Step 1: List the First 10 Prime Numbers
The first 10 prime numbers are:
\( 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 \)
Step 2: Find the Median
Since there are 10 numbers (even count), the median is the average of the 5th and 6th terms:
\[ \text{Median} = \frac{11 + 13}{2} = \frac{24}{2} = 12 \]
Final Answer: \( \mathbf{12} \)
The first step in solving the problem is to identify the first 10 prime numbers.
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers.
The first 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
To find the median of these numbers, we must first arrange them in order; however, as they are already in ascending order, we can proceed to find the median. Since there are 10 numbers, an even count, the median will be the average of the 5th and 6th numbers.
In this list, the 5th number is 11, and the 6th number is 13.
Therefore, the median is calculated as follows:
Median = (11 + 13) / 2 = 24 / 2 = 12
Thus, the median of the first 10 prime numbers is 12.
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).