Expected Value and Variance of a Discrete Random Variable · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

One number for a whole distribution

Mathematics · Probability · ages 16-17
Name ______________________   Date ____________

  1. What does the variance of a distribution measure?

    • the value the variable is most likely to take on one go
    • how far the values spread out around the expected value
    • the largest value the variable is able to produce at all
    • the number of different values that appear in the table

  2. E(X) always has to be one of the values the variable can take.

    Circle one:   True   False

  3. The variance is worked out as E(X squared) minus the square of E(X).

    Circle one:   True   False

  4. A variable takes values 2 and 4 with probabilities 0.75 and 0.25. What is E(X)?

    Answer: ______________

  5. Why does a rare large prize pull the expected value up less than its size suggests?

    • because its contribution is the prize multiplied by a small probability
    • because large prizes are worth less than the sign claims they are
    • because the calculation caps how much any one single value is able to contribute
    • because rare prizes are left out of the expected value calculation

  6. A stall charges £2 a go and the expected payout is £1.40. What is the player's expected net result per go?

    • an average gain of 60p, found by subtracting the payout from the stake
    • zero average gain, because £1.40 is close enough to the £2 stake
    • an average loss of 60p, found by subtracting the stake from the payout
    • no expected result can be calculated without knowing each actual payout

  7. X takes values -2, 0 and 3 with probabilities 0.25, 0.5 and 0.25. Calculate E(X²).

    Answer: ______________

  8. A variable takes values 1, 2 and 3 with probabilities 0.5, 0.3 and 0.2. What is E(X squared)?

    Answer: ______________

  9. Sam uses [E(X)]² - E(X²) instead of E(X²) - [E(X)]² and gets a negative result. What explains his result?

    • a negative possible value of X makes every variance calculation negative
    • reversing the subtraction changes the sign of the nonzero variance
    • a negative variance is allowed whenever values extend below the mean
    • rounding has made the true variance negative, even though the formula is correct

  10. A stall's expected payout is £0.90 and its stake is £1. Dee says 90p is mathematically fair. The stall says its higher stake gives a margin before costs. Who is right?

    • only the stall, because a mathematically fair stake must cover running costs
    • only Dee, because every stall is forbidden to charge above the expected payout
    • both: 90p is the fair stake and £1 gives an expected margin of 10p
    • neither, because an expected payout and a fixed stake cannot be compared
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Answer key

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

One number for a whole distribution W1-mt_PFdo0Nm3FP-s1

  1. how far the values spread out around the expected value · E(X) says where the distribution sits and the variance says how widely it is scattered around that point. The two answer different questions about the same table.
  2. False · A spinner paying £0, £2 or £6 with probabilities 0.5, 0.3 and 0.2 has E(X) = 0 × 0.5 + 2 × 0.3 + 6 × 0.2 = £1.80. That is not a possible payout.
  3. True · Square each value, weight it by its probability and add for E(X squared), then take off the square of the mean.
  4. 2.5 · Weight and add: 2 × 0.75 + 4 × 0.25 = 1.5 + 1 = 2.5.
  5. because its contribution is the prize multiplied by a small probability · A prize of 100 at a probability of 0.01 contributes 1 to the total, the same as a prize of 2 at a probability of 0.5. Size and rarity trade off against each other exactly.
  6. an average loss of 60p, found by subtracting the stake from the payout · Expected net gain is expected payout minus stake: £1.40 - £2 = -£0.60. This is a mean loss of 60p per go, not the loss on every individual go.
  7. 3.25 · Square each value: (-2)² = 4, 0² = 0 and 3² = 9. Then E(X²) = 4 × 0.25 + 0 × 0.5 + 9 × 0.25 = 3.25.
  8. 3.5 · Square each value first, then weight: 1 × 0.5 + 4 × 0.3 + 9 × 0.2 = 0.5 + 1.2 + 1.8 = 3.5.
  9. reversing the subtraction changes the sign of the nonzero variance · Sam has reversed the subtraction. The correct variance is E(X²) - [E(X)]². His negative result is its opposite, so the correct variance is positive.
  10. both: 90p is the fair stake and £1 gives an expected margin of 10p · A mathematically fair stake equals the expected payout, so it is 90p. Charging £1 gives the stall an expected margin of £1 - £0.90 = £0.10 before costs.
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