Question

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

• A. 90
• B. 110
• C. 100
• D. 120

Hint:

Use the properties of variance and mean under linear transformations to solve problems involving such transformations.

Answer 1

The Correct Option is C

$\sum_{i=1}^{10} x_i = 50, \therefore \text{mean} = 5$

$\text{Variance} = \frac{4}{5} = \frac{\sum x_i^2}{10} - \left(\frac{\sum x_i}{10}\right)^2$

$\frac{4}{5} = \frac{\sum x_i^2}{10} - 25$

$\Rightarrow \sum x_i^2 = 258 \quad ....(1)$

$\text{Now } \sum_{i=1}^{10} (x_i - \beta)^2 = 98$

$\sum_{i=1}^{10} (x_i^2 - 2\beta x_i + \beta^2) = 98$

$258 - 2\beta(50) + 10\beta^2 = 98$

$(\beta - 8)(\beta - 2) = 0$

$\beta = 8 \text{ or } \beta = 2 \text{ (as } \beta > 2)$

$\therefore \beta = 8 \quad ....(2)$

Now as per the question

$2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2(x_{10} - 1) + 4\beta$

can be simplified to

$2x_1 + 30, 2x_2 + 30, ..., 2x_{10} + 30$

using eq. (2)

$\mu = 2(5) + 30 = 40 \quad .....(3)$

$\sigma^2 = 2^2\left(\frac{4}{5}\right) = \frac{16}{5}$

$\therefore \frac{\beta\mu}{\sigma^2} = \frac{8 \times 40}{\frac{16}{5}} = 100$