Calculating Moments of Distribution: Finding the Moment About 10

[kidia]

I have a question here I will appreciate for any idea,The mean of distribution is 5.The second and third moments about the mean are 20 and 140 respectively.Find the moment of the distribution about 10.

 

[mathman]

All you need to do is use the binomial theorem and take averages.
For the second moment we have
(x-b)²=(x-a+a-b)²
=(x-a)²+2(x-a)(a-b)+(a-b)²

Now assume a is the mean and b is some other value, and take averages.
We the get:
Second moment around b=second moment around a +(a-b)²
(Note that x average =a).

For the third moment, carry out a similar expansion.

 

[tacman]

To find the moment of a distribution about a specific value, we can use the formula:

Moment about x = (x - mean)^n * frequency

In this case, we are looking for the moment about 10, so x = 10. The mean of the distribution is given as 5, so we have:

Moment about 10 = (10 - 5)^n * frequency

To find the frequency, we need to use the second and third moments about the mean. We can use the formula:

Second moment about mean = mean^2 + variance

Third moment about mean = mean^3 + 3 * mean * variance + skewness

In this case, we have the second moment about the mean as 20 and the third moment about the mean as 140. We can solve for the variance and skewness using these values:

Variance = 20 - 5^2 = 20 - 25 = -5
Skewness = 140 - 5^3 - 3 * 5 * (-5) = 140 - 125 + 75 = 90

Now, we can plug these values into our formula for frequency:

Frequency = (x - mean)^n * (mean^2 + variance) = (10 - 5)^n * (5^2 - 5) = 5^n * 20

Substituting this into our formula for the moment about 10, we have:

Moment about 10 = (10 - 5)^n * frequency = 5^n * (5^n * 20) = 100 * 5^n

Therefore, the moment of the distribution about 10 is 100 * 5^n.