Calculate expectation and variance from a probability table
Weight values and squared values separately, then distinguish a long-run mean from a possible individual outcome.
Section 5.4 · Discrete random variables
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.4. A probability table needs non-negative probabilities summing to one. E(X) is a probability-weighted mean; Var(X) = E(X²) − [E(X)]², with squared units. A binomial model B(n,p) counts successes in a fixed non-negative integer number n of independent two-outcome trials with a constant success probability p. A geometric model Geo(p) counts the trial of the first success, starting at 1, with independent trials and a constant 0 < p ≤ 1. Formula proofs are not required. The expectation of the geometric distribution is required; its variance is not required by section 5.4. Normal approximations belong to section 5.5 and are not used here. Exact model probabilities do not establish real-world independence or constant rates without evidence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Read a valid probability table.
- Square values and calculate weighted sums.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How is E(X²) calculated?
What you will learn
- Calculate E(X), E(X²), Var(X) and standard deviation.
- Interpret weighted means and the units of each measure.
Use probability weights twice, for different quantities
E(X) = ΣxP(X = x) and E(X²) = Σx²P(X = x). Then Var(X) = E(X²) − [E(X)]². For X values −1, 0, 2 with probabilities 1/4, 1/2, 1/4, E(X) = −1/4 + 2/4 = 1/4, while E(X²) = 1/4 + 4/4 = 5/4.
The variance is 5/4 − (1/4)² = 19/16 and standard deviation √19/4. Squaring the mean is different from averaging squared values. A negative value of X can contribute negatively to the mean, but its square is non-negative. A genuine variance cannot be negative; a negative result signals an invalid table, arithmetic error or premature rounding.
Expectation is not a forecast for one observation
Two fair coins give head-count mean 1 and variance 1/2. A distribution on values 0, 1, 3 with probabilities 0.20, 0.50, 0.30 has E(X) = 1.40 and E(X²) = 3.20, so variance 1.24 and standard deviation √31/5. The expected value 1.40 is not a possible single outcome in that table.
Expectation describes the probability-weighted average, consistent with averages over many repetitions of the stated experiment. It is not a guarantee about any one trial or a claim that a small sample must equal it. If X measures points, its mean and standard deviation are in points while variance is in points squared. Do not use a sample-estimator divisor n − 1 for these exact distribution moments.
Pause and explain
E(X) = 1/4 and E(X²) = 5/4. What is Var(X)?
Worked example
X is −1, 0, 2 with probabilities 1/4, 1/2, 1/4. Find E(X), E(X²) and Var(X).
Show the worked solution
- E(X) = (−1)(1/4) + 0 + 2(1/4) = 1/4.
- E(X²) = 1(1/4) + 0 + 4(1/4) = 5/4.
- Var(X) = 5/4 − 1/16 = 19/16.
Answer E(X) = 1/4; E(X²) = 5/4; Var(X) = 19/16.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
The two-fair-coin head count has values 0, 1, 2 and probabilities 1/4, 1/2, 1/4. Find mean and variance.
Give me a hint
Calculate the first and second weighted moments separately.
Compare my reasoning
- E(X) = 0 + 1/2 + 2/4 = 1.
- E(X²) = 0 + 1/2 + 4/4 = 3/2.
- Var(X) = 3/2 − 1² = 1/2.
Mean 1; variance 1/2.
Look for these in your work
- Use the probability weights.
- Subtract the square of the mean.
X takes −1, 0, 2 with probabilities 1/4, 1/2, 1/4. Find its mean, variance and standard deviation.
Give me a hint
Negative values retain their sign in the first moment but become positive when squared.
Compare my reasoning
- E(X) = 1/4 and E(X²) = 5/4.
- Var(X) = 5/4 − (1/4)² = 19/16.
- Standard deviation is √19/4.
Mean 1/4; variance 19/16; SD √19/4.
Look for these in your work
- Keep the negative contribution to the mean.
- Take a non-negative square root for standard deviation.
A point variable takes 0, 1, 3 with probabilities 0.20, 0.50, 0.30. Find mean and variance, then explain whether the mean can occur on one trial.
Give me a hint
Compute both weighted moments and compare the mean with the actual support.
Compare my reasoning
- E(X) = 0.50 + 0.90 = 1.40 points.
- E(X²) = 0.50 + 2.70 = 3.20, so variance 3.20 − 1.96 = 1.24 points squared.
- The support is 0, 1, 3; mean 1.40 is not a possible single observation.
Mean 1.40 points; variance 1.24 points squared; the mean is not an attainable outcome.
Look for these in your work
- State the correct units.
- Distinguish expectation from a predicted single result.
Common mistakes
- Do not weight columns equally unless their probabilities are equal.
- E(X²) is different from [E(X)]².
- An expected value need not be an attainable outcome.
What can you explain now?
Does E(X) = 1.4 mean the next observation will be 1.4?
Compare with the explanation
No; it is a probability-weighted average.
The listed outcomes can exclude the expected value entirely.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Build xP(X) and x²P(X) columns, sum them, subtract the squared mean and interpret the result in its proper units.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.