1. Calculate Moment about mean from the following grouped data

X	Frequency
0	1
1	5
2	10
3	6
4	3

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

`=55/25`

`=2.2`

`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)`
0	1	0	-2.2	-2.2	4.84	-10.648	23.4256
1	5	5	-1.2	-6	7.2	-8.64	10.368
2	10	20	-0.2	-2	0.4	-0.08	0.016
3	6	18	0.8	4.8	3.84	3.072	2.4576
4	3	12	1.8	5.4	9.72	17.496	31.4928
---	---	---	---	---	---	---	---
--	`n=25`	`sum f*x=55`	--	`=0`	`=26`	`=1.2`	`=67.76`

Now, calculate Central Moments

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

`=(0)/(25)`

`=0`

---

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

`=(26)/(25)`

`=1.04`

---

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

`=(1.2)/(25)`

`=0.048`

---

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

`=(67.76)/(25)`

`=2.7104`

---

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

`=(0.048)^2/(1.04)^3`

`=(0.0023)/(1.1249)`

`=0.002`

---

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

`=(2.7104)/(1.04)^2`

`=(2.7104)/(1.0816)`

`=2.5059`

---

Moment coefficient of skewness
`beta_1>0` : The distribution is positively skewed (a longer tail to the right).

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

---

2. 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)

---

3. Calculate Moment about mean from the following grouped data

Class	Frequency
2 - 4	3
4 - 6	4
6 - 8	2
8 - 10	1

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

`=52/10`

`=5.2`

Class `(1)`	Mid value (`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)`
2 - 4	3	3	9	-2.2	-6.6	14.52	-31.944	70.2768
4 - 6	5	4	20	-0.2	-0.8	0.16	-0.032	0.0064
6 - 8	7	2	14	1.8	3.6	6.48	11.664	20.9952
8 - 10	9	1	9	3.8	3.8	14.44	54.872	208.5136
---	---	---	---	---	---	---	---	---
--	--	`n=10`	`sum f*x=52`	--	`=0`	`=35.6`	`=34.56`	`=299.792`

Now, calculate Central Moments

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

`=(0)/(10)`

`=0`

---

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

`=(35.6)/(10)`

`=3.56`

---

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

`=(34.56)/(10)`

`=3.456`

---

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

`=(299.792)/(10)`

`=29.9792`

---

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

`=(3.456)^2/(3.56)^3`

`=(11.9439)/(45.118)`

`=0.2647`

---

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

`=(29.9792)/(3.56)^2`

`=(29.9792)/(12.6736)`

`=2.3655`

---

Moment coefficient of skewness
`beta_1>0` : The distribution is positively skewed (a longer tail to the right).

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

---

4. Calculate Moment about mean from the following grouped data

Class	Frequency
0 - 2	5
2 - 4	16
4 - 6	13
6 - 8	7
8 - 10	5
10 - 12	4

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

`=256/50`

`=5.12`

Class `(1)`	Mid value (`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)`
0 - 2	1	5	5	-4.12	-20.6	84.872	-349.6726	1440.6513
2 - 4	3	16	48	-2.12	-33.92	71.9104	-152.45	323.1941
4 - 6	5	13	65	-0.12	-1.56	0.1872	-0.0225	0.0027
6 - 8	7	7	49	1.88	13.16	24.7408	46.5127	87.4439
8 - 10	9	5	45	3.88	19.4	75.272	292.0554	1133.1748
10 - 12	11	4	44	5.88	23.52	138.2976	813.1899	4781.5565
---	---	---	---	---	---	---	---	---
--	--	`n=50`	`sum f*x=256`	--	`=0`	`=395.28`	`=649.6128`	`=7766.0233`

Now, calculate Central Moments

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

`=(0)/(50)`

`=0`

---

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

`=(395.28)/(50)`

`=7.9056`

---

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

`=(649.6128)/(50)`

`=12.9923`

---

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

`=(7766.0233)/(50)`

`=155.3205`

---

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

`=(12.9923)^2/(7.9056)^3`

`=(168.7987)/(494.0882)`

`=0.3416`

---

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

`=(155.3205)/(7.9056)^2`

`=(155.3205)/(62.4985)`

`=2.4852`

---

Moment coefficient of skewness
`beta_1>0` : The distribution is positively skewed (a longer tail to the right).

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

---

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