Question

Find the mean and variance, if they exist, of each of the following distributions. (a) \(p(x)=\frac{3 !}{x !(3-x) !}\left(\frac{1}{2}\right)^{3}, x=0,1,2,3\), zero elsewhere. (b) \(f(x)=6 x(1-x), 0

Short answer

The solutions will be numerical values derived from the formulas provided in the steps. For the actual numerical solutions, these calculations should be executed.

Step-by-step solution

Calculate Mean and Variance for the first distribution

The first distribution is a discrete distribution function since it only takes particular values. The mean for \(p(x)\) can be calculated by summing \(x\) multiplied by its probability \(p(x)\) for all \(x\) values, i.e., \(E[X] = \sum x*p(x)\). Similarly, the variance can be determined using the formula \(Var[X] = E[X^2] - (E[X])^2\).

Calculate Mean and Variance for the second distribution

The second distribution \(f(x) = 6x(1-x)\) is a continuous distribution occurring between 0 and 1. In this case, the mean is calculated as \(E[X] = \int xf(x) dx\) from 0 to 1. The variance for this distribution is again calculated as \(Var[X] = E[X^2] - (E[X])^2\). The integration can be made easier by using integration by parts or substituting.

Calculate Mean and Variance for the third distribution

The third distribution \(f(x) = 2/x^3\) is another continuous distribution with domain (1,∞). The formulas for calculating mean and variance stay the same, which are \(E[X] = \int xf(x) dx\) for the range 1 to infinity and \(Var[X] = E[X^2] - (E[X])^2\). Complex integral calculations are expected in this part. To ease integrations, substitute variables or apply integral techniques if necessary.

Interpret results

Interpret the results for each distribution. The mean represents the average expected value while the variance quantifies the amount of variation or variability from the average.