Question

Determine the mean and standard deviation for the following distribution :

Marks	Frequency
2	1
3	6
4	6
5	8
6	8
7	2
8	2
9	3
10	0
11	2
12	1
13	0
14	0
15	0
16	1

Answer 1

$x_i$	$f_i$	$f_i{x}_i$
2	1	2
3	6	18
4	6	24
5	8	40
6	8	48
7	2	14
8	2	16
9	3	27
10	0	0
11	2	22
12	1	12
13	0	0
14	0	0
15	0	0
16	1	16

$\sum f_i=40$

$\sum f_i{x}_i=239$

Mean $\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}$

$\left(\overline{x}\right)=\frac{239}{40}$

$\left(\overline{x}\right)=5.975\approx 6$

$x_i$	$f_i$	$d_i$
2	1	-4
3	6	-3
4	6	-2
5	8	-1
6	8	0
7	2	1
8	2	2
9	3	3
10	0	4
11	2	5
12	1	6
13	0	7
14	0	8
15	0	9
16	1	10

$\sum f_i=40$ $\overline{x}\approx 6$

$d_i=x_i-\overline{x}$

$f_i$	$d_i$	$f_i{d}_i$
1	-4	-4
6	-3	-18
6	-2	-12
8	-1	-8
8	0	0
2	1	2
2	2	4
3	3	9
0	4	0
2	5	10
1	6	6
0	7	0
0	8	0
0	9	0
1	10	10

$\sum f_i=40$ $\overline{x}\approx 6$

$d_i=x_i-\overline{x}$

$\sum f_i{d}_i=-1$

$f_i$	$d_i$	$f_i{d}_i^2$
1	-4	16
6	-3	54
6	-2	24
8	-1	8
8	0	0
2	1	2
2	2	8
3	3	27
0	4	0
2	5	50
1	6	36
0	7	0
0	8	0
0	9	0
1	10	100

$\sum f_i=40$ $\overline{x}\approx 6$

$d_i=x_i-\overline{x}$

$\sum f_i{d}_i=-1$

$\sum f_i{d}_i^2=325$

$\sigma =\sqrt{\frac{\sum f_i{d}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)}^2}$

$\sigma =\sqrt{\frac{325}{40}-{\left(\frac{-1}{40}\right)}^2}$

$\sigma =2.85$

Hence, mean and standard deviation of the given distribution are 5.975 $(\approx 6)$ & 2.85 respectively.