2017　滋賀大学　0106解答

【４解答】(1)　$4$回の試行で$1$回も当たりが出ない確率は${\left({\displaystyle \frac{3}{4}}\right)}^4={\displaystyle \frac{81}{256}}$

　求める確率は，余事象の確率を考えて，$1-{\displaystyle \frac{81}{256}}={\displaystyle \frac{\mathbf{175}}{\mathbf{256}}}$

(2)　$m=10$より，$p={\displaystyle \frac{1}{10}}\text{．}$これと$n=100$より，$X$は正規分布$N(100\cdot {\displaystyle \frac{1}{10}},100\cdot {\displaystyle \frac{1}{10}}\cdot (1-{\displaystyle \frac{1}{10}}\left)\right)\text{，}$すなわち，$N(10,9)$に従う．このとき，$Z={\displaystyle \frac{X-10}{\sqrt{9}}}={\displaystyle \frac{X-10}{3}}$は標準正規分布$N(0,1)$に従う．

$P(X\geqq 10)=P({\displaystyle \frac{X-10}{3}}\geqq {\displaystyle \frac{10-10}{3}})$$=P(Z\geqq 0)=\mathbf{0.5}$

(3)　$X$は二項分布$B(100,{\displaystyle \frac{1}{10}})$に従うので，

$P(X\geqq 10)\fallingdotseq P(Z\geqq {\displaystyle \frac{10-0.5-100\cdot {\displaystyle \frac{1}{10}}}{\sqrt{100\cdot {\displaystyle \frac{1}{10}}(1-{\displaystyle \frac{1}{10}})}}})$

$=P(Z\geqq {\displaystyle \frac{-0.5}{3}})$$=P(Z\geqq -{\displaystyle \frac{1}{6}})$$=P(Z\leqq {\displaystyle \frac{1}{6}})$$=P(Z\leqq 0)+P(0\leqq Z\leqq {\displaystyle \frac{1}{6}})$$=0.5+0.066=\mathbf{0.566}$

解答はDYさんによる(2023/12/10)