Question

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

The objective is to compute the value of $\frac{\beta \mu}{\sigma^2}$, where $\mu$ and $\sigma^2$ represent the mean and variance, respectively, of the transformed data points given by \(y_i = 2(x_i - 1) + 4\beta\). The following information is provided:

• \(\sum_{i=1}^{10} (x_i - 2) = 30\)
• \(\sum_{i=1}^{10} (x_i - \beta)^2 = 98\), with the condition \(\beta > 2\)
• The variance of the observations \(x_1, x_2, \ldots, x_{10}\) is \(\frac{4}{5}\)

The mean \(\bar{x}\) of \(x_1, x_2, \ldots, x_{10}\) is derived from the first condition:

\(\sum_{i=1}^{10} x_i - 20 = 30 \implies \sum_{i=1}^{10} x_i = 50\).

Therefore, \(\bar{x} = \frac{50}{10} = 5\).

The variance formula is applied as follows:

\(\text{Variance}(x) = \frac{1}{10}\sum_{i=1}^{10}(x_i - \bar{x})^2 = \frac{4}{5}\).

Substituting \(\bar{x} = 5\):

\(\frac{1}{10}\sum_{i=1}^{10}(x_i - 5)^2 = \frac{4}{5} \implies \sum_{i=1}^{10}(x_i - 5)^2 = 8\).

Using the given \(\sum_{i=1}^{10} (x_i - \beta)^2 = 98\) and the identity:

\(\sum_{i=1}^{10} (x_i - \beta)^2 = \sum_{i=1}^{10} (x_i - 5)^2 + 10(\beta - 5)^2\).

Substituting the value of \(\sum_{i=1}^{10} (x_i - 5)^2\):

\(98 = 8 + 10(\beta - 5)^2 \implies 90 = 10(\beta - 5)^2\).

Solving for \(\beta\):

\((\beta - 5)^2 = 9 \implies \beta - 5 = \pm 3\).

Considering \(\beta > 2\), we find \(\beta = 8\).

The transformed observations are \(y_i = 2(x_i - 1) + 4\beta\). Substituting \(\beta = 8\):

\(y_i = 2x_i - 2 + 4(8) = 2x_i - 2 + 32 = 2x_i + 30\).

The mean \(\mu\) of \(y_i\) is calculated as:

\(\mu = \frac{1}{10} \sum_{i=1}^{10} y_i = 2\bar{x} + 30 = 2(5) + 30 = 10 + 30 = 40\).

The variance \(\sigma^2\) of \(y_i\) is:

\(\sigma^2 = \text{Var}(2x_i + 30) = 2^2 \times \text{Var}(x_i) = 4 \times \frac{4}{5} = \frac{16}{5}\).

Finally, the expression $\frac{\beta \mu}{\sigma^2}$ is computed:

\(\frac{\beta \mu}{\sigma^2} = \frac{8 \times 40}{\frac{16}{5}} = \frac{320}{\frac{16}{5}} = \frac{320 \times 5}{16} = 20 \times 5 = 100\).

The result is \(\boxed{100}\).