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One number for a whole distribution

Mathematics · Probability · ages 16-17

Name ______________________ Date ____________

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
E(X) always has to be one of the values the variable can take.
Circle one: True False
The variance is worked out as E(X squared) minus the square of E(X).
Circle one: True False
A variable takes values 2 and 4 with probabilities 0.75 and 0.25. What is E(X)?
Answer: ______________
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
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
X takes values -2, 0 and 3 with probabilities 0.25, 0.5 and 0.25. Calculate E(X²).
Answer: ______________
A variable takes values 1, 2 and 3 with probabilities 0.5, 0.3 and 0.2. What is E(X squared)?
Answer: ______________
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
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

Answer key

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

One number for a whole distribution W1-mt_PFdo0Nm3FP-s1

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.
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.
True · Square each value, weight it by its probability and add for E(X squared), then take off the square of the mean.
2.5 · Weight and add: 2 × 0.75 + 4 × 0.25 = 1.5 + 1 = 2.5.
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.
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.
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.
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.
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.
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.