Aparat wykonuje w takich samych warunkach serie niezależnych fotograficznych pewnego obiektu. Wiadomo , ze tylko 10% zdjęć spełnia wymagania techniczne. Stosując centralne twierdzenie graniczne oszacować, ile zdjęć należy wykonać, aby z prawdopodobieństwem 0,9 co najmniej 10 zdjęć spełniało wymagania techniczne.
z góry dziękuje za pomoc .
centralne twierdzenie graniczne
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bstq
- Użytkownik
- Posty: 319
- Rejestracja: 7 lut 2008, o 12:45
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- Pomógł: 67 razy
centralne twierdzenie graniczne
\(\displaystyle{ P\left(X_{k}=1\right)=10\%=\frac{1}{10}}\)
\(\displaystyle{ P\left(\sum_{k=1}^{n}X_{k}\ge10\right)=0.9}\) - z tego trzeba wyliczyć \(\displaystyle{ n}\)
\(\displaystyle{ P\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\ge\frac{10}{n}\right)=0.9}\)
\(\displaystyle{ P\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)\ge\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)\right)=0.9}\)
\(\displaystyle{ P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\ge\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=0.9}\)
\(\displaystyle{ 1-P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=0.9}\)
\(\displaystyle{ P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=1-0.9=0.1}\)
\(\displaystyle{ \frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\rightarrow Z\sim N\left(0,1\right)\Rightarrow}\)
\(\displaystyle{ \Rightarrow P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)\to\Phi\left(\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)}\)
Czyli patrzymy kiedy
\(\displaystyle{ \Phi\left(\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=0.1}\)
\(\displaystyle{ \frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}=\Phi{}^{-1}\left(0.1\right)}\)
\(\displaystyle{ \frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)=\Phi{}^{-1}\left(0.1\right)\cdot\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\)
\(\displaystyle{ \mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)=\frac{1}{n}\cdot n\cdot\mathbb{E}X_{1}=1\cdot\frac{1}{10}+0\cdot\frac{9}{10}=\frac{1}{10}}\)
\(\displaystyle{ Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)=\frac{1}{n^{2}}Var\left(\sum_{k=1}^{n}X_{k}\right)=\frac{1}{n^{2}}\sum_{k=1}^{n}Var\left(X_{k}\right)=\frac{1}{n^{2}}\cdot n\cdot Var\left(X_{k}\right)=\frac{1}{n}\cdot\left[\mathbb{E}X_{1}^{2}-\left(\mathbb{E}X_{1}\right)^{2}\right]=\frac{1}{n}\cdot\left[1^{2}\cdot\frac{1}{10}+0^{2}\cdot\frac{1}{10}-\frac{1}{100}\right]=\frac{1}{n}\cdot\left[\frac{1}{10}-\frac{1}{100}\right]=\frac{1}{n}\cdot\frac{90}{100}}\)
\(\displaystyle{ \Phi{}^{-1}\left(0.1\right)=z_{0.1}=u_{0.1}=}\)kwantyl rozkladu normalnego rzedu 0.1}=argument dla ktorego dystrybuanta przyjmuje wartosc 0.1\(\displaystyle{ =-1.285}\)
Z tego, że \(\displaystyle{ 0.1<0.5}\) wynika, że \(\displaystyle{ -x=\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<0}\) , zatem możemy zastosować wzór: \(\displaystyle{ \Phi\left(-x\right)=1-\Phi(x)=0.1\Rightarrow\Phi(x)=0.9\Leftrightarrow x\approx1,285}\), czyli \(\displaystyle{ -x=-1.285}\)
\(\displaystyle{ \frac{10}{n}-\frac{1}{10}=-1.285\cdot\sqrt{\frac{1}{n}\cdot\frac{90}{100}}}\)
\(\displaystyle{ \frac{\sqrt{n}10}{n}-\frac{\sqrt{n}}{10}=-1.285\cdot\sqrt{\frac{90}{100}}\approx-1,2191}\)
\(\displaystyle{ \frac{10}{\sqrt{n}}-\frac{\sqrt{n}}{10}=-1.285\cdot\sqrt{\frac{90}{100}}\approx-1,2191}\)
\(\displaystyle{ \frac{10}{\sqrt{n}}=\frac{\sqrt{n}}{10}-1,2191=\frac{\sqrt{n}-12,191}{10}}\)
\(\displaystyle{ \sqrt{n}\left(\sqrt{n}-12,191\right)=100}\)
\(\displaystyle{ x^{2}-12,191x-100=0}\)
\(\displaystyle{ \Delta=12,191^{2}-4\cdot(-100)=548,62}\)
\(\displaystyle{ \sqrt{\Delta}=23,42}\)
\(\displaystyle{ x_{1}=\frac{12,191+23,42}{2}=17,81}\)
\(\displaystyle{ x_{2}=\frac{12,191-23,42}{2}<0}\)
\(\displaystyle{ x_{1}=\sqrt{n}=17,81\Rightarrow n=\left\lceil 317,08...\right\rceil =318}\)
\(\displaystyle{ P\left(\sum_{k=1}^{n}X_{k}\ge10\right)=0.9}\) - z tego trzeba wyliczyć \(\displaystyle{ n}\)
\(\displaystyle{ P\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\ge\frac{10}{n}\right)=0.9}\)
\(\displaystyle{ P\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)\ge\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)\right)=0.9}\)
\(\displaystyle{ P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\ge\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=0.9}\)
\(\displaystyle{ 1-P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=0.9}\)
\(\displaystyle{ P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=1-0.9=0.1}\)
\(\displaystyle{ \frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\rightarrow Z\sim N\left(0,1\right)\Rightarrow}\)
\(\displaystyle{ \Rightarrow P\left(\frac{\frac{1}{n}\sum_{k=1}^{n}X_{k}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)\to\Phi\left(\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)}\)
Czyli patrzymy kiedy
\(\displaystyle{ \Phi\left(\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\right)=0.1}\)
\(\displaystyle{ \frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}=\Phi{}^{-1}\left(0.1\right)}\)
\(\displaystyle{ \frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)=\Phi{}^{-1}\left(0.1\right)\cdot\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}\)
\(\displaystyle{ \mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)=\frac{1}{n}\cdot n\cdot\mathbb{E}X_{1}=1\cdot\frac{1}{10}+0\cdot\frac{9}{10}=\frac{1}{10}}\)
\(\displaystyle{ Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)=\frac{1}{n^{2}}Var\left(\sum_{k=1}^{n}X_{k}\right)=\frac{1}{n^{2}}\sum_{k=1}^{n}Var\left(X_{k}\right)=\frac{1}{n^{2}}\cdot n\cdot Var\left(X_{k}\right)=\frac{1}{n}\cdot\left[\mathbb{E}X_{1}^{2}-\left(\mathbb{E}X_{1}\right)^{2}\right]=\frac{1}{n}\cdot\left[1^{2}\cdot\frac{1}{10}+0^{2}\cdot\frac{1}{10}-\frac{1}{100}\right]=\frac{1}{n}\cdot\left[\frac{1}{10}-\frac{1}{100}\right]=\frac{1}{n}\cdot\frac{90}{100}}\)
\(\displaystyle{ \Phi{}^{-1}\left(0.1\right)=z_{0.1}=u_{0.1}=}\)kwantyl rozkladu normalnego rzedu 0.1}=argument dla ktorego dystrybuanta przyjmuje wartosc 0.1\(\displaystyle{ =-1.285}\)
Z tego, że \(\displaystyle{ 0.1<0.5}\) wynika, że \(\displaystyle{ -x=\frac{\frac{10}{n}-\mathbb{E}\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}{\sqrt{Var\left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)}}<0}\) , zatem możemy zastosować wzór: \(\displaystyle{ \Phi\left(-x\right)=1-\Phi(x)=0.1\Rightarrow\Phi(x)=0.9\Leftrightarrow x\approx1,285}\), czyli \(\displaystyle{ -x=-1.285}\)
\(\displaystyle{ \frac{10}{n}-\frac{1}{10}=-1.285\cdot\sqrt{\frac{1}{n}\cdot\frac{90}{100}}}\)
\(\displaystyle{ \frac{\sqrt{n}10}{n}-\frac{\sqrt{n}}{10}=-1.285\cdot\sqrt{\frac{90}{100}}\approx-1,2191}\)
\(\displaystyle{ \frac{10}{\sqrt{n}}-\frac{\sqrt{n}}{10}=-1.285\cdot\sqrt{\frac{90}{100}}\approx-1,2191}\)
\(\displaystyle{ \frac{10}{\sqrt{n}}=\frac{\sqrt{n}}{10}-1,2191=\frac{\sqrt{n}-12,191}{10}}\)
\(\displaystyle{ \sqrt{n}\left(\sqrt{n}-12,191\right)=100}\)
\(\displaystyle{ x^{2}-12,191x-100=0}\)
\(\displaystyle{ \Delta=12,191^{2}-4\cdot(-100)=548,62}\)
\(\displaystyle{ \sqrt{\Delta}=23,42}\)
\(\displaystyle{ x_{1}=\frac{12,191+23,42}{2}=17,81}\)
\(\displaystyle{ x_{2}=\frac{12,191-23,42}{2}<0}\)
\(\displaystyle{ x_{1}=\sqrt{n}=17,81\Rightarrow n=\left\lceil 317,08...\right\rceil =318}\)
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nutcracker
- Użytkownik
- Posty: 6
- Rejestracja: 17 wrz 2010, o 22:48
- Płeć: Mężczyzna
- Lokalizacja: warszawa
centralne twierdzenie graniczne
Czy tutaj zastosowałeś twierdzenie lindberga-levy'ego które jest szczegolnym przypadkiem CTG?
centralne twierdzenie graniczne
Tak, to to twierdzenie. Wydaje mi się, że wariancja jest źle policzona, powinno wychodzi 0,09 (błąd w odejmowaniu;). Wtedy liczba zdjęć to około 147 (do kwadratu podnosilem 12,1).