1. Calculate Moment about mean from the following data
`10,50,30,20,10,20,70,30`

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

`=(10+50+30+20+10+20+70+30)/8`

`=240/8`

`=30`

`x`	`(x-bar x)` `=(x-30)`	`(x-bar x)^2` `=(x-30)^2`	`(x-bar x)^3` `=(x-30)^3`	`(x-bar x)^4` `=(x-30)^4`
10	-20	400	-8000	160000
50	20	400	8000	160000
30	0	0	0	0
20	-10	100	-1000	10000
10	-20	400	-8000	160000
20	-10	100	-1000	10000
70	40	1600	64000	2560000
30	0	0	0	0
---	---	---	---	---
`240`	`0`	`3000`	`54000`	`3060000`

Now, calculate Central Moments

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

`=(0)/(8)`

`=0`

---

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

`=(3000)/(8)`

`=375`

---

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

`=(54000)/(8)`

`=6750`

---

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

`=(3060000)/(8)`

`=382500`

---

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

`=(6750)^2/(375)^3`

`=(45562500)/(52734375)`

`=0.864`

---

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

`=(382500)/(375)^2`

`=(382500)/(140625)`

`=2.72`

---

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 data
`85,96,76,108,85,80,100,85,70,95`

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

`=(85+96+76+108+85+80+100+85+70+95)/10`

`=880/10`

`=88`

`x`	`(x-bar x)` `=(x-88)`	`(x-bar x)^2` `=(x-88)^2`	`(x-bar x)^3` `=(x-88)^3`	`(x-bar x)^4` `=(x-88)^4`
85	-3	9	-27	81
96	8	64	512	4096
76	-12	144	-1728	20736
108	20	400	8000	160000
85	-3	9	-27	81
80	-8	64	-512	4096
100	12	144	1728	20736
85	-3	9	-27	81
70	-18	324	-5832	104976
95	7	49	343	2401
---	---	---	---	---
`880`	`0`	`1216`	`2430`	`317284`

Now, calculate Central Moments

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

`=(0)/(10)`

`=0`

---

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

`=(1216)/(10)`

`=121.6`

---

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

`=(2430)/(10)`

`=243`

---

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

`=(317284)/(10)`

`=31728.4`

---

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

`=(243)^2/(121.6)^3`

`=(59049)/(1798045.696)`

`=0.0328`

---

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

`=(31728.4)/(121.6)^2`

`=(31728.4)/(14786.56)`

`=2.1458`

---

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)

---

3. Calculate Moment about mean from the following data
`3,23,13,11,15,5,4,2`

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

`=(3+23+13+11+15+5+4+2)/8`

`=76/8`

`=9.5`

`x`	`(x-bar x)` `=(x-9.5)`	`(x-bar x)^2` `=(x-9.5)^2`	`(x-bar x)^3` `=(x-9.5)^3`	`(x-bar x)^4` `=(x-9.5)^4`
3	-6.5	42.25	-274.625	1785.0625
23	13.5	182.25	2460.375	33215.0625
13	3.5	12.25	42.875	150.0625
11	1.5	2.25	3.375	5.0625
15	5.5	30.25	166.375	915.0625
5	-4.5	20.25	-91.125	410.0625
4	-5.5	30.25	-166.375	915.0625
2	-7.5	56.25	-421.875	3164.0625
---	---	---	---	---
`76`	`0`	`376`	`1719`	`40559.5`

Now, calculate Central Moments

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

`=(0)/(8)`

`=0`

---

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

`=(376)/(8)`

`=47`

---

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

`=(1719)/(8)`

`=214.875`

---

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

`=(40559.5)/(8)`

`=5069.9375`

---

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

`=(214.875)^2/(47)^3`

`=(46171.2656)/(103823)`

`=0.4447`

---

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

`=(5069.9375)/(47)^2`

`=(5069.9375)/(2209)`

`=2.2951`

---

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)

---

---