5- Em um escritório, o tempo por dia (em minutos), no horário de trabalho, que a Internet na área que João trabalha fica fora do ar em virtude de uma falha da prestadora de serviços, está apresentado na distribuição seguir.
Tempo parado em minutos 0 – 10 10 – 20 20 – 30 30 -- 40 40 – 50 total
Fi 2 15 17 13 3 50
a) Calcule a média, moda e mediana;
b) Calcule o Q3, D3, P63
Soluções para a tarefa
Respondido por
16
Boa tarde Cleber!
Solução!
Para resolver esse exercicio é importante saber o que é limite inferior e superior,classes,amplitude.
![Classe~~~~~~~~~~~Minutos~~~~~~~~~fi~~~~Xi~~~~~~~fi.Xi~~~~~~Fi\\
-------------------------------\\\
~~~1~~~~~~~~~~~~~~~~~~[0,10[~~~~~~~~~~~~~2~~~~~5~~~~~~~~~~~10~~~~~~~~~2~\\\\\\
~~~2~~~~~~~~~~~~~~~~~~[10,20[~~~~~~~~~~~~15~~~~15~~~~~~~~~225~~~~~~~~17\\\\\\
~~~3~~~~~~~~~~~~~~~~~~[20,30[~~~~~~~~~~~~17~~~~25~~~~~~~~~425~~~~~~~~34\\\\\\
~~~4~~~~~~~~~~~~~~~~~~[30,40[~~~~~~~~~~~~13~~~~35~~~~~~~~~455~~~~~~~~47\\\\\\
~~~~~5~~~~~~~~~~~~~~~~~~
[40,50[~~~~~~~~~~~~~3~~~~45~~~~~~~~~135~\\\\
~~~~~~~~](https://tex.z-dn.net/?f=Classe%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7EMinutos%7E%7E%7E%7E%7E%7E%7E%7E%7Efi%7E%7E%7E%7EXi%7E%7E%7E%7E%7E%7E%7Efi.Xi%7E%7E%7E%7E%7E%7EFi%5C%5C%0A-------------------------------%5C%5C%5C%0A%7E%7E%7E1%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5B0%2C10%5B%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E2%7E%7E%7E%7E%7E5%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E10%7E%7E%7E%7E%7E%7E%7E%7E%7E2%7E%5C%5C%5C%5C%5C%5C%0A%7E%7E%7E2%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5B10%2C20%5B%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E15%7E%7E%7E%7E15%7E%7E%7E%7E%7E%7E%7E%7E%7E225%7E%7E%7E%7E%7E%7E%7E%7E17%5C%5C%5C%5C%5C%5C%0A%7E%7E%7E3%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5B20%2C30%5B%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E17%7E%7E%7E%7E25%7E%7E%7E%7E%7E%7E%7E%7E%7E425%7E%7E%7E%7E%7E%7E%7E%7E34%5C%5C%5C%5C%5C%5C%0A%7E%7E%7E4%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%5B30%2C40%5B%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E13%7E%7E%7E%7E35%7E%7E%7E%7E%7E%7E%7E%7E%7E455%7E%7E%7E%7E%7E%7E%7E%7E47%5C%5C%5C%5C%5C%5C%0A%7E%7E%7E%7E%7E5%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%0A%5B40%2C50%5B%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E3%7E%7E%7E%7E45%7E%7E%7E%7E%7E%7E%7E%7E%7E135%7E%5C%5C%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%7E%0A "Classe~~~~~~~~~~~Minutos~~~~~~~~~fi~~~~Xi~~~~~~~fi.Xi~~~~~~Fi\\
-------------------------------\\\
~~~1~~~~~~~~~~~~~~~~~~[0,10[~~~~~~~~~~~~~2~~~~~5~~~~~~~~~~~10~~~~~~~~~2~\\\\\\
~~~2~~~~~~~~~~~~~~~~~~[10,20[~~~~~~~~~~~~15~~~~15~~~~~~~~~225~~~~~~~~17\\\\\\
~~~3~~~~~~~~~~~~~~~~~~[20,30[~~~~~~~~~~~~17~~~~25~~~~~~~~~425~~~~~~~~34\\\\\\
~~~4~~~~~~~~~~~~~~~~~~[30,40[~~~~~~~~~~~~13~~~~35~~~~~~~~~455~~~~~~~~47\\\\\\
~~~~~5~~~~~~~~~~~~~~~~~~
[40,50[~~~~~~~~~~~~~3~~~~45~~~~~~~~~135~\\\\
~~~~~~~~")
![Media!\\\\\\
\overline{X}=\dfrac{\sum~fi.xi}{n}\\\\\\\
\overline{X}=\dfrac{1250}{50}\\\\\\\
\boxed{\overline{X}=25}](https://tex.z-dn.net/?f=Media%21%5C%5C%5C%5C%5C%5C%0A%5Coverline%7BX%7D%3D%5Cdfrac%7B%5Csum%7Efi.xi%7D%7Bn%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Coverline%7BX%7D%3D%5Cdfrac%7B1250%7D%7B50%7D%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7B%5Coverline%7BX%7D%3D25%7D%0A "Media!\\\\\\
\overline{X}=\dfrac{\sum~fi.xi}{n}\\\\\\\
\overline{X}=\dfrac{1250}{50}\\\\\\\
\boxed{\overline{X}=25}")
![Moda!\\\\\
Moda=li+(\dfrac{D1}{D1+D2)}).h\\\\\\\\\
D1=fi-Fi(ant)\\\\\\\\
D2=fi-Fant(post)\\\\\\\\\\
D1=17-15\\\\\
\boxed{D1=2}\\\\\\
D2=17-13$\\\\\
\boxed{D2=4}\\\\\\\\
Moda=20+(\dfrac{2}{6}).10\\\\\\\
Moda=20+(\dfrac{20}{6})\\\\\\\
Moda=20+3,3\\\\\\\
\boxed{Moda=23,3}](https://tex.z-dn.net/?f=Moda%21%5C%5C%5C%5C%5C%0A%0AModa%3Dli%2B%28%5Cdfrac%7BD1%7D%7BD1%2BD2%29%7D%29.h%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AD1%3Dfi-Fi%28ant%29%5C%5C%5C%5C%5C%5C%5C%5C%0AD2%3Dfi-Fant%28post%29%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0AD1%3D17-15%5C%5C%5C%5C%5C%0A%5Cboxed%7BD1%3D2%7D%5C%5C%5C%5C%5C%5C%0AD2%3D17-13%24%5C%5C%5C%5C%5C%0A%5Cboxed%7BD2%3D4%7D%5C%5C%5C%5C%5C%5C%5C%5C%0AModa%3D20%2B%28%5Cdfrac%7B2%7D%7B6%7D%29.10%5C%5C%5C%5C%5C%5C%5C%0AModa%3D20%2B%28%5Cdfrac%7B20%7D%7B6%7D%29%5C%5C%5C%5C%5C%5C%5C%0AModa%3D20%2B3%2C3%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BModa%3D23%2C3%7D "Moda!\\\\\
Moda=li+(\dfrac{D1}{D1+D2)}).h\\\\\\\\\
D1=fi-Fi(ant)\\\\\\\\
D2=fi-Fant(post)\\\\\\\\\\
D1=17-15\\\\\
\boxed{D1=2}\\\\\\
D2=17-13$\\\\\
\boxed{D2=4}\\\\\\\\
Moda=20+(\dfrac{2}{6}).10\\\\\\\
Moda=20+(\dfrac{20}{6})\\\\\\\
Moda=20+3,3\\\\\\\
\boxed{Moda=23,3}")
Medidas separatrizes!
Quartis: Divide a amostra em 4 partes.
Decis:Divide a amostra em 10 partes.
Percentis: Divide a amostra em 100 partes.
Vamos usar essa formula para calcular todas!
![Q=li+ \dfrac{[ \frac{\sum fi}{n}-Fi(ant)].h }{fi}](https://tex.z-dn.net/?f=Q%3Dli%2B+%5Cdfrac%7B%5B+%5Cfrac%7B%5Csum+fi%7D%7Bn%7D-Fi%28ant%29%5D.h+%7D%7Bfi%7D+ "Q=li+ \dfrac{[ \frac{\sum fi}{n}-Fi(ant)].h }{fi}")
![Q_{3}=li+ \dfrac{[ \frac{\sum fi}{4}-Fi(ant)].h }{fi}\\\\\\\\\
Posica\~o=3. \dfrac{50}{4}=37,5\\\\\\
Q_{3}=30+ \dfrac{[37,5-34].10 }{13}\\\\\\\\\
Q_{3}=30+ \dfrac{[3,5].10 }{13}\\\\\\\\\
Q_{3}=30+ \dfrac{35 }{13}\\\\\\\\\
Q_{3}=30+ 2,69\\\\\\\\\
\boxed{Q_{3}=32,69}](https://tex.z-dn.net/?f=Q_%7B3%7D%3Dli%2B+%5Cdfrac%7B%5B+%5Cfrac%7B%5Csum+fi%7D%7B4%7D-Fi%28ant%29%5D.h+%7D%7Bfi%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0APosica%5C%7Eo%3D3.+%5Cdfrac%7B50%7D%7B4%7D%3D37%2C5%5C%5C%5C%5C%5C%5C%0AQ_%7B3%7D%3D30%2B+%5Cdfrac%7B%5B37%2C5-34%5D.10+%7D%7B13%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AQ_%7B3%7D%3D30%2B+%5Cdfrac%7B%5B3%2C5%5D.10+%7D%7B13%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%0AQ_%7B3%7D%3D30%2B+%5Cdfrac%7B35+%7D%7B13%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C++%0AQ_%7B3%7D%3D30%2B+2%2C69%5C%5C%5C%5C%5C%5C%5C%5C%5C++%0A%5Cboxed%7BQ_%7B3%7D%3D32%2C69%7D "Q_{3}=li+ \dfrac{[ \frac{\sum fi}{4}-Fi(ant)].h }{fi}\\\\\\\\\
Posica\~o=3. \dfrac{50}{4}=37,5\\\\\\
Q_{3}=30+ \dfrac{[37,5-34].10 }{13}\\\\\\\\\
Q_{3}=30+ \dfrac{[3,5].10 }{13}\\\\\\\\\
Q_{3}=30+ \dfrac{35 }{13}\\\\\\\\\
Q_{3}=30+ 2,69\\\\\\\\\
\boxed{Q_{3}=32,69}")
![D=li+ \dfrac{[ \frac{\sum fi}{n}-Fi(ant)].h }{fi}\\\\\\\\
D_{3} =li+ \dfrac{[ \frac{\sum fi}{10}-Fi(ant)].h }{fi}\\\\\\
Posic\~ao=3. \dfrac{50}{10}=15\\\\\\\
D_{3} =10+ \dfrac{[15-2].10 }{15}\\\\\\
D_{3} =10+ \dfrac{[13].10 }{15}\\\\\\
D_{3} =10+ \dfrac{130 }{15}\\\\\\
D_{3} =10+ 8,6\\\\\\
\boxed{D_{3} =18,6}\\\\\\](https://tex.z-dn.net/?f=D%3Dli%2B+%5Cdfrac%7B%5B+%5Cfrac%7B%5Csum+fi%7D%7Bn%7D-Fi%28ant%29%5D.h+%7D%7Bfi%7D%5C%5C%5C%5C%5C%5C%5C%5C%0AD_%7B3%7D+%3Dli%2B+%5Cdfrac%7B%5B+%5Cfrac%7B%5Csum+fi%7D%7B10%7D-Fi%28ant%29%5D.h+%7D%7Bfi%7D%5C%5C%5C%5C%5C%5C%0APosic%5C%7Eao%3D3.+%5Cdfrac%7B50%7D%7B10%7D%3D15%5C%5C%5C%5C%5C%5C%5C%0AD_%7B3%7D+%3D10%2B+%5Cdfrac%7B%5B15-2%5D.10+%7D%7B15%7D%5C%5C%5C%5C%5C%5C%0AD_%7B3%7D+%3D10%2B+%5Cdfrac%7B%5B13%5D.10+%7D%7B15%7D%5C%5C%5C%5C%5C%5C%0AD_%7B3%7D+%3D10%2B+%5Cdfrac%7B130+%7D%7B15%7D%5C%5C%5C%5C%5C%5C+%0AD_%7B3%7D+%3D10%2B+8%2C6%5C%5C%5C%5C%5C%5C+%0A%5Cboxed%7BD_%7B3%7D+%3D18%2C6%7D%5C%5C%5C%5C%5C%5C++%0A+ "D=li+ \dfrac{[ \frac{\sum fi}{n}-Fi(ant)].h }{fi}\\\\\\\\
D_{3} =li+ \dfrac{[ \frac{\sum fi}{10}-Fi(ant)].h }{fi}\\\\\\
Posic\~ao=3. \dfrac{50}{10}=15\\\\\\\
D_{3} =10+ \dfrac{[15-2].10 }{15}\\\\\\
D_{3} =10+ \dfrac{[13].10 }{15}\\\\\\
D_{3} =10+ \dfrac{130 }{15}\\\\\\
D_{3} =10+ 8,6\\\\\\
\boxed{D_{3} =18,6}\\\\\\")
![P=li+ \dfrac{[ \frac{\sum fi}{n}-Fi(ant)].h }{fi}\\\\\\
P_{63} =li+ \dfrac{[ \frac{\sum fi}{100}-Fi(ant)].h }{fi}\\\\\\
Posic\~ao=63. \dfrac{50}{100}=31,5\\\\\\
P_{63} =20+ \dfrac{[ 31,5-17].10 }{17}\\\\\\
P_{63} =20+ \dfrac{[ 14,5].10 }{17}\\\\\\
P_{63} =20+ \dfrac{145}{17}\\\\\\
P_{63} =20+ \dfrac{145}{17}\\\\\\
P_{63} =20+ 8,52\\\\\\
\boxed{P_{63} =28,52}](https://tex.z-dn.net/?f=P%3Dli%2B+%5Cdfrac%7B%5B+%5Cfrac%7B%5Csum+fi%7D%7Bn%7D-Fi%28ant%29%5D.h+%7D%7Bfi%7D%5C%5C%5C%5C%5C%5C%0AP_%7B63%7D+%3Dli%2B+%5Cdfrac%7B%5B+%5Cfrac%7B%5Csum+fi%7D%7B100%7D-Fi%28ant%29%5D.h+%7D%7Bfi%7D%5C%5C%5C%5C%5C%5C%0APosic%5C%7Eao%3D63.+%5Cdfrac%7B50%7D%7B100%7D%3D31%2C5%5C%5C%5C%5C%5C%5C%0AP_%7B63%7D+%3D20%2B+%5Cdfrac%7B%5B+31%2C5-17%5D.10+%7D%7B17%7D%5C%5C%5C%5C%5C%5C%0AP_%7B63%7D+%3D20%2B+%5Cdfrac%7B%5B+14%2C5%5D.10+%7D%7B17%7D%5C%5C%5C%5C%5C%5C%0AP_%7B63%7D+%3D20%2B+%5Cdfrac%7B145%7D%7B17%7D%5C%5C%5C%5C%5C%5C%0AP_%7B63%7D+%3D20%2B+%5Cdfrac%7B145%7D%7B17%7D%5C%5C%5C%5C%5C%5C%0AP_%7B63%7D+%3D20%2B+8%2C52%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BP_%7B63%7D+%3D28%2C52%7D%0A "P=li+ \dfrac{[ \frac{\sum fi}{n}-Fi(ant)].h }{fi}\\\\\\
P_{63} =li+ \dfrac{[ \frac{\sum fi}{100}-Fi(ant)].h }{fi}\\\\\\
Posic\~ao=63. \dfrac{50}{100}=31,5\\\\\\
P_{63} =20+ \dfrac{[ 31,5-17].10 }{17}\\\\\\
P_{63} =20+ \dfrac{[ 14,5].10 }{17}\\\\\\
P_{63} =20+ \dfrac{145}{17}\\\\\\
P_{63} =20+ \dfrac{145}{17}\\\\\\
P_{63} =20+ 8,52\\\\\\
\boxed{P_{63} =28,52}")
Boa tarde!
Bons estudos!
Solução!
Para resolver esse exercicio é importante saber o que é limite inferior e superior,classes,amplitude.
Medidas separatrizes!
Quartis: Divide a amostra em 4 partes.
Decis:Divide a amostra em 10 partes.
Percentis: Divide a amostra em 100 partes.
Vamos usar essa formula para calcular todas!
Boa tarde!
Bons estudos!
Anexos:
Usuário anônimo:
Qualquer dúvida manda uma mensagem,a tabela não esta legal.
Vou colocar um anexo da tabela!
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