Question #153614

With equal probability, the observations 5, 10, 8, 2, and 7 show the number of defective

units found during five inspections in a laboratory. Find (a) the first four central moments,

(b) central moments from the moments about origin and (c) central moments from the

moments about an arbitrary value 8.


Expert's answer

a)

Mean value  is  m=2+5+7+8+105=6.4μ1=∑(X−m)15=0μ2=∑(X−m)25=(2−6.4)2+(5−6.4)2+(7−6.4)2+(8−6.4)2+(10−6.4)25=7.44μ3=∑(X−m)35=−7.392μ4=∑(X−m)45=110.6592Mean~value~~is~~m = \frac{2+5+7+8+10}{5}=6.4\\ \mu_1=\frac{\sum{(X-m)^1}}{5}=0\\ \mu_2=\frac{\sum{(X-m)^2}}{5}=\\\frac{(2-6.4)^2+(5-6.4)^2+(7-6.4)^2+(8-6.4)^2+(10-6.4)^2}{5}=7.44\\ \mu_3=\frac{\sum{(X-m)^3}}{5}=-7.392\\ \mu_4=\frac{\sum{(X-m)^4}}{5}=110.6592

b)

m1=∑(X−0)15=6.4m2=∑(X−0)25=48.4m3=∑(X−0)35=397.6m4=∑(X−0)45=3427.6m_1=\frac{\sum{(X-0)^1}}{5}=6.4\\ m_2=\frac{\sum{(X-0)^2}}{5}=48.4\\ m_3=\frac{\sum{(X-0)^3}}{5}=397.6\\ m_4=\frac{\sum{(X-0)^4}}{5}=3427.6

c)

m1=∑(X−8)15=−1.6m2=∑(X−8)25=10m3=∑(X−8)35=−47.2m4=∑(X−8)45=278.8m_1=\frac{\sum{(X-8)^1}}{5}=-1.6\\ m_2=\frac{\sum{(X-8)^2}}{5}=10\\ m_3=\frac{\sum{(X-8)^3}}{5}=-47.2\\ m_4=\frac{\sum{(X-8)^4}}{5}=278.8


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