Consider 10 observations x_1,x_2, ,x_10 such that _i=1^10(x_i- )=2 and _i=1^10(x_i- )^2=40…

Question

Consider 10 observations ${x}_{1},{x}_{2},\ldots ,{x}_{10}$ such that $\sum _{i=1}^{10}({x}_{i}-\alpha )=2$ and $\sum _{i=1}^{10}{({x}_{i}-\beta )}^{2}=40$, where $\alpha ,\beta $ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{\beta }{\alpha }$ is equal to :

Accepted Answer

We have given $\vec{x}($ mean $)=\frac{6}{5}$

$\begin{gathered} =\frac{84}{25} \ \sum _{i=1}^{10}({x}_{i}-\alpha )=2 \ \Rightarrow {x}_{1}+{x}_{2}+\ldots +{x}_{10}-10\alpha =2 \ \Rightarrow \frac{{x}_{1}+{x}_{2}+\ldots +{x}_{10}}{10}-\alpha =\frac{2}{10} \ \Rightarrow \frac{6}{5}-\alpha =\frac{2}{10} \ \Rightarrow \alpha =1 \end{gathered}$

$\begin{gathered} \sum _{i=1}^{10}{({x}_{i}-\beta )}^{2}=40 \ {({x}_{1}-\beta )}^{2}+{({x}_{2}-\beta )}^{2}+\ldots +{({x}_{10}-\beta )}^{2}=40 \ {x}_{2}^{1}+{x}_{2}^{2}+\ldots +{x}_{2}^{10}+10{\beta }^{2}-2\beta ({x}_{1}+{x}_{2}+\ldots +{x}_{10})=40 \ \Rightarrow \frac{{x}_{2}^{1}+{x}_{2}^{2}+\ldots +{x}_{2}^{10}}{10}+{\beta }^{2}-\frac{2\beta ({x}_{1}+{x}_{2}+\ldots +{x}_{10})}{10}=4 \ \Rightarrow \frac{{x}_{2}^{1}+{x}_{2}^{2}+\ldots +{x}_{2}^{10}}{10}-\frac{36}{25}+\frac{36}{25}+{\beta }^{2}-2\beta 	imes \frac{6}{5}=4 \end{gathered}$

$[$ Variance $=\frac{\sum _{n}^{i=1}{x}_{2}^{i}}{n}-(\vec{x}{)}^{2}]$

$\begin{gathered} \Rightarrow \frac{84}{25}+\frac{36}{25}+{\beta }^{2}-\frac{12\beta }{5}-4=0 \ \Rightarrow \frac{120}{25}+{\beta }^{2}-\frac{12\beta }{5}-4=0 \ \Rightarrow 25{\beta }^{2}-60\beta +20=0 \ \Rightarrow 5{\beta }^{2}-12\beta +4=0 \ \Rightarrow \beta =2,\frac{2}{5} \end{gathered}$

Take $\beta =2$

$\frac{\beta }{\alpha }=\frac{2}{1}=2$

Solutions · 1