If X has E[X] = 5 and Var(X) = 4, what are E[2X + 3] and Var(2X + 3)?
AE = 13, Var = 11
BE = 13, Var = 16
CE = 13, Var = 8
DE = 10, Var = 16
E[2X + 3] = 2·E[X] + 3 = 2(5) + 3 = 13. Var(2X + 3) = 2²·Var(X) = 4·4 = 16. The constant +3 shifts the distribution without changing spread (variance unaffected), while scaling by 2 multiplies all distances from the mean by 2, so squared distances (variance) multiply by 4.
Question 2 True / False
The expected value E[X] of a discrete random variable is typically the value X is most likely to take.
TTrue
FFalse
Answer: False
E[X] is the long-run average (weighted mean), not the mode (most likely value). For example, if X = 0 with probability 0.9 and X = 100 with probability 0.1, then E[X] = 10 — but X is most likely to equal 0. The expected value can even be a value X never actually takes (like E[die roll] = 3.5).
Question 3 Short Answer
A fair six-sided die is rolled. What is E[X]? What does this value mean if the die is rolled many times?
Think about your answer, then reveal below.
Model answer: E[X] = (1+2+3+4+5+6)/6 = 3.5. If the die is rolled a large number of times, the average of all outcomes will converge to 3.5.
Expected value is the long-run average. Even though 3.5 is never rolled, averaging thousands of rolls will produce a value very close to 3.5 by the law of large numbers. The formula weights each outcome by its probability: E[X] = Σ x·p(x) = Σ x·(1/6) = 21/6 = 3.5.