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.


1
Expert's answer
2021-01-04T19:56:02-0500

a)

Mean value  is  m=2+5+7+8+105=6.4μ1=(Xm)15=0μ2=(Xm)25=(26.4)2+(56.4)2+(76.4)2+(86.4)2+(106.4)25=7.44μ3=(Xm)35=7.392μ4=(Xm)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=(X0)15=6.4m2=(X0)25=48.4m3=(X0)35=397.6m4=(X0)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=(X8)15=1.6m2=(X8)25=10m3=(X8)35=47.2m4=(X8)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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