Question

Solve problem 4 by step-deviation method.

Answer 1

Take A= 55, h = 10

Marks (Class interval)	Class Marks $\left(x_i\right)$	$$\begin{gathered} d_i=x_i-A \\ {d}_i=x_i-55 \end{gathered}$$	$u_i=\frac{d_i}{h}=\frac{d_i}{10}$	No. of students$\left(f_i\right)$	$f_i{u}_i$
0-10	5	$-$50	$-$5	3	$-$15
10-20	15	$-$40	$-$4	5	$-$20
20-30	25	$-$30	$-$3	7	$-$21
30-40	35	$-$20	$-$2	10	$-$20
40-50	45	$-$10	$-$1	12	$-$12
50-60	$55\to A$	0	0	15	0
60-70	65	10	1	12	12
70-80	75	20	2	6	12
80-90	85	30	3	2	6
90-100	95	40	4	8	32
Total	-	-	-	$\sum _{}^{}{f}_i=80$	$\sum _{}^{}{f}_i{u}_i=-26$

Now, mean u_i s is can be calculated thus:
$$\begin{gathered} \overline{u}=\frac{{\displaystyle \sum _{}^{}}{f}_i{u}_i}{{\displaystyle \sum _{}^{}}{f}_i}=\frac{-26}{80}=-0.325 \\ \mathrm{Mean}=\overline{x}=A+h\overline{u} \\ =55+10(-0.325) \\ =55-3.25 \\ =51.75 \\ \therefore \mathrm{Mean}\mathrm{marks}\mathrm{scored}\mathrm{by}\mathrm{a}\mathrm{student}\mathrm{is}51.75. \end{gathered}$$