The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are $\mathrm{5,7},\mathrm{10,12,14,15}$ , then the absolute difference of the remaining two observations is:

Let the mean and the variance of 5 observations $x_1,x_2,x_3,x_4,x_{5}be\frac{24}{5}and\frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and a respectively, then (4a + x₅) is equal to:

The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38{\sigma }^2$ is equal to………

Let the mean and the variance of 20 observations x1, x2,…….x20 be 15 and 9, respectively. For $\in R$ , if the mean of ${\left(x_1+\alpha \right)}^2,{\left(x_2+\alpha \right)}^2,....,{\left(x_{20}+\alpha \right)}^2$ is 178, then the square of the maximum value of is equal to………….

If the variance of the following frequency distribution :

$\begin{array}{lllll}\text{Class}& :& 10-20& 20-30& 30-40\\ \text{Frequency}& :& 2& x& 2\end{array}$ is 50 , then $x$ is equal to:

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $\frac{9}{4}$ respectively. Then $x^4+y^2$ is equal to.