1. Calculate Moment about mean from the following grouped data

X	Frequency
10	3
11	12
12	18
13	12
14	3

Solution:
Moments :
Mean `bar x=(sum f x)/n`

`=576/48`

`=12`

`x` `(2)`	`f` `(3)`	`f*x` `(4)=(2)xx(3)`	`(x-bar x)` `(5)`	`f*(x-bar x)` `(6)=(3)xx(5)`	`f*(x-bar x)^2` `(7)=(5)xx(6)`	`f*(x-bar x)^3` `(8)=(5)xx(7)`	`f*(x-bar x)^4` `(9)=(5)xx(8)`
10	3	30	-2	-6	12	-24	48
11	12	132	-1	-12	12	-12	12
12	18	216	0	0	0	0	0
13	12	156	1	12	12	12	12
14	3	42	2	6	12	24	48
---	---	---	---	---	---	---	---
--	`n=48`	`sum f*x=576`	--	`=0`	`=48`	`=0`	`=120`

Now, calculate Central Moments

First Central Moment
`m_1=(sum f*(x-bar x))/n`

`=(0)/(48)`

`=0`

---

Second Central Moment
`m_2=(sum f*(x-bar x)^2)/n`

`=(48)/(48)`

`=1`

---

Third Central Moment
`m_3=(sum f*(x-bar x)^3)/n`

`=(0)/(48)`

`=0`

---

Fourth Central Moment
`m_4=(sum f*(x-bar x)^4)/n`

`=(120)/(48)`

`=2.5`

---

Skewness `beta_1=(m_3)^2/(m_2)^3`

`=(0)^2/(1)^3`

`=(0)/(1)`

`=0`

---

Kurtosis `beta_2=(m_4)/(m_2)^2`

`=(2.5)/(1)^2`

`=(2.5)/(1)`

`=2.5`

---

Moment coefficient of skewness
`beta_1=0` : The distribution is perfectly symmetrical (like a normal distribution).

Moment coefficient of kurtosis
`beta_2<3` : platykurtic (flatter with lighter tails)