Question

If there are 60 students in a class and the following is the frequency distribution of the marks obtained by the students in a test.

Marks	0	1	2	3	4	5
Frequency	x - 2	x	$x^2$	$(x+1{)}^2$	2x	x + 1

Where, $x$ is a positive integer.
Which of the following are true?
I. Mean of the marks is 3.
II. Mean of the marks is $2.8$.
III. Variance of the marks is $1.12$.
IV. Variance of the marks is $1.14$.

• A. I and III are true
• B. II and III are true
• C. I and IV are true
• D. II and IV are true

Answer 1

Answer: (B)

First find $x$.
As given total number of students is $60$ i.e.,
$x-2+x+x^2+(x+1{)}^2+2x+x+1=60$
$\Rightarrow 5x-2+x^2+x^2+1+2x+1=60$
$\Rightarrow 2x^2+7x=60$
$\Rightarrow 2x^2+7x-60=0$
$\Rightarrow 2x^2+15x-8x-60=0$
$\Rightarrow x(7x+15)-4(2x+15)=0$
$\Rightarrow (2x+15)(x-4)=0$
$x=-\frac{15}{2},x=4$
$x=4(\because x$ is positive )

Marks$(x_i)$	Frequency $(f_i)$	$x_i^2$	$f_i{x}_i$	$f_i{x}_i^2$
0	2	0	0	0
1	4	1	4	4
2	16	4	32	64
3	25	9	75	225
4	8	16	32	128
5	5	25	25	125
Total	60		168	546

Mean $(\bar{x})=\frac{\sum f_i{x}_i}{N}=\frac{168}{60}=2.8$
$\therefore$ Variance $\left({\sigma }^2\right)=\sqrt{\frac{\Sigma f_i{x}_i^2}{N}-(\bar{x}{)}^2}=\sqrt{\frac{546}{60}-(2.8{)}^2}$
$=\sqrt{9.1-7.84}=\sqrt{1.26}=1.12$