| Abdullah Al Mahmud | www.statmania.info |
Can uniquely characterize a distribution
Central Moments: μr=∑(xi−ˉx)rn
μ′r=∑(xi−a)rn; a is arbitrary number
μ′1=∑(xi−a)n=∑xin−nan=ˉx−a
How to Remember??
↓
(a−b)3=a3−3a2b+3ab2−b3
(a−b)4=a4−4a3b+6a2b2−4ab3+b4 (Pascal triangle can be used)
ar=μ′r
br=μ′r1
Now,
μ3=μ′3−3μ′2μ′1+2μ′31
From μ′r(a) to μ′r(k)
Assume, ar=μ′r(a), b=a−k
Binomial Formulae
Use k=ˉx
That’s it! DO NOT MEMORIZE!
Lack of symmetry
Normal distribution →
Pearson’s Coefficient: SKP=Mean−Modeσ=3(Mean−Median)σ;(−3,3)
(Mode=3Me−2ˉX)
Estimate Skewness
4, 23, 55, 70, 74, 78, 86, 89
γ2=β2−3
γ2=0 or β2=3)→ Mesokurtik γ2>0 or β2>3)→ Leptokurtik γ2<0 or β2<3)→ Platykurtik
9, 7, 8, 6 → Find kurtosis