Contabilidade, perguntado por islanlira9, 8 meses atrás

A DISTRIBUIÇÃO DE IDADE DE UMA COMUNIDADE É FORNECIDA PELA TABELA: (MÉDIA, MEDIANA E MODA, FREQUÊNCIA)

Anexos:

Soluções para a tarefa

Respondido por Usuário anônimo
8

Explicação:

a) Variável quantitativa

b)

\sf \Big|~~idade~~~~\Big|~~~~~freq.~~~~~\Big|~~~~freq.~ab.~~~\Big|~~~~~~freq.~~~~~~\Big|~~~~freq.~rel~~~~</p><p>\Big|

\sf \Big|~~(anos)~~\Big|~~absoluta~~\Big|~~acumulada~\Big|~~~~relativa~~~\Big|~~acumulada~~\Big|

\sf \Big|~~~~~~0~~~~~~\Big|~~~~~~~~12~~~~~~~\Big|~~~~~~~~~12~~~~~~~~\Big|~~~~~7,5\%~~~~~\Big|~~~~~~~7,5\%~~~~~~\Big|

\sf \Big|~~~~~~1~~~~~~\Big|~~~~~~~~13~~~~~~~\Big|~~~~~~~~~25~~~~~~~~\Big|~~~8,125\%~~~\Big|~~~~15,625\%~~~\Big|

\sf \Big|~~~~~~2~~~~~~\Big|~~~~~~~~22~~~~~~~\Big|~~~~~~~~~47~~~~~~~~\Big|~~~13,75\%~~~\Big|~~~29,375\%~~~~\Big|

\sf \Big|~~~~~~3~~~~~~\Big|~~~~~~~~50~~~~~~~\Big|~~~~~~~~~97~~~~~~~~\Big|~~~31,25\%~~~\Big|~~~60,625\%~~~~\Big|

\sf \Big|~~~~~~4~~~~~~\Big|~~~~~~~~31~~~~~~~\Big|~~~~~~~~128~~~~~~~\Big|~~19,375\%~~\Big|~~~~~~~80\%~~~~~~~\Big|

\sf \Big|~~~~~~5~~~~~~\Big|~~~~~~~~22~~~~~~~\Big|~~~~~~~~150~~~~~~~\Big|~~~13,75\%~~~\Big|~~~~93,75\%~~~~\Big|

\sf \Big|~~~~~~6~~~~~~\Big|~~~~~~~~10~~~~~~~\Big|~~~~~~~~160~~~~~~~\Big|~~~~6,25\%~~~~\Big|~~~~~~100\%~~~~~~\Big|

c)

=> Média

\sf M=\dfrac{0\cdot12+1\cdot13+2\cdot22+3\cdot50+4\cdot31+5\cdot22+6\cdot10}{12+13+22+50+31+22+10}

\sf M=\dfrac{0+13+44+150+124+110+60}{160}

\sf M=\dfrac{501}{160}

\sf \red{M=3,13125}

=> Mediana

A mediana é valor central (ou a média dos valores centrais) quando colocamos os valores em ordem crescente

Como há 160 valores, os valores centrais são o 80° e o 81°

Temos \sf 12+13+22=47. Assim, 47 crianças tem menos de 3 anos.

E \sf 12+13+22+50=97. Note que já passamos de 81. Os valores centrais são 3 e 3

Logo, a mediana é \sf \dfrac{3+3}{2}=\dfrac{6}{2}=3

A mediana é 3

=> Moda

É o valor que mais se repete.

A moda é 3

d)

=> Variância

\sf \sigma^2=\dfrac{12\cdot(0-3,13125)^2+13\cdot(1-3,13125)^2+22\cdot(2-3,13125)^2+50\cdot(3-3,13125)^2+31\cdot(4-3,13125)^2+22\cdot(5-3,13125)^2+10\cdot(6-3,13125)^2}{12+13+22+50+31+22+10}

\sf \sigma^2=\dfrac{12\cdot(-3,13125)^2+13\cdot(-2,13125)^2+22\cdot(-1,13125)^2+50\cdot(0,13125)^2+31\cdot(-1,13125)^2+22\cdot2,13125^2+10\cdot3,13125^2}{160}

\sf \sigma^2=\dfrac{12\cdot9,8047+13\cdot4,5422+22\cdot1,2797+50\cdot0,0172+31\cdot1,2797+22\cdot4,5422+10\cdot9,8047}{160}

\sf \sigma^2=\dfrac{117,6564+59,0486+28,1534+0,86+39,6707+99,9284+98,047}{160}

\sf \sigma^2=\dfrac{443,3645}{160}

\sf \red{\sigma^2=2,771}

=> Desvio padrão

\sf \sigma=\sqrt{\sigma^2}

\sf \sigma=\sqrt{2,771}

\sf \red{\sigma=1,6646}

=> Coeficiente de variação

\sf CV\%=\dfrac{\sigma}{M}\cdot100

\sf CV\%=\dfrac{1,6646}{3,13125}\cdot100

\sf CV\%=\dfrac{166,46}{3,13125}

\sf CV\%=\dfrac{16646000}{313125}

\sf \red{CV\%=53,16\%}

Alta dispersão, dados heterogêneos

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