Question

Find the mean marks of the students from the following cumulative frequency distribution:

Marks:	Below 10	Below 20	Below 30	Below 40	Below 50	Below 60	Below 70	Below 80	Below 90	Below 100
Number of students:	5	9	17	29	45	60	70	78	83	85

Answer 1

Here we have, cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution. We observe that the number of students getting marks less than $10$ is $5$ and $9$ students have secured marks less than $20$. Therefore, number of students getting marks between $10$ and $20$ ( inclusive $0$ and exclusive 10) is $9-5=4$. Similarly, the number of students getting marks between $20$ and $30$ is $17-9=8$ and so on. Thus, we have the following distribution frequency distribution:

Marks:	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Number of students:	$5$	$4$	$8$	$12$	$16$	$15$	$10$	$8$	$5$	$2$

Let us Let us now compute arithmetic mean by taking $55$ and the assumed mean.
Computation of Mean

Marks	Mid-value	Frequency	$u_i\frac{x_i-55}{10}$	$f_i{u}_i$
$0-10$	$5$	$5$	$-5$	$-25$
$10-20$	$10$	$4$	$-4$	$-16$
$20-30$	$25$	$8$	$-3$	$-24$
$30-40$	$35$	$12$	$-2$	$-24$
$40-50$	$45$	$16$	$-1$	$-16$
$50-60$	$55$	$15$	$0$	$0$
$60-70$	$65$	$10$	$1$	$10$
$70-80$	$75$	$8$	$2$	$16$
$80-90$	$85$	$5$	$3$	$15$
$90-100$	$95$	$2$	$4$	$8$
Total		$N=\sum f_i=85$		$\sum f_i{u}_i=-56$

We have,
N=$N=\sum f_i=85,\sum f_i{u}_i=-56,h=10andA=55$
$\therefore \overline{X}=A+h\left(\frac{1}{N}\sum f_i{u}_i\right)$
$\Rightarrow \overline{X}=55+10\times \frac{-56}{85}=55-6.59=48.41$Marks.
Hence, mean marks scored by the students = $48.41$.