Let x_1, x_2, ..., x_{10} be ten observations such that \sum\limits_{i=1}^{10} (x_i - 2) = 30, \sum\limits_{i=1}^{10} (x_i - \beta)^2 = 98, \beta 2, and their variance is \frac{4}{5}. If \mu and \sigma^2 are respectively the mean and the variance of 2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2

Question

Let x_1, x_2, ..., x_{10} be ten observations such that \sum\limits_{i=1}^{10} (x_i - 2) = 30, \sum\limits_{i=1}^{10} (x_i - \beta)^2 = 98, \beta > 2, and their variance is \frac{4}{5}. If \mu and \sigma^2 are respectively the mean and the variance of 2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2(x_{10} - 1) + 4\beta, then \frac{\beta\mu}{\sigma^2} is equal to :

Accepted Answer

Answer: A. 100 \begin{aligned} & \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=50, \quad 	herefore 	ext { mean }=5 \ & 	ext { Variance }=\frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}ight)^2 \ & \frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-25 \ & \Rightarrow \sum \mathrm{x}_{\mathrm{i}}^2=258 \quad	ext{.... (1)}\ & 	ext { Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\betaight)^2=98 \ & \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}^2-2 \beta \cdot \mathrm{x}_{\mathrm{i}}+\beta^2ight)=98 \ & 258-2 \beta(50)+10 \beta^2=98 \ & (\beta-8)(\beta-2)=0 \ & \beta=	ext { or } \beta=2 \quad(	ext { as } \beta>2) \ & 	herefore \beta=8\quad	ext{..... (2)} \end{aligned} Now as per the question \begin{aligned} &2\left(\mathrm{x}_1-1ight)+4 \beta, 2\left(\mathrm{x}_2-1ight)+4 \beta, \ldots .2\left(\mathrm{x}_{10}-1ight)+4 \beta\ &	ext { can be simplified to }\ &2 \mathrm{x}_1+30,2 \mathrm{x}_2+30, \ldots . .2 \mathrm{x}_{10}+30 	ext { using eq. (2) }\ &\mu=2(5)+30=40\quad	ext{..... (3)}\ &\sigma^2=2^2\left(\frac{4}{5}ight)=\frac{16}{5}\ &\because \frac{\beta \mu}{\sigma^2}=\frac{8 	imes 40}{16 / 5}=100 \end{aligned}

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