Question ID: #529
Let $x_{1}, x_{2}, \dots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_{i}-2)=30$, $\sum_{i=1}^{10}(x_{i}-\beta)^{2}=98$, where $\beta>2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^{2}$ are respectively the mean and the variance of $2(x_1-1)+4\beta, \dots, 2(x_{10}-1)+4\beta$, then $\frac{\beta\mu}{\sigma^{2}}$ is equal to:
- (1) 100
- (2) 110
- (3) 120
- (4) 90
Solution:
Given $\sum (x_i – 2) = 30$.
$\sum x_i – 20 = 30 \Rightarrow \sum x_i = 50$.
Mean of $x$, $\bar{x} = \frac{50}{10} = 5$.
Given Variance of $x$, $\text{Var}(x) = \frac{4}{5}$.
$\frac{\sum x_i^2}{10} – (\bar{x})^2 = \frac{4}{5} \Rightarrow \frac{\sum x_i^2}{10} – 25 = 0.8 \Rightarrow \sum x_i^2 = 258$.
Also given $\sum (x_i – \beta)^2 = 98$.
Expand: $\sum (x_i^2 – 2\beta x_i + \beta^2) = 98$.
$\sum x_i^2 – 2\beta \sum x_i + 10\beta^2 = 98$.
$258 – 2\beta(50) + 10\beta^2 = 98$.
$10\beta^2 – 100\beta + 160 = 0$.
Divide by 10: $\beta^2 – 10\beta + 16 = 0$.
$(\beta – 8)(\beta – 2) = 0$.
Since $\beta > 2$, we have $\beta = 8$.
Now, consider the new observations: $y_i = 2(x_i – 1) + 4\beta$.
$y_i = 2x_i – 2 + 32 = 2x_i + 30$.
Mean $\mu$ of $y$: $\mu = 2\bar{x} + 30 = 2(5) + 30 = 40$.
Variance $\sigma^2$ of $y$: $\text{Var}(ax+b) = a^2 \text{Var}(x)$.
$\sigma^2 = 2^2 \times \frac{4}{5} = 4 \times 0.8 = 3.2 = \frac{16}{5}$.
We need to find $\frac{\beta\mu}{\sigma^{2}}$.
$\frac{8 \times 40}{16/5} = \frac{320 \times 5}{16} = 20 \times 5 = 100$.
Ans. (1)
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