Expectation and Variance by Integration · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Means and spreads by integration

Mathematics · Probability · ages 19-20
Name ______________________   Date ____________

  1. X has density f(x) = x/8 on [0, 4] and zero elsewhere. Find E[X] by integration. Round to two decimal places.

    Answer: ______________

  2. X is uniform on [0, 6], so its density is 1/6 there. Find E[X] by integration.

    Answer: ______________

  3. Which formula gives the variance of a continuous variable?

    • E[X²] minus the square of E[X]
    • E[X²] plus the square of E[X]
    • The square root of the mean of X²

  4. For variables with finite expectations, E[X + Y] = E[X] + E[Y] requires independence.

    Circle one:   True   False

  5. X is uniform on [-2, 2]. Find E[3X² + 2] by integrating against its density.

    Answer: ______________

  6. X and Y are dependent with finite expectations E[X] = 2 and E[Y] = 4. Find E[3X - 2Y + 5].

    • 13
    • 3
    • -2

  7. X is uniform on [0, 6], with E[X] = 3 and E[X²] = 12. Find Var(X).

    Answer: ______________

  8. X has density f(x) = 3x² on [0, 1] and zero elsewhere. Compute Var(X) from its moment integrals. Give an exact decimal.

    Answer: ______________

  9. Pat has E[X] = 2 and E[X²] = 12, but reports Var(X) = 12. Which correction completes the calculation?

    • Divide 12 by the mean to obtain 6
    • Subtract the mean from 12 to obtain 10
    • Subtract the squared mean to obtain 8

  10. X has density 4e^(-4x) for x ≥ 0 and zero otherwise. Use its first and second moments to find Var(X). Give a decimal.

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Means and spreads by integration W1-mt_kRSaKgp3bs-s1

  1. 2.67 · E[X] = ∫₀⁴ x²/8 dx = 8/3, which rounds to 2.67.
  2. 3 · Integrating x times 1/6 from 0 to 6 gives 36/12 = 3.
  3. E[X²] minus the square of E[X] · Spread is the mean square minus the square of the mean.
  4. False · Linearity holds for variables with finite expectations, whether or not they are independent.
  5. 6 · The density is 1/4. Integrating (3x² + 2)/4 from -2 to 2 gives 6.
  6. 3 · Linearity gives 3E[X] - 2E[Y] + 5 = 6 - 8 + 5 = 3. Dependence does not prevent this calculation.
  7. 3 · 12 minus the squared mean 9 leaves 3.
  8. 0.0375 · E[X] = 3/4 and E[X²] = 3/5, so Var(X) = 3/5 - 9/16 = 3/80 = 0.0375.
  9. Subtract the squared mean to obtain 8 · Variance is E[X²] - (E[X])² = 12 - 4 = 8.
  10. 0.0625 · The rate is 4, so E[X] = 1/4 and E[X²] = 2/16. Subtracting (1/4)² gives 1/16 = 0.0625.
Worksheet · LightMySky