Question

Consider 10 observations $x_1, x_2, \ldots, x_{10}$ such that \[\sum_{i=1}^{10} (x_i - \alpha) = 2 \quad \text{and} \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 40,\]where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$, respectively. The value of $\frac{\beta}{\alpha}$ is equal to:

• A. 2
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. 1

Answer 1

The Correct Option is A

To address the problem, we must analyze the 10 observations \(x_1, x_2, \ldots, x_{10}\) concerning their mean and variance, and their relationships with \(\alpha\) and \(\beta\).

1. Given:

\[\sum_{i=1}^{10} (x_i - \alpha) = 2\]

1. and

\[\sum_{i=1}^{10} (x_i - \beta)^2 = 40\]

1. .
2. The mean of the observations is

\[\bar{x} = \frac{6}{5}\]

1. . From the definition of the mean,

\[\bar{x} = \frac{1}{10} \sum_{i=1}^{10} x_i = \frac{6}{5}\]

1. , it follows that

\[\sum_{i=1}^{10} x_i = 12\]

1. .
2. Using the property of summation with respect to the mean,

\[\sum_{i=1}^{10} (x_i - \alpha) = \sum_{i=1}^{10} x_i - 10\alpha = 2\]

1. . Substituting \(\sum_{i=1}^{10} x_i = 12\):

\[12 - 10\alpha = 2\]

1. . Solving for \(\alpha\):

\[10\alpha = 10 \Rightarrow \alpha = 1\]

1. .
2. The variance of the observations is

\[\text{Variance} = \frac{84}{25}\]

1. . The variance formula is

\[\frac{1}{10} \sum_{i=1}^{10} (x_i - \bar{x})^2 = \frac{84}{25}\]

1. . This implies

\[\sum_{i=1}^{10} (x_i - \bar{x})^2 = \frac{84}{2.5} = 33.6\]

1. .
2. We are also given

\[\sum_{i=1}^{10} (x_i - \beta)^2 = 40\]

1. . This can be expressed as

\[\sum_{i=1}^{10} ((x_i - \bar{x}) + (\bar{x} - \beta))^2 = 40\]

1. . Expanding this, we use the relationship between the sum of squared deviations from \(\bar{x}\) and \(\beta\):

\[\sum_{i=1}^{10} (x_i - \bar{x})^2 + 2 \sum_{i=1}^{10} (x_i - \bar{x})(\bar{x} - \beta) + \sum_{i=1}^{10} (\bar{x} - \beta)^2 = 40\]

1. . Since \(\sum_{i=1}^{10} (x_i - \bar{x}) = 0\), the middle term is zero.

\[33.6 + 10 (\bar{x} - \beta)^2 = 40\]

1. .
2. Solving for \(\beta\):

\[10 (\beta - \bar{x})^2 = 40 - 33.6 = 6.4\]

1. .

\[(\beta - \frac{6}{5})^2 = 0.64\]

1. .

\[\beta - \frac{6}{5} = \pm \sqrt{0.64} = \pm \frac{4}{5}\]

1. .
2. This gives two possible values for \(\beta\):

\[\beta = \frac{6}{5} \pm \frac{4}{5}\]

1. . The possible values are \(\beta = \frac{10}{5} = 2\) and \(\beta = \frac{2}{5}\).
2. Given that \(\alpha\) is an integer (\(\alpha=1\)) and assuming \(\beta\) is also an integer, we select \(\beta = 2\).
3. The ratio \(\frac{\beta}{\alpha}\) is therefore

\[\frac{\beta}{\alpha} = \frac{2}{1} = 2\]

1. .

The final result is: 2.