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`

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

`=(3000)/(8)`

`=375`

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

`=(54000)/(8)`

`=6750`

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

`=(3060000)/(8)`

`=382500`

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

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

`=(45562500)/(52734375)`

`=0.864`

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

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

`=(382500)/(140625)`

`=2.72`

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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)