For 10 observations x_1,x_2, ,x_10, if _i=1^10(x_i+2)^2=180 and _i=1^10(x_i-1)^2=90, then …

Question

For 10 observations ${x}_{1},{x}_{2},\ldots ,{x}_{10}$, if $\sum _{i=1}^{10}{({x}_{i}+2)}^{2}=180$ and $\sum _{i=1}^{10}{({x}_{i}-1)}^{2}=90$, then their standard deviation is :

Accepted Answer

$\sum _{i=1}^{10}({x}_{i}+2{)}^{2}=180$

$\sum _{i=1}^{10}{x}_{2}^{i}+4\sum _{i=1}^{10}{x}_{i}+\sum _{i=1}^{10}4=180$

$\sum _{i=1}^{10}{x}_{2}^{i}+4\sum _{i=1}^{10}{x}_{i}=180-40\, \ldots (1)$

Also $\sum _{i=1}^{10}({x}_{i}-1{)}^{2}=90$

$\sum _{i=1}^{10}{x}_{2}^{i}-2\sum _{i=1}^{10}{x}_{i}+\sum _{i=1}^{10}1=90$

$\sum _{i=1}^{10}{x}_{2}^{i}-2\sum _{i=1}^{10}{x}_{i}=90-10\, \ldots (2)$

$\sum _{i=1}^{10}{x}_{2}^{i}+4\sum _{i=1}^{10}{x}_{i}=140$

$\sum _{i=1}^{10}{x}_{2}^{i}-2\sum _{i=1}^{10}{x}_{i}=80$

$\sum _{i=1}^{10}{x}_{2}^{i}=100\, \, \sum _{i=1}^{10}{x}_{i}=10$

${\sigma }^{2}=\frac{\sum _{i=1}^{10}{x}_{2}^{i}}{N}-{((|\frac{\sum _{i=1}^{10}{x}_{i}}{N}))|}^{2}$

$=\frac{100}{10}-{(\frac{10}{10})}^{2}$

${\sigma }^{2}=10-1=9$

$\Rightarrow \sigma =3$

Solutions · 1