boa dia podem me ajudar com as contas abaixo de estatística preciso da conta mais resoluçao , obrigado
Anexos:
Soluções para a tarefa
Respondido por
1
Bom dia Marcelo!
Solução!
![~~y~~~~~~~~1~~~~~~~~~2~~~~~~~~~~~~3\\\\\\
p(y)~~~~~~ \dfrac{1}{4}~~~~~~~~~ \dfrac{1}{4}~~~~~~~~~~ \dfrac{~1}{2}\\\\\\\
yp(y)~~~~ \dfrac{~1}{4} ~~~~~~~ \dfrac{~1}{2}~~~~~~~~~~ \dfrac{~3}2}\\\\\\\
y^{2}~~~~~~~~~1~~~~~~~~4~~~~~~~~~~~~9\\\\\\\\
y^{2}p(y)~~~~ \dfrac{1}{4}~~~~~~1~~~~~~~~~~~~ \dfrac{9}{2}](https://tex.z-dn.net/?f=++%7E%7Ey%7E%7E%7E%7E%7E%7E%7E%7E1%7E%7E%7E%7E%7E%7E%7E%7E%7E2%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E3%5C%5C%5C%5C%5C%5C%0Ap%28y%29%7E%7E%7E%7E%7E%7E+%5Cdfrac%7B1%7D%7B4%7D%7E%7E%7E%7E%7E%7E%7E%7E%7E+%5Cdfrac%7B1%7D%7B4%7D%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E+%5Cdfrac%7B%7E1%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0Ayp%28y%29%7E%7E%7E%7E+%5Cdfrac%7B%7E1%7D%7B4%7D+%7E%7E%7E%7E%7E%7E%7E+%5Cdfrac%7B%7E1%7D%7B2%7D%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E+%5Cdfrac%7B%7E3%7D2%7D%5C%5C%5C%5C%5C%5C%5C%0Ay%5E%7B2%7D%7E%7E%7E%7E%7E%7E%7E%7E%7E1%7E%7E%7E%7E%7E%7E%7E%7E4%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E9%5C%5C%5C%5C%5C%5C%5C%5C%0Ay%5E%7B2%7Dp%28y%29%7E%7E%7E%7E+%5Cdfrac%7B1%7D%7B4%7D%7E%7E%7E%7E%7E%7E1%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E+%5Cdfrac%7B9%7D%7B2%7D++++%0A++++ "~~y~~~~~~~~1~~~~~~~~~2~~~~~~~~~~~~3\\\\\\
p(y)~~~~~~ \dfrac{1}{4}~~~~~~~~~ \dfrac{1}{4}~~~~~~~~~~ \dfrac{~1}{2}\\\\\\\
yp(y)~~~~ \dfrac{~1}{4} ~~~~~~~ \dfrac{~1}{2}~~~~~~~~~~ \dfrac{~3}2}\\\\\\\
y^{2}~~~~~~~~~1~~~~~~~~4~~~~~~~~~~~~9\\\\\\\\
y^{2}p(y)~~~~ \dfrac{1}{4}~~~~~~1~~~~~~~~~~~~ \dfrac{9}{2}")
Esperança matemática é dada por:
Variância!
![\sigma^{2}(y)=E(y^{2})-[E(y)]^{2}\\\\\\
ou\\\\\
\sigma^{2}(y)=E(y^{2})-\mu(y)\\\\\\
\sigma(y^{2})= \frac{1}{4}+1+ \frac{9}{2}- \frac{5}{4}\\\\\
\sigma(y^{2})= \dfrac{1+4+18-5}{4} \\\\\\
\sigma(y^{2})= \dfrac{18}{5}\\\\\\
\sigma(y^{2})=3,6](https://tex.z-dn.net/?f=%5Csigma%5E%7B2%7D%28y%29%3DE%28y%5E%7B2%7D%29-%5BE%28y%29%5D%5E%7B2%7D%5C%5C%5C%5C%5C%5C%0Aou%5C%5C%5C%5C%5C+%0A%5Csigma%5E%7B2%7D%28y%29%3DE%28y%5E%7B2%7D%29-%5Cmu%28y%29%5C%5C%5C%5C%5C%5C%0A%5Csigma%28y%5E%7B2%7D%29%3D+%5Cfrac%7B1%7D%7B4%7D%2B1%2B+%5Cfrac%7B9%7D%7B2%7D-+%5Cfrac%7B5%7D%7B4%7D%5C%5C%5C%5C%5C%0A%0A%5Csigma%28y%5E%7B2%7D%29%3D+%5Cdfrac%7B1%2B4%2B18-5%7D%7B4%7D+%5C%5C%5C%5C%5C%5C%0A%0A%5Csigma%28y%5E%7B2%7D%29%3D+%5Cdfrac%7B18%7D%7B5%7D%5C%5C%5C%5C%5C%5C%0A+%5Csigma%28y%5E%7B2%7D%29%3D3%2C6%0A++++ "\sigma^{2}(y)=E(y^{2})-[E(y)]^{2}\\\\\\
ou\\\\\
\sigma^{2}(y)=E(y^{2})-\mu(y)\\\\\\
\sigma(y^{2})= \frac{1}{4}+1+ \frac{9}{2}- \frac{5}{4}\\\\\
\sigma(y^{2})= \dfrac{1+4+18-5}{4} \\\\\\
\sigma(y^{2})= \dfrac{18}{5}\\\\\\
\sigma(y^{2})=3,6")
Para saber o valor do desvio padrão basta fazer a raiz quadrada da variância.
Boa tarde!
Bons estudos!
Solução!
Esperança matemática é dada por:
Variância!
Para saber o valor do desvio padrão basta fazer a raiz quadrada da variância.
Boa tarde!
Bons estudos!
MarceloGrossi:
vc me indica alguma video aula para poder aprender mais sobre essa materia ?
Sim! Estatística Autores Ermes medeiros ,Elio Medeiros,Valter Gonçalves. Tem exatamente o que você quer.
obrigado !
Perguntas interessantes