The first four moments of a distribution about the origin are -1.5, 17, -30 and 108. The third moment about the mean is:
39.75
The third central moment in terms of moments about the origin is:
\[\mu_3=\nu_3-3\nu_2\nu_1+2\nu_1^{3}\]
Step 1 — Substitute \(\nu_1=-1.5,\ \nu_2=17,\ \nu_3=-30\):
\[\mu_3=-30-3(17)(-1.5)+2(-1.5)^{3}\]
Step 2 — Evaluate each term:
\[3(17)(-1.5)=-76.5,\quad 2(-1.5)^{3}=2(-3.375)=-6.75\]
Step 3 — Combine:
\[\mu_3=-30+76.5-6.75=39.75\]
Hence the third moment about the mean is 39.75.
Similar Questions
The first four moments of a distribution about the value 2 are 1, 2.5, 5.5 and 16. What is the 4th moment about zero?
Given below are moments about an arbitrary origin 5.
If μ'1 = -4, μ'2 = 22, μ'3 = -117 and μ'4 = 560, then μ3 is equal to:The third order central moment of normal distribution is:
If μ'r and μr, respectively, denote rth order moments about origin and mean, then:
The standard deviation of a symmetrical distribution is 3. What must be the value of the fourth moment about the mean for the distribution to be mesokurtic?
Important Questions from Central Moments
The first four moments of a distribution about the value 2 are 1, 2.5, 5.5 and 16. What is the 4th moment about zero?
Given below are moments about an arbitrary origin 5.
If μ'1 = -4, μ'2 = 22, μ'3 = -117 and μ'4 = 560, then μ3 is equal to:The third order central moment of normal distribution is:
If μ'r and μr, respectively, denote rth order moments about origin and mean, then:
The standard deviation of a symmetrical distribution is 3. What must be the value of the fourth moment about the mean for the distribution to be mesokurtic?
