To solve this problem, we will use the empirical relationship between mean, median, and mode typically observed in a moderately skewed distribution.
This relationship is given by: Mode = 3 \(\times\) Median - 2 \(\times\) Mean
Let's denote:
According to the problem, the difference between mode and mean is k times the difference between median and mean:
(Mo - X̄) = k(Me - X̄)
Substitute the empirical relationship into the equation:
(3Me - 2X̄ - X̄) = k(Me - X̄)
3Me - 3X̄ = kMe - kX̄
Rearranging terms gives us:
3Me - kMe = 3X̄ - kX̄
Me(3 - k) = X̄(3 - k)
Since Me and X̄ are not equal in a skewed distribution, we cannot cancel (3 - k).
Therefore, if 3 - k = 0, k = 3.
Thus, the value of k is 3.
Given: The difference between mode and mean of a data is \( k \) times the difference between median and mean.
Step 1: Recall the Empirical Formula
The empirical relationship between mode, mean, and median is:
\[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \]
Rearranging this equation:
\[ \text{Mode} - \text{Mean} = 3 \times (\text{Median} - \text{Mean}) \]
Comparing with the given condition:
\[ k = 3 \]
Final Answer: \( \mathbf{3} \)
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).