Question

A) Show that the expected value (E) of a random variable X with a uniform distribution on the set {1, 2, …, k} is equal to (2k + 1).

b) Show that the variance (Var) of X is equal to (12k^2 - 1).

Answer 1

The student's question contains a typographical error; the correct expected value for a uniform distribution on the set {1, 2, ..., k} is (k+1)/2 and the variance is ((k^2-1)/12), not the values provided in the question.

Step-by-step explanation:

The question seems to have some confusion, as some parts of it contain incorrect information for a uniform distribution. The student is supposed to show that the expected value (E) of a random variable X with a uniform distribution on the set \{1, 2, …, k\} is (k+1)/2 and not (2k+1) as initially stated in the question.

To find the expected value for a uniform distribution where the random variable X can take on the values \{1, 2,…,k\}, we use the formula:
E(X) = (1+2+3+…+k)/k = (k*(k+1))/(2*k) = (k+1)/2.

Furthermore, the variance (Var) for the uniform distribution is calculated using the formula:
Var(X) = ((k^2-1)/12) and not (12k^2 - 1) as incorrectly mentioned in the question.

Note that for a uniform distribution, the probabilities are all equal, and thus the calculation of the mean and variance takes into account the symmetrical nature of the distribution.