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 $.........$

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

If the mean and the variance of $6, 4, a, 8, b, 12, 10, 13$ are $9$ and $9.25$ respectively,then $a+b+ab$ is equal to:

Let $n \geq 3$. $A$ list of numbers $x_1, x_2, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. $A$ new list of numbers $y_1, y_2, \ldots, y_n$ is made as follows: $y_1 = \frac{x_1+x_2}{2}$,$y_2 = \frac{x_1+x_2}{2}$ and $y_j = x_j$ for $j = 3, 4, \ldots, n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?

Which one of the following is false?

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	$3$	$\lambda$	$4$	$7$