Respuesta :
None of the alternatives are correct: the expected value of a random variable is a weighted average of all the possible values of the variable, where the weight of each value is its probability.
If the variable X assumes the values [tex]x_1, x_2,\ldots x_n[/tex] with probability [tex]p_1, p_2,\ldots p_n[/tex], then the expected value is
[tex]\mathbb{E}(X)=\displaystyle\sum_{i=1}^n x_ip_i[/tex]
Here are counterexamples that show why the other options are wrong:
- If X is a fair die, each number from 1 to 6 appears with probability 1/6. The expected value is 3.5, which isn't even one of the possible outcome, let alone the one that occurs most frequently!
- If X assumes two opposite values a and -a with the same probability, its expected value is 0, but its variance is positive. So, the square root of the variance won't be zero. By the way, the square root of the variance is the standard deviation
- The example of the die is still valid: the expected value is not one of the possible outcomes, so you can't expect it to be the next value observed.
