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

$A$ problem in statistics is given to three students $P, Q$ and $R$. Their chances of solving the problem are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. If all of them try independently,then the probability that the problem is solved 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	$3$	$\lambda$	$4$	$7$

Consider the frequency distribution of the given numbers. If the mean is known to be $3$,then the value of $f$ is:

Value	$1$	$2$	$3$	$4$
Frequency	$5$	$4$	$6$	$f$

Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$ (where $\alpha < \beta$) be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is:

In a Mathematics test,the average marks of boys is $x \%$ and the average marks of girls is $y \%$ with $x \neq y$. If the average marks of all students is $z \%$,the ratio of the number of girls to the total number of students is