A fair six-sided die has expected value E[X] = 3.5. Which best explains why 3.5 is the expected value even though a die can never actually land on 3.5?
AIt is a rounding artifact of the calculation
BExpected value is the long-run average over many rolls, not a value the variable must be able to take
C3.5 is the median of the outcomes 1 through 6
DExpected value always equals the midpoint between the minimum and maximum
Expected value is a weighted average of all possible outcomes — it describes the center of the distribution in the long run. It need not be a value the random variable can actually take. The median of {1,2,3,4,5,6} is also 3.5 here, but that's a coincidence — for asymmetric distributions, mean and median differ.
Question 2 True / False
The expected value of the sum of two random variables always equals the sum of their individual expected values, even when the variables are not independent.
TTrue
FFalse
Answer: True
Linearity of expectation holds unconditionally: E[X + Y] = E[X] + E[Y] for any X and Y, regardless of dependence. Independence is only required for the stronger property E[XY] = E[X]E[Y]. This makes linearity an unusually powerful tool — you can decompose complicated sums without worrying about dependence.
Question 3 Short Answer
If X is a random variable with E[X] = 5, what is E[3X − 2]?
Think about your answer, then reveal below.
Model answer: 13
By linearity of expectation, E[aX + b] = aE[X] + b. Substituting a = 3, b = -2, and E[X] = 5: E[3X − 2] = 3(5) − 2 = 13. This works because expectation is a linear operator — it distributes over addition and pulls constants through.