Given that the mean \(\bar{x} = 56\).
Given that the variance \(\sigma^2 = 66.2\).
Using the formula for variance with mean and variance values:
\[ \frac{\alpha^2 + \beta^2 + 25678}{10} - (56)^2 = 66.2 \]
Rearranging, we find:
\[ \alpha^2 + \beta^2 = 6344 \]
So, the correct answer is: 6344
Step 1: Use the formula for the mean.
Mean \( \bar{x} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
There are 10 observations: \[ \frac{65 + 68 + 58 + 44 + 48 + 45 + 60 + \alpha + \beta + 60}{10} = 56 \]
Simplify: \[ (65 + 68 + 58 + 44 + 48 + 45 + 60 + 60) + \alpha + \beta = 560 \] \[ 448 + \alpha + \beta = 560 \] \[ \boxed{\alpha + \beta = 112} \]
Variance \( \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \)
Given: \[ \sigma^2 = 66.2, \quad \bar{x} = 56, \quad n = 10 \] Substitute: \[ 66.2 = \frac{\sum x_i^2}{10} - 56^2 \] \[ \frac{\sum x_i^2}{10} = 66.2 + 3136 = 3202.2 \] \[ \sum x_i^2 = 32022 \]
Known data: \( 65, 68, 58, 44, 48, 45, 60, 60 \) \[ \sum x_i^2 = 65^2 + 68^2 + 58^2 + 44^2 + 48^2 + 45^2 + 60^2 + 60^2 + \alpha^2 + \beta^2 \]
Compute the known squares: \[ 65^2 = 4225, \quad 68^2 = 4624, \quad 58^2 = 3364, \quad 44^2 = 1936, \] \[ 48^2 = 2304, \quad 45^2 = 2025, \quad 60^2 = 3600 \] There are two 60’s → \( 2 \times 3600 = 7200 \)
Sum of known squares: \[ 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 7200 = 25678 \]
Now: \[ 25678 + \alpha^2 + \beta^2 = 32022 \] \[ \boxed{\alpha^2 + \beta^2 = 32022 - 25678 = 6344} \]
\[ \boxed{\alpha^2 + \beta^2 = 6344} \]
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 $x_1, x_2, \ldots, x_{10}$ 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:
Let \( \{ W(t) : t \geq 0 \} \) be a standard Brownian motion. Then \[ E\left( (W(2) + W(3))^2 \right) \] equals _______ (answer in integer).

Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
If \( z \) is a complex number and \( k \in \mathbb{R} \), such that \( |z| = 1 \), \[ \frac{2 + k^2 z}{k + \overline{z}} = kz, \] then the maximum distance from \( k + i k^2 \) to the circle \( |z - (1 + 2i)| = 1 \) is: