Step 1: Understanding Ogive Intersection
In statistics, the intersection of the less than ogive and the more than ogive gives the median of the data.
Step 2: Identify the Median
From the given data, the two ogives intersect at \( (15.5, 20) \). The x-coordinate of this point represents the median.
\[ \text{Median} = 15.5 \]
Final Answer: \( \mathbf{15.5} \)
The point where the 'less than ogive' and 'more than ogive' intersect gives the median of the data on the x-axis.
The y-coordinate represents the cumulative frequency at the median.
Given that the intersection point is (15.5, 20), the x-coordinate of this point is 15.5.
Therefore, the median of the data is 15.5.
Final Answer: The final answer is $\boxed{15.5}$
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).