【高校数学B】2-3-1 母集団と標本｜問題集

\(1\)回投げた時の出た目\(X\)の確率分布は

確率分布

\(X\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	計
\(P\)	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{1}{6}\)	\(1\)

\(X\)の平均\(E(X)\)は
\(E(X)\)
\(\displaystyle =(1+2+3+4+5+6)\times\frac{1}{6}\)
\(\displaystyle =\frac{7}{2}\)
\(X\)の標準偏差\(\sigma(X)\)は
\(\sigma(X)\)
\(=\sqrt{E(X^2)-{E(X)}^2}\)
\(\displaystyle =\sqrt{(1^2+2^2+3^2+4^2+5^2+6^2)\times\frac{1}{6}-\left(\frac{7}{2}\right)^2}\)
\(\displaystyle =\sqrt{\frac{91}{6}-\frac{49}{4}}\)
\(\displaystyle =\sqrt{\frac{35}{12}}\)
\(\displaystyle =\frac{\sqrt{105}}{6}\)
よって、大きさ\(3\)の標本の
標本平均\(E(\bar{X})\)は
\(E(\bar{X})\)=(母平均)=\(\displaystyle \frac{7}{2}\)
標準偏差\(\sigma(\bar{X})\)は
\(\displaystyle \sigma(\bar{X})=\frac{\sigma(X)}{\sqrt{3}}=\frac{\sqrt{35}}{6}\)

母集団分布が正規分布\(N(270,30^2)\)であるから、標本平均\(\bar{X}\)は正規分布\(\displaystyle N\left(270,\frac{30^2}{36}\right)\)に従う。
標準化した確率変数\(\displaystyle Z=\frac{\bar{X}-270}{\frac{30}{\sqrt{36}}}=\frac{\bar{X}-270}{5}\)は、標準正規分布\(N(0,1)\)に従う。
\(X=264\)のとき、\(\displaystyle Z=\frac{264-270}{5}=-1.2\)
よって、
\(P(\bar{X}\leqq 264)\)
\(=P(Z\leqq -1.2)\)
\(=P(Z\geqq 0)-P(0\leqq Z\leqq 1.2)\)
\(=0.5-0.3849\)
\(=0.1151\)