The average marks of boys in a class is $40$ and that of girls is $45$. The average marks of both boys and girls combined is $42$. Then the percentage of boys in the class is (in $\%$)

The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35$. If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for the correct data,then $(\alpha, \beta)$ is:

The mean of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. If an observation $x_{q}$ is replaced by $x_{q}^{\prime}$, then the new mean is:

An analysis of monthly wages paid to workers in two firms $A$ and $B$,belonging to the same industry,gives the following results:

\text{Parameter}	\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}	$A: 586, B: 648$
\text{Mean of monthly wages}	$Rs. 5253$
\text{Variance of wages}	$A: 100, B: 121$

Which firm,$A$ or $B$,pays a larger total amount as monthly wages?

The following data was collected from a newspaper: (percentage distribution). It is an example of

$Country$	$Agriculture$	$Industries$	$Services$	$Others$
$India$	$45$	$19$	$28$	$4$
$U.K.$	$3$	$40$	$44$	$13$
$Japan$	$6$	$48$	$43$	$3$
$U.S.A.$	$3$	$35$	$61$	$1$

The mean and standard deviation of $100$ observations are $40$ and $5.1$,respectively. By mistake,one observation is taken as $50$ instead of $40$. If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively,then $10(\mu+\sigma)$ is equal to