Question
Download Solution PDFFor a distribution, the first four moments about the origin are -1.3, 15, -26 and 102. The third moment about the mean is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe first, second, third and fourth moments about the mean are given as \(\mu_1 =(-1.3), \mu_2=15, \mu_3=(-26), \mu_4=102\); where \(\mu_1 = 0\), \(\mu_2 = \mu'_2 - (\mu'_1)^2\), and, \(\mu_3 = \mu'_3 - 3\mu'_2\mu'_1 + 2(\mu'_1)^3\). Therefore, plugging the information into the equation of the third moment about the mean, we have
\(\mu_3 = \mu'_3 - 3\mu'_2\mu'_1 + 2(\mu'_1)^3\)
or, \(\mu_3 = (-26) - 3 \times 15 \times (-1.3) + 2(-1.3)^3\)
or, \(\mu_3 = -26 + 58.5 -4.394\)
or, \(\mu_3 \approx 28.1\)
Hence, the third moment about the mean is 28.1
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