\(\displaystyle{ N(-1,5)}\)
Wyznaczyć \(\displaystyle{ x}\):
\(\displaystyle{ 1.}\) \(\displaystyle{ P(|X+1|>x)=0,4}\)
\(\displaystyle{ 2.}\) \(\displaystyle{ P(X>x)=0,25}\)
\(\displaystyle{ 1.}\)
\(\displaystyle{ N(-1,5)}\)
\(\displaystyle{ P(X>x)=0,25}\)
\(\displaystyle{ P(\frac{X+1}{5} > \frac{x+1}{5}) = 0,25}\)
\(\displaystyle{ 1 - P(\frac{X+1}{5} \le \frac{x+1}{5}) = 0,25}\)
\(\displaystyle{ 1 - \phi (\frac{x+1}{5}) = \phi (-0,68)}\)
\(\displaystyle{ 1- \frac{x+1}{5} = -0,68}\)
\(\displaystyle{ \frac{x+1}{5} = 1,68}\)
\(\displaystyle{ x=7,4}\)
\(\displaystyle{ 2.}\)
\(\displaystyle{ P(|X+1|>x)=0,4}\)
\(\displaystyle{ P(X+1>x)=0,4 \qquad v \qquad P(X+1<-x)=0,4}\)
\(\displaystyle{ 1-P(\frac{X+1}{5} < \frac{x}{5})=0,4 \qquad v \qquad P(\frac{X+1}{5}) < \frac{-x}{5})=0,4}\)
\(\displaystyle{ 1- \phi(\frac{x}{5}) = \phi(-0,26) \qquad v \qquad \phi(\frac{-x}{5})=\phi(-0,26))}\)
\(\displaystyle{ 1 - \frac{x}{5}= -0,26 \qquad v \qquad \frac{-x}{5}= -0,26}\)
\(\displaystyle{ \frac{x}{5}=1,26 \qquad v \qquad x=1,3}\)
\(\displaystyle{ x=6,3 \qquad v \qquad x=1,3}\)
Czy tak jest dobrze?
Rozkład normalny
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sushi
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Rozkład normalny
co do pierwszego
\(\displaystyle{ 1- P ( \frac{X+1}{5} < \frac{x+1}{5} ) = 0.25}\) to po rpsotu robi sie porzadki i mamy
\(\displaystyle{ 0.75= P ( \frac{X+1}{5} < \frac{x+1}{5} )}\)
\(\displaystyle{ 0.75= \Phi ( \frac{x+1}{5} )}\) i teraz w ztablic odczytujemy
\(\displaystyle{ \Phi(0.67449)= 0.75}\)-- 22 czerwca 2010, 16:08 --do 2
\(\displaystyle{ P(|X+1|>x)= 0.4}\)
\(\displaystyle{ 1- P(|X+1| \le x)= 0.4}\)
\(\displaystyle{ P(|X+1| \le x)= 0.6}\)
\(\displaystyle{ P(-x \le X+1 \le x)= 0.6}\)
\(\displaystyle{ P(-x-1 \le X \le x -1)= 0.6}\)
\(\displaystyle{ P( \frac{ -x -1+1}{5} \le Y \le \frac{ x -1 +1}{5})= 0.6}\)
\(\displaystyle{ \Phi( \frac{ x -1 +1}{5})- \Phi ( \frac{ -x -1 +1}{5})= 0.6}\)
\(\displaystyle{ \Phi( \frac{ x }{5})- \Phi ( \frac{ -x }{5})= 0.6}\)
\(\displaystyle{ \Phi(-x)= 1- \Phi(x)}\)
\(\displaystyle{ \Phi( \frac{ x }{5})- (1- \Phi ( \frac{ x }{5}) )= 0.6}\)
\(\displaystyle{ 2 \Phi( \frac{ x }{5})- 1 = 0.6}\)
\(\displaystyle{ 2 \Phi( \frac{ x }{5}) = 1.6}\)
\(\displaystyle{ \Phi( \frac{ x }{5}) = 0.8}\)
potem z tablic odczytujemy ile musi byc "y" dla \(\displaystyle{ \Phi(y)= 0.8}\)
\(\displaystyle{ 1- P ( \frac{X+1}{5} < \frac{x+1}{5} ) = 0.25}\) to po rpsotu robi sie porzadki i mamy
\(\displaystyle{ 0.75= P ( \frac{X+1}{5} < \frac{x+1}{5} )}\)
\(\displaystyle{ 0.75= \Phi ( \frac{x+1}{5} )}\) i teraz w ztablic odczytujemy
\(\displaystyle{ \Phi(0.67449)= 0.75}\)-- 22 czerwca 2010, 16:08 --do 2
\(\displaystyle{ P(|X+1|>x)= 0.4}\)
\(\displaystyle{ 1- P(|X+1| \le x)= 0.4}\)
\(\displaystyle{ P(|X+1| \le x)= 0.6}\)
\(\displaystyle{ P(-x \le X+1 \le x)= 0.6}\)
\(\displaystyle{ P(-x-1 \le X \le x -1)= 0.6}\)
\(\displaystyle{ P( \frac{ -x -1+1}{5} \le Y \le \frac{ x -1 +1}{5})= 0.6}\)
\(\displaystyle{ \Phi( \frac{ x -1 +1}{5})- \Phi ( \frac{ -x -1 +1}{5})= 0.6}\)
\(\displaystyle{ \Phi( \frac{ x }{5})- \Phi ( \frac{ -x }{5})= 0.6}\)
\(\displaystyle{ \Phi(-x)= 1- \Phi(x)}\)
\(\displaystyle{ \Phi( \frac{ x }{5})- (1- \Phi ( \frac{ x }{5}) )= 0.6}\)
\(\displaystyle{ 2 \Phi( \frac{ x }{5})- 1 = 0.6}\)
\(\displaystyle{ 2 \Phi( \frac{ x }{5}) = 1.6}\)
\(\displaystyle{ \Phi( \frac{ x }{5}) = 0.8}\)
potem z tablic odczytujemy ile musi byc "y" dla \(\displaystyle{ \Phi(y)= 0.8}\)