Answer to Question #153614 in Statistics and Probability for nitin sher

Question

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.

Answer 1

a)

$Mean~value~~is~~m = \frac{2+5+7+8+10}{5}=6.4\\ \mu_1=\frac{\sum{(X-m)^1}}{5}=0\\ \mu_2=\frac{\sum{(X-m)^2}}{5}=\\\frac{(2-6.4)^2+(5-6.4)^2+(7-6.4)^2+(8-6.4)^2+(10-6.4)^2}{5}=7.44\\ \mu_3=\frac{\sum{(X-m)^3}}{5}=-7.392\\ \mu_4=\frac{\sum{(X-m)^4}}{5}=110.6592$

b)

$m_1=\frac{\sum{(X-0)^1}}{5}=6.4\\ m_2=\frac{\sum{(X-0)^2}}{5}=48.4\\ m_3=\frac{\sum{(X-0)^3}}{5}=397.6\\ m_4=\frac{\sum{(X-0)^4}}{5}=3427.6$

c)

$m_1=\frac{\sum{(X-8)^1}}{5}=-1.6\\ m_2=\frac{\sum{(X-8)^2}}{5}=10\\ m_3=\frac{\sum{(X-8)^3}}{5}=-47.2\\ m_4=\frac{\sum{(X-8)^4}}{5}=278.8$