The number of observations in a group is 40. | vedclass

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(B) Let the sum of the first $10$ observations be $S_1$ and the sum of the remaining $30$ observations be $S_2$.Given that the average of the first $1

Vedclass · Accepted Answer

$\frac{15}{4}$

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(B) Let the sum of the first $10$ observations be $S_1$ and the sum of the remaining $30$ observations be $S_2$.Given that the average of the first $10$ observations is $4.5$,we have:$S_1 = 10 	imes 4.5 = 45$Given that the average of the remaining $30$ observations is $3.5$,we have:$S_2 = 30 	imes 3.5 = 105$The total sum of all $40$ observations is $S = S_1 + S_2 = 45 + 105 = 150$.The average of the whole group is $\frac{S}{40} = \frac{150}{40} = \frac{15}{4}$.

Marks obtained	No. of students
$0-10$	$2$
$10-20$	$18$
$20-30$	$30$
$30-40$	$45$
$40-50$	$35$
$50-60$	$20$
$60-70$	$6$
$70-80$	$3$

Class	Frequency
$10-20$	$180$
$20-30$	$82$
$30-40$	$34$
$40-50$	$180$
$50-60$	$136$
$60-70$	$23$
$70-80$	$50$

The number of observations in a group is $40$. If the average of the first $10$ observations is $4.5$ and the average of the remaining $30$ observations is $3.5$,then the average of the whole group is:

Similar Questions

Compute the median from the following table:

Marks obtained No. of students

$0-10$ $2$

$10-20$ $18$

$20-30$ $30$

$30-40$ $45$

$40-50$ $35$

$50-60$ $20$

$60-70$ $6$

$70-80$ $3$

If the mean of $3, 4, x, 7, 10$ is $6$,then the value of $x$ is

Find the median for the following distribution:
Class Frequency
$10-20$ $180$
$20-30$ $82$
$30-40$ $34$
$40-50$ $180$
$50-60$ $136$
$60-70$ $23$
$70-80$ $50$

$A$ student scores $75\%, 80\%,$ and $85\%$ in three subjects. If the marks of a fourth subject are added,what is the minimum possible average percentage?

If ${d_i}$ is the deviation of a class mark ${y_i}$ from $a$,the assumed mean,and ${f_i}$ is the frequency,if ${M_g} = x + \frac{1}{{\sum {f_i}}}(\sum {f_i}{d_i})$,then $x$ is:

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