Question

Dado o rol de medidas das alturas (dadas em cm na imagem em anexo) de uma amostra de 100 indivíduos de uma faculdade, calcule:

Desvio médio;
variância;
desvio padrão;
coeficiente de variação.

9bc656de78c0a819165f20501bb3a17b.pdf

Answer 1

Resposta:

Vide abaixo!

Explicação:

Boa tarde!

Para calcular coloquei os dados em uma tabela:

$\begin{tabular}{c|c|c|c|c|c|c}x&f&fx&\left|x-media\right|&f\left|x-media\right|&(x-media)^2&f(x-media)^2\\151&1&151&20,40&20,40&416,16&416,16\\152&1&152&19,40&19,40&376,36&376,36\\154&1&154&17,40&17,40&302,76&302,76\\155&1&155&16,40&16,40&268,96&268,96\\158&1&158&13,40&13,40&179,56&179,56\\159&2&318&12,40&24,80&153,76&307,52\\160&1&160&11,40&11,40&129,96&129,96\\161&3&483&10,40&31,20&108,16&324,48\\162&1&162&9,40&9,40&88,36&88,36\\163&3&489&8,40&25,20&70,56&211,68\\164&1&164&7,40&7,40&54,76&54,76\\165&3&495&6,40&19,20&40,96&122,88\\166&4&664&5,40&21,60&29,16&116,64\\167&5&835&4,40&22,00&19,36&96,80\\168&10&1\,680&3,40&34,00&11,56&115,60\\169&7&1\,183&2,40&16,80&5,76&40,32\\170&7&1\,190&1,40&9,80&1,96&13,72\\171&4&684&0,40&1,60&0,16&0,64\\172&3&516&0,60&1,80&0,36&1,08\\173&3&519&1,60&4,80&2,56&7,68\\174&3&522&2,60&7,80&6,76&20,28\\175&4&700&3,60&14,40&12,96&51,84\\176&4&704&4,60&18,40&21,16&84,64\\177&4&708&5,60&22,40&31,36&125,44\\178&3&534&6,60&19,80&43,56&130,68\\179&2&358&7,60&15,20&57,76&115,52\\180&4&720&8,60&34,40&73,96&295,84\\181&3&543&9,60&28,80&92,16&276,48\\182&3&546&10,60&31,80&112,36&337,08\\183&1&183&11,60&11,60&134,56&134,56\\184&1&184&12,60&12,60&158,76&158,76\\185&1&185&13,60&13,60&184,96&184,96\\186&1&186&14,60&14,60&213,16&213,16\\187&1&187&15,60&15,60&243,36&243,36\\188&1&188&16,60&16,60&275,56&275,56\\190&2&380&18,60&37,20&345,96&691,92\\\Sigma&100&17\,140&-&642,80&-&6\,516\end{tabular}$

Agora podemos calcular o que se pede:

Média:

$\overline{x}=\dfrac{\sum\;fx}{\sum\;f}=\dfrac{17\,140}{100}=171,40$

Desvio-médio:

$DM=\dfrac{\sum\;f|x-\overline{x}|}{\sum\;f}=\dfrac{642,80}{100}=6,428$

Variância:

$\sigma^2=\dfrac{\sum\;f(x-\overline{x})^2}{\sum\;f}=\dfrac{6\,516}{100}=65,16$

Desvio-padrão:

$\sigma=\sqrt{65,16}\\\sigma\approx 8,072$

Coeficiente de variação:

$CV=\dfrac{\sigma}{\overline{x}}=\dfrac{8,072}{171,4}\\CV\approx 4,709\%$

Espero ter ajudado!

Answer 2

Resposta:

Como vc achou esses valores

Explicação:valores da tabela como vc chegou 20,40