X X X a a a a > 0 a>0 a > 0 X ∼ N ( 220 , a 2 ) X\sim N(220,a^2) X ∼ N ( 220 , a 2 )
표본 크기가 n n n
X ‾ ∼ N ( 220 , ( a n ) 2 ) \overline{X}\sim N\!\left(220,\left(\dfrac{a}{\sqrt{n}}\right)^2\right) X ∼ N ( 220 , ( n a ) 2 ) 이에요. 표준화하면
Z = X ‾ − 220 a n Z=\dfrac{\overline{X}-220}{\dfrac{a}{\sqrt{n}}} Z = n a X − 220 일 때 Z ∼ N ( 0 , 1 ) Z\sim N(0,1) Z ∼ N ( 0 , 1 )
주어진 확률을 표준정규분포로 쓰면
P ( X ‾ ≤ 215 ) = P ( Z ≤ 215 − 220 a n ) = P ( Z ≤ − 5 n a ) = P ( Z ≥ 5 n a ) = 1 2 − P ( 0 ≤ Z ≤ 5 n a ) \begin{aligned}
P(\overline{X}\leq 215)
&=P\!\left(Z\leq \dfrac{215-220}{\dfrac{a}{\sqrt{n}}}\right)
=P\!\left(Z\leq -\dfrac{5\sqrt{n}}{a}\right) \\
&=P\!\left(Z\geq \dfrac{5\sqrt{n}}{a}\right)
=\dfrac{1}{2}-P\!\left(0\leq Z\leq \dfrac{5\sqrt{n}}{a}\right)
\end{aligned} P ( X ≤ 215 ) = P Z ≤ n a 215 − 220 = P ( Z ≤ − a 5 n ) = P ( Z ≥ a 5 n ) = 2 1 − P ( 0 ≤ Z ≤ a 5 n ) 이에요. (마지막 등호는 Z Z Z y y y P ( Z ≤ − t ) = P ( Z ≥ t ) \,P(Z\leq -t)=P(Z\geq t)\, P ( Z ≤ − t ) = P ( Z ≥ t ) P ( Z ≥ t ) = 1 2 − P ( 0 ≤ Z ≤ t ) \,P(Z\geq t)=\dfrac{1}{2}-P(0\leq Z\leq t)\, P ( Z ≥ t ) = 2 1 − P ( 0 ≤ Z ≤ t )
이 값이 0.1587 0.1587 0.1587
1 2 − P ( 0 ≤ Z ≤ 5 n a ) = 0.1587 , P ( 0 ≤ Z ≤ 5 n a ) = 0.3413 \dfrac{1}{2}-P\!\left(0\leq Z\leq \dfrac{5\sqrt{n}}{a}\right)=0.1587,
\qquad
P\!\left(0\leq Z\leq \dfrac{5\sqrt{n}}{a}\right)=0.3413 2 1 − P ( 0 ≤ Z ≤ a 5 n ) = 0.1587 , P ( 0 ≤ Z ≤ a 5 n ) = 0.3413 이에요. 표준정규분포표에서 P ( 0 ≤ Z ≤ z ) = 0.3413 P(0\leq Z\leq z)=0.3413 P ( 0 ≤ Z ≤ z ) = 0.3413 z = 1.0 z=1.0 z = 1.0
5 n a = 1 , a n = 5 ⋯ ⋯ ㉠ \dfrac{5\sqrt{n}}{a}=1,
\qquad
\dfrac{a}{\sqrt{n}}=5
\quad\cdots\cdots\text{㉠} a 5 n = 1 , n a = 5 ⋯⋯ ㉠ 이에요.
다음으로 (나)에 따라 Y Y Y 1.5 a = 3 a 2 1.5a=\dfrac{3a}{2} 1.5 a = 2 3 a Y ∼ N ( 240 , ( 3 a 2 ) 2 ) Y\sim N\!\left(240,\left(\dfrac{3a}{2}\right)^2\right) Y ∼ N ( 240 , ( 2 3 a ) 2 ) 9 n 9n 9 n
Y ‾ ∼ N ( 240 , ( 3 a 2 9 n ) 2 ) = N ( 240 , ( a 2 n ) 2 ) \overline{Y}\sim N\!\left(240,\left(\dfrac{\dfrac{3a}{2}}{\sqrt{9n}}\right)^2\right)
=N\!\left(240,\left(\dfrac{a}{2\sqrt{n}}\right)^2\right) Y ∼ N 240 , 9 n 2 3 a 2 = N ( 240 , ( 2 n a ) 2 ) 이에요. ㉠을 쓰면 a 2 n = 5 2 \dfrac{a}{2\sqrt{n}}=\dfrac{5}{2} 2 n a = 2 5
표준화하면 Z = Y ‾ − 240 5 2 Z=\dfrac{\overline{Y}-240}{\dfrac{5}{2}} Z = 2 5 Y − 240 Z ∼ N ( 0 , 1 ) Z\sim N(0,1) Z ∼ N ( 0 , 1 )
P ( Y ‾ ≥ 235 ) = P ( Z ≥ 235 − 240 5 2 ) = P ( Z ≥ − 2 ) = P ( − 2 ≤ Z ≤ 0 ) + 1 2 = P ( 0 ≤ Z ≤ 2 ) + 1 2 = 0.4772 + 0.5 = 0.9772 \begin{aligned}
P(\overline{Y}\geq 235)
&=P\!\left(Z\geq \dfrac{235-240}{\dfrac{5}{2}}\right)
=P(Z\geq -2) \\
&=P(-2\leq Z\leq 0)+\dfrac{1}{2}
=P(0\leq Z\leq 2)+\dfrac{1}{2} \\
&=0.4772+0.5=0.9772
\end{aligned} P ( Y ≥ 235 ) = P Z ≥ 2 5 235 − 240 = P ( Z ≥ − 2 ) = P ( − 2 ≤ Z ≤ 0 ) + 2 1 = P ( 0 ≤ Z ≤ 2 ) + 2 1 = 0.4772 + 0.5 = 0.9772 이에요. (표에서 P ( 0 ≤ Z ≤ 2 ) = 0.4772 P(0\leq Z\leq 2)=0.4772 P ( 0 ≤ Z ≤ 2 ) = 0.4772
따라서 정답은 ⑤ 예요.
