Statistics Question 2
Question 2 - 2024 (01 Feb Shift 2)
Consider 10 observation $x _1, \mathbf{x} _{2, \ldots}, \mathbf{x} _{10}$. such that $\sum _{i=1}^{10}\left(x _i-\alpha\right)=2$ and $\sum _{i=1}^{10}\left(x _i-\beta\right)^{2}=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
(1) 2
(2) $\frac{3}{2}$
(3) $\frac{5}{2}$
(4) 1
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Answer (1)
Solution
$x _1, x _2 \ldots \ldots x _{10}$
$\sum _{i=1}^{10}\left(x _i-\alpha\right)=2 \Rightarrow \sum _{i=1}^{10} x _i-10 \alpha=2$
Mean $\mu=\frac{6}{5}=\frac{\sum x _i}{10}$
$\therefore \quad \sum x _i=12$
$10 \alpha+2=12 \therefore \alpha=1$
Now $\sum _{i=1}^{10}\left(x _i-\beta\right)^{2}=40 \quad$ Let $y _i=x _i-\beta$
$\therefore \sigma _y^{2}=\frac{1}{10} \sum y _i^{2}-(\overline{y})^{2}$
$\sigma _x^{2}=\frac{1}{10} \sum\left(x _i-\beta\right)^{2}-\left(\frac{\sum _{i=1}^{10}\left(x _i-\beta\right)}{10}\right)^{2}$
$\frac{84}{25}=4-\left(\frac{12-10 \beta}{10}\right)^{2}$
$\therefore\left(\frac{6-5 \beta}{5}\right)^{2}=4-\frac{84}{25}=\frac{16}{25}$
$6-5 \beta= \pm 4 \Rightarrow \beta=\frac{2}{5}$ (not possible) or $\beta=2$
Hence $\frac{\beta}{\alpha}=2$
