+20점 · P ( 4 < X 1 + X 2 < 8 ) \displaystyle \mathrm{P}(4<X_1+X_2<8) P ( 4 < X 1 + X 2 < 8 ) (1-1) P ( 2 < X ‾ < 4 ) = P ( 4 < X 1 + X 2 < 8 ) \displaystyle \mathrm{P}(2<\overline X<4)=\mathrm{P}(4<X_1+X_2<8) P ( 2 < X < 4 ) = P ( 4 < X 1 + X 2 < 8 )
+20점 · P ( X 1 = 2 , X 2 = 3 ) + P ( X 1 = 3 , X 2 = 2 ) + P ( X 1 = 3 , X 2 = 3 ) \displaystyle \mathrm{P}(X_1=2,X_2=3)+P(X_1=3,X_2=2)+\mathrm{P}(X_1=3,X_2=3) P ( X 1 = 2 , X 2 = 3 ) + P ( X 1 = 3 , X 2 = 2 ) + P ( X 1 = 3 , X 2 = 3 ) = P ( X 1 = 2 , X 2 = 3 ) + P ( X 1 = 3 , X 2 = 2 ) + P ( X 1 = 3 , X 2 = 3 ) \displaystyle =\mathrm{P}(X_1=2,X_2=3)+\mathrm{P}(X_1=3,X_2=2)+\mathrm{P}(X_1=3,X_2=3) = P ( X 1 = 2 , X 2 = 3 ) + P ( X 1 = 3 , X 2 = 2 ) + P ( X 1 = 3 , X 2 = 3 )
+30점 · 답: 1 4 a ( 4 − 3 a ) \displaystyle \frac14a(4-3a) 4 1 a ( 4 − 3 a ) = ( 1 − a ) × a 2 + a 2 × ( 1 − a ) + a 2 × a 2 = 1 4 a ( 4 − 3 a ) \displaystyle =(1-a)\times\frac a2+\frac a2\times(1-a)+\frac a2\times\frac a2=\frac14a(4-3a) = ( 1 − a ) × 2 a + 2 a × ( 1 − a ) + 2 a × 2 a = 4 1 a ( 4 − 3 a )
+30점 (1-2) 방의 개수에 대한 기댓값은 1 × a 2 + 2 × ( 1 − a ) + 3 × a 2 = 2 \displaystyle 1\times\frac a2+2\times(1-a)+3\times\frac a2=2 1 × 2 a + 2 × ( 1 − a ) + 3 × 2 a = 2 따라서 E ( X k ) = 2 , k = 1 , 2 , … , 100 \displaystyle \mathrm{E}(X_k)=2,k=1,2,\ldots,100 E ( X k ) = 2 , k = 1 , 2 , … , 100
+30점 · E ( T ) = 1 100 ∑ k = 1 100 k E ( X k ) \displaystyle \mathrm{E}(T)=\frac1{100}\sum_{k=1}^{100}k\mathrm{E}(X_k) E ( T ) = 100 1 k = 1 ∑ 100 k E ( X k ) +20점 · 답: 101
그러므로 E ( T ) = 1 100 ∑ k = 1 100 k E ( X k ) = 2 100 ∑ k = 1 100 k = 2 100 × 100 × 101 2 = 101 \displaystyle \mathrm{E}(T)=\frac1{100}\sum_{k=1}^{100}k\mathrm{E}(X_k)=\frac2{100}\sum_{k=1}^{100}k=\frac2{100}\times\frac{100\times101}2=101 E ( T ) = 100 1 k = 1 ∑ 100 k E ( X k ) = 100 2 k = 1 ∑ 100 k = 100 2 × 2 100 × 101 = 101
+20점 · 방의 개수가 1개 또는 3개인 아파트의 수를 Y \displaystyle Y Y Y ∼ B ( 100 , p ) \displaystyle Y\sim\mathrm{B}(100,p) Y ∼ B ( 100 , p ) Correction. 채점기준의 Y ∼ B ( 100 , p ) \displaystyle Y\sim\mathrm{B}(100,p) Y ∼ B ( 100 , p ) p \displaystyle p p a \displaystyle a a 100 a \displaystyle 100a 100 a
(1-3) 방의 개수가 1개 또는 3개인 아파트의 수를 Y \displaystyle Y Y Y \displaystyle Y Y B ( 100 , a ) \displaystyle \mathrm{B}(100,a) B ( 100 , a )
+20점 · Y − 100 a 100 a ( 1 − a ) \displaystyle \frac{Y-100a}{\sqrt{100a(1-a)}} 100 a ( 1 − a ) Y − 100 a a ^ − a a ( 1 − a ) 100 ∼ N ( 0 , 1 ) \displaystyle \frac{\hat a-a}{\sqrt{\frac{a(1-a)}{100}}}\sim\mathrm{N}(0,1) 100 a ( 1 − a ) a ^ − a ∼ N ( 0 , 1 ) 따라서 a ^ = Y 100 \displaystyle \hat a=\frac Y{100} a ^ = 100 Y Y − 100 × a 100 × a ( 1 − a ) = Y 100 − a a ( 1 − a ) 100 \displaystyle \frac{Y-100\times a}{\sqrt{100\times a(1-a)}}=\frac{\frac Y{100}-a}{\sqrt{\frac{a(1-a)}{100}}} 100 × a ( 1 − a ) Y − 100 × a = 100 a ( 1 − a ) 100 Y − a N ( 0 , 1 ) \displaystyle \mathrm{N}(0,1) N ( 0 , 1 )
+20점 그러므로 a \displaystyle a a ( 1 5 − 1.96 × 1 / 5 × 4 / 5 100 , 1 5 + 1.96 × 1 / 5 × 4 / 5 100 ) \displaystyle \left(\frac15-1.96\times\sqrt{\frac{1/5\times4/5}{100}},\frac15+1.96\times\sqrt{\frac{1/5\times4/5}{100}}\right) ( 5 1 − 1.96 × 100 1/5 × 4/5 , 5 1 + 1.96 × 100 1/5 × 4/5 )
+20점 · 답: ( 76 625 , 174 625 ) \displaystyle \left(\frac{76}{625},\frac{174}{625}\right) ( 625 76 , 625 174 ) ( 0.1216 , 0.2784 ) \displaystyle (0.1216,0.2784) ( 0.1216 , 0.2784 ) = ( 125 − 49 625 , 125 + 49 625 ) = ( 76 625 , 174 625 ) = ( 0.1216 , 0.2784 ) \displaystyle =\left(\frac{125-49}{625},\frac{125+49}{625}\right)=\left(\frac{76}{625},\frac{174}{625}\right)=(0.1216,0.2784) = ( 625 125 − 49 , 625 125 + 49 ) = ( 625 76 , 625 174 ) = ( 0.1216 , 0.2784 )
수리논술, 이게 다다.