為抽出するとき，その標本平均$\bar{Y}$について，

$E(\bar{Y})=E(Y)=17\text{，}$$V(\bar{Y})={\displaystyle \frac{V(Y)}{117}}$$={\displaystyle \frac{234}{117}}=2$

(イ)　標本平均$\bar{Y}$は近似的に正規分布$N(17,2)$に従う．$Z={\displaystyle \frac{\bar{Y}-17}{\sqrt{2}}}$とおくと，確率変数$Z$は標準正規分布$N(0,1)$に従う．

$16\leqq \bar{Y}\leqq 18$$⟺{\displaystyle \frac{16-17}{\sqrt{2}}}\leqq {\displaystyle \frac{\bar{Y}-17}{\sqrt{2}}}\leqq {\displaystyle \frac{18-17}{2}}$$⟺-{\displaystyle \frac{\sqrt{2}}{2}}\leqq Z\leqq {\displaystyle \frac{\sqrt{2}}{2}}$

より，求める確率$P(16\leqq \bar{Y}\leqq 18)$は，

$P(16\leqq \bar{Y}\leqq 18)$$=P(-{\displaystyle \frac{\sqrt{2}}{2}}\leqq Z\leqq {\displaystyle \frac{\sqrt{2}}{2}})$$=2P(0\leqq Z\leqq {\displaystyle \frac{\sqrt{2}}{2}})$$\fallingdotseq 2P(0\leqq Z\leqq 0.71)$$=2\cdot 0.2611$$\text{【正規分布表より】}$

$\fallingdotseq \mathbf{0.52}$

解答はDYさんによる(2023/11/12)