Question

The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are

• A. (2, 16)
• B. (2, 15)
• C. (3, 15)
• D. (1, 16)

Hint:

When given moments about a point \( A \), remember the conversion formulas to find the mean and variance: \[\begin{array}{rl} \bullet & \text{Mean \( \bar{x} = A + \mu'_1 \) (the first moment about A)} \\ \bullet & \text{Variance \( \sigma^2 = \mu_2 = \mu'_2 - (\mu'_1)^2 \) (the second moment about A minus the square of the first moment about A)} \\ \end{array}\]

Answer 1

The Correct Option is C

We are given the first three raw moments about an arbitrary point \( A=2 \). Let's denote them by \( \mu'_1, \mu'_2, \mu'_3 \). So, \( \mu'_1 = 1 \), \( \mu'_2 = 16 \), and \( \mu'_3 = -40 \).

Step 1: Calculate the Mean. The mean of the distribution, \( \bar{x} \), is the first moment about the origin. We can find it using the first moment about point A. The formula is: Mean \( (\bar{x}) = A + \mu'_1 \). \[ \bar{x} = 2 + 1 = 3 \]

Step 2: Calculate the Variance. The variance, \( \sigma^2 \), is the second central moment (the second moment about the mean), denoted by \( \mu_2 \). The formula to convert the second raw moment to the second central moment is: \( \mu_2 = \mu'_2 - (\mu'_1)^2 \). \[ \sigma^2 = \mu_2 = 16 - (1)^2 = 16 - 1 = 15 \] So, the mean of the distribution is 3 and the variance is 15.