Question

Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests:

Ravi:	25	50	45	30	70	42	36	48	35	60
Hashina:	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent? [NCERT EXEMPLAR]

Answer 1

For Ravi

Marks $\left(x_i\right)$	$d_i=x_i-45$	$d_i^2$
25	−20	400
50	5	25
45	0	0
30	−15	225
70	25	625
42	−3	9
36	−9	81
48	3	9
35	−10	100
60	15	225
	$\sum _{}{d}_i=-9$	$\sum _{}{d}_i^2=1699$

Mean, ${\overline{X}}_R=A+\frac{\sum _{}{d}_i}{10}=45+\frac{\left(-9\right)}{10}=44.1$

Standard deviation, ${\sigma }_R=\sqrt{\frac{\sum _{}{d}_i^2}{10}-{\left(\frac{\sum _{}{d}_i}{10}\right)}^2}=\sqrt{\frac{1699}{10}-{\left(\frac{-9}{10}\right)}^2}=\sqrt{169.09}=13.003$

Coefficicent of variation = $\frac{{\sigma }_R}{{\overline{X}}_R}\times 100=\frac{13.003}{44.1}\times 100=29.49$

For Hashina

Marks $\left(x_i\right)$	$d_i=x_i-55$	$d_i^2$
10	−45	2025
70	15	625
50	−5	25
20	−35	1225
95	40	1600
55	0	0
42	−13	169
60	5	25
48	−7	49
80	25	625
	$\sum _{}{d}_i=-20$	$\sum _{}{d}_i^2=6368$

Mean, ${\overline{X}}_H=A+\frac{\sum _{}{d}_i}{10}=55+\frac{\left(-20\right)}{10}=53$

Standard deviation, ${\sigma }_H=\sqrt{\frac{\sum _{}{d}_i^2}{10}-{\left(\frac{\sum _{}{d}_i}{10}\right)}^2}=\sqrt{\frac{6368}{10}-{\left(\frac{-20}{10}\right)}^2}=\sqrt{632.8}=25.16$

Coefficicent of variation = $\frac{{\sigma }_H}{{\overline{X}}_H}\times 100=\frac{25.16}{53}\times 100=47.47$

Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.