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 $4$ observations are $\frac{7}{2}$ and $a$ respectively,then $(4a + x_{5})$ is equal to

For the set $A = \{x_1, x_2, x_3, x_4, x_5\}$,the variance is $4$ and the mean is $2$. For the set $B = \{y_1, y_2, y_3, y_4, y_5\}$,the variance is $5$ and the mean is $4$. Then,the variance of $A \cup B$ is

For two data sets,each of size $5$,the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$,respectively. The variance of the combined data set is

If the mean and the variance of the data are $\mu$ and $19$ respectively, then the value of $\lambda+\mu$ is:

Class	$4-8$	$8-12$	$12-16$	$16-20$
Frequency	$3$	$\lambda$	$4$	$7$

Marks obtained by all the students of class $12$ are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be $14$ with median class interval $12-18$ and median class frequency $12$. If the number of students whose marks are less than $12$ is $18$,then the total number of students is

What is the weighted mean of the first $n$ natural numbers when the weights are equal to the corresponding natural numbers?