Q.
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 $1.14$.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | x - 2 | x | $x^2$ | $(x+1)^2$ | 2x | x + 1 |
Statistics
Solution:
First find $x$.
As given total number of students is $60$ i.e.,
$x-2+x+x^2+(x+1)^2+2 x+x+1=60$
$\Rightarrow 5 x-2+x^2+x^2+1+2 x+1=60 $
$ \Rightarrow 2 x^2+7 x=60 $
$\Rightarrow 2 x^2+7 x-60=0$
$ \Rightarrow 2 x^2+15 x-8 x-60=0$
$ \Rightarrow x(7 x+15)-4(2 x+15)=0$
$ \Rightarrow (2 x+15)(x-4)=0 $
$ x=-\frac{15}{2}, x=4 $
$x=4 (\because x $ is positive )
Marks$(x_i)$
Frequency $(f_i)$
$x_i^2$
$f_ix_i$
$f_ix_i^2$
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$
| Marks$(x_i)$ | Frequency $(f_i)$ | $x_i^2$ | $f_ix_i$ | $f_ix_i^2$ |
|---|---|---|---|---|
| 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 |