Let the mean and the variance of $6$ observations $a, b, 68, 44, 48, 60$ be $55$ and $194$,respectively. If $a > b$,then the value of $a + 3b$ is:

In a data set with $15$ observations $x_1, x_2, x_3, \ldots, x_{15}$,we are given $\sum_{i=1}^{15} x_i^2 = 3600$ and $\sum_{i=1}^{15} x_i = 175$. If the value of one observation $20$ was found to be incorrect and was replaced by its correct value $40$,then the corrected variance of the data is:

The mean of the numbers $a, b, 8, 5$ and $10$ is $6$ and the variance is $6.80$. Then the possible values of $a$ and $b$ are:

The standard deviation of the following distribution is:

Class interval	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$1$	$3$	$4$	$2$

Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_i-2)=30$,$\sum_{i=1}^{10}(x_i-\beta)^2=98$,$\beta > 2$ 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:

If the mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.80$,then which of the following is a possible value for $a$ and $b$?