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:
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:
Show Hint
- 90
- 110
- 100
- 120
The Correct Option is C
Solution and Explanation
The problem requires us to determine the value of $\frac{\beta \mu}{\sigma^2}$, where $\mu$ and $\sigma^2$ are the mean and variance of the transformed observations given by \(2(x_i - 1) + 4\beta\). We are provided with:
- \(\sum_{i=1}^{10} (x_i - 2) = 30\)
- \(\sum_{i=1}^{10} (x_i - \beta)^2 = 98\), where \(\beta > 2\)
- Variance of \(x_1, x_2, \ldots, x_{10}\) is \(\frac{4}{5}\)
First, we find the mean \(\bar{x}\) of \((x_1, x_2, \ldots, x_{10})\) using the first condition:
\(\sum_{i=1}^{10} x_i - 20 = 30 \implies \sum_{i=1}^{10} x_i = 50\).
Thus, \(\bar{x} = \frac{50}{10} = 5\).
Next, we use the provided variance formula:
\(\text{Variance} = \frac{1}{10}\sum_{i=1}^{10}(x_i - \bar{x})^2 = \frac{4}{5}\).
Since \(\bar{x} = 5\), we have:
\(\frac{1}{10}\sum_{i=1}^{10}(x_i - 5)^2 = \frac{4}{5} \implies \sum_{i=1}^{10}(x_i - 5)^2 = 8\).
Now, using \(\sum_{i=1}^{10} (x_i - \beta)^2 = 98\), we apply the identity:
\(\sum_{i=1}^{10} (x_i - \beta)^2 = \sum_{i=1}^{10} (x_i - 5)^2 + 10(\beta - 5)^2\).
Substitute for \(\sum_{i=1}^{10} (x_i - 5)^2 = 8\):
\(98 = 8 + 10(\beta - 5)^2 \implies 90 = 10(\beta - 5)^2\).
Solving for \(\beta\):
\((\beta - 5)^2 = 9 \implies \beta - 5 = \pm 3\).
Given \(\beta > 2\), we have \(\beta = 8\).
The transformed observations are \(y_i = 2(x_i - 1) + 4\beta\). Thus:
\(y_i = 2x_i - 2 + 32 = 2x_i + 30\).
The mean \(\mu\) of \(y_i\) is:
\(\mu = \frac{1}{10} \sum_{i=1}^{10} y_i = 2\bar{x} + 30 = 2(5) + 30 = 40\).
The variance of \(y_i\) is:
\(\sigma^2 = (2^2) \times \frac{4}{5} = \frac{16}{5}\).
Finally, calculate \(\frac{\beta \mu}{\sigma^2}\):
\(\frac{8 \times 40}{\frac{16}{5}} = \frac{320}{\frac{16}{5}} = \frac{320 \times 5}{16} = 100\).
Thus, the answer is \(\boxed{100}\).
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