To solve the problem, we need to recall the empirical relationship that connects the mode, median, and mean of a moderately skewed distribution.
1. Understanding the Empirical Relation:
In statistics, for a moderately skewed distribution, the following empirical relationship holds true:
\( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)
2. Rearranging the Equation:
We can rearrange the above formula to express the relation in a different form:
\( \text{Mode} - \text{Median} = 2 \times (\text{Median} - \text{Mean}) \)
3. Identifying the Correct Option:
This matches option (2) exactly:
\( \text{Mode} - \text{Median} = 2 \times (\text{Median} - \text{Mean}) \)
Final Answer:
The correct empirical relation is (B): \({\text{Mode} - \text{Median} = 2 (\text{Median} - \text{Mean})} \).
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:
For a statistical data \( x_1, x_2, \dots, x_{10} \) of 10 values, a student obtained the mean as 5.5 and \[ \sum_{i=1}^{10} x_i^2 = 371. \] He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively.
The variance of the corrected data is:
Let \( \{ W(t) : t \geq 0 \} \) be a standard Brownian motion. Then \[ E\left( (W(2) + W(3))^2 \right) \] equals _______ (answer in integer).