Statistics Question 2
Question 2 - 2024 (01 Feb Shift 2)
Consider 10 observation $\mathrm{x}{1}, \mathbf{x}{2, \ldots}, \mathbf{x}{10}$. 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
$\mathrm{x}{1}, \mathrm{x}{2} \ldots \ldots \mathrm{x}_{10}$
$\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}{\mathrm{i}}-\alpha\right)=2 \Rightarrow \sum{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}-10 \alpha=2$
Mean $\mu=\frac{6}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}$
$\therefore \quad \sum \mathrm{x}_{\mathrm{i}}=12$
$10 \alpha+2=12 \therefore \alpha=1$
Now $\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}{\mathrm{i}}-\beta\right)^{2}=40 \quad$ Let $\mathrm{y}{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\beta$
$\therefore \sigma_{\mathrm{y}}^{2}=\frac{1}{10} \sum \mathrm{y}_{\mathrm{i}}^{2}-(\overline{\mathrm{y}})^{2}$
$\sigma_{x}^{2}=\frac{1}{10} \sum\left(x_{i}-\beta\right)^{2}-\left(\frac{\sum_{i=1}^{10}\left(x_{i}-\beta\right)}{10}\right)^{2}$
$\frac{84}{25}=4-\left(\frac{12-10 \beta}{10}\right)^{2}$
$\therefore\left(\frac{6-5 \beta}{5}\right)^{2}=4-\frac{84}{25}=\frac{16}{25}$
$6-5 \beta= \pm 4 \Rightarrow \beta=\frac{2}{5}$ (not possible) or $\beta=2$
Hence $\frac{\beta}{\alpha}=2$
