NCEA Level 2Expected valueAS 91267

Expected value: 7 exam questions solved step by step

What this lesson covers

  • An expected number is the probability multiplied by the number of trials.
  • Expected value is a long-run average, so it need not be a possible single outcome.
Key idea

7 questions, two formulas

  • 1. Expected number = n × p.
  • 2. E(X) = Σ x × P(X = x).
  • 3. Subtract any cost to play.
  • 4. Round and explain in context.
Your turn

Question 1 · Achieved, in context

  • A packhouse grades 2500 kiwifruit.
  • P(undersized) = 0.08
  • How many are expected to be undersized?
Worked example

Q1: expected number

  • Trials × probability.
  • Answer with units.
  1. expected number = n × p
  2. = 2500 × 0.08
  3. = 200
  4. about 200 undersized kiwifruit
  5. not undersized: 2500 − 200 = 2300
Your turn

Question 2 · Achieved

  • A fair spinner pays these prizes.
  • Find the expected prize, E(X).
Prize, xP(X = x)
$00.5
$20.3
$50.15
$200.05
Worked example

Q2: E(X) for the spinner

  • Check the total is 1.
  • Each prize × its probability.
  1. 0.5 + 0.3 + 0.15 + 0.05 = 1
  2. E(X) = Σ x × P(X = x)
  3. = 0 × 0.5 + 2 × 0.3
  4. + 5 × 0.15 + 20 × 0.05
  5. = 0 + 0.6 + 0.75 + 1
  6. = 2.35
  7. expected prize: $2.35 a spin
Your turn

Question 3 · Merit, in context

  • The school charges $3 a spin.
  • Find the player's expected net gain.
  • Expected profit from 200 spins?
Prize, xP(X = x)
$00.5
$20.3
$50.15
$200.05
Worked example

Q3: subtract the cost

  • E(X) is not the profit.
  • Net = prize − cost.
  1. E(net) = E(X) − cost
  2. = 2.35 − 3
  3. = −0.65
  4. player: −$0.65 a spin on average
  5. school: +$0.65 a spin
  6. 200 × 0.65 = 130
  7. expected profit: $130
Your turn

Question 4 · Merit

  • X = people in a car on the bridge.
  • Find k, then E(X).
  • Explain what E(X) means.
People, xP(X = x)
10.1
20.35
3k
40.2
Worked example

Q4: find k, then E(X)

  • Find k first.
  • E(X) need not be whole.
  1. 0.1 + 0.35 + k + 0.2 = 1total is 1
  2. 0.65 + k = 1add
  3. k = 0.35subtract 0.65
  4. E(X) = 1 × 0.1 + 2 × 0.35
  5. + 3 × 0.35 + 4 × 0.2
  6. = 0.1 + 0.7 + 1.05 + 0.8
  7. = 2.65 people
  8. a long-run average per car
Your turn

Question 5 · Merit, in context

  • A rugby club has 80 players.
  • 60% are forwards, 40% backs.
  • Injured: 15% of forwards, 10% of backs.
  • How many injured players are expected?
Question 5 · Merit, in context
Worked example

Q5: tree, then n × p

  • Find P(injured) first.
  • Round to whole people.
  1. P(F and I) = 0.6 × 0.15 = 0.09
  2. P(B and I) = 0.4 × 0.1 = 0.04
  3. P(I) = 0.09 + 0.04 = 0.13add across
  4. expected = 80 × 0.13
  5. = 10.4
  6. about 10 injured playerswhole people
  7. 48 × 0.15 + 32 × 0.1 = 10.4
Your turn

Question 6 · Excellence

  • A fair stall game costs $4 to play.
  • Find x so the game is fair.
  • Then find x for $1 profit a game.
ResultProbabilityWin
Top prize0.1$x
Money back0.25$4
Nothingthe rest$0
Worked example

Q6: a fair game

  • Fair: E(winnings) = cost.
  • Form an equation in x.
  1. P(nothing) = 1 − 0.1 − 0.25 = 0.65
  2. E(W) = 0.1x + 0.25 × 4 + 0.65 × 0
  3. = 0.1x + 1
  4. 0.1x + 1 = 4fair: E(W) = cost
  5. 0.1x = 3subtract 1
  6. x = 30divide by 0.1
  7. fair top prize: $30
Worked example

Q6: $1 profit a game

  • Profit = cost − E(W).
  • Check the new prize.
  1. E(W) = 0.1x + 1
  2. E(W) = 4 − 1 = 3school keeps $1
  3. 0.1x + 1 = 3
  4. 0.1x = 2, x = 20
  5. check: 0.1 × 20 + 1 = 3
  6. profit: 4 − 3 = $1 a game
  7. top prize: $20
Your turn

Question 7 · Excellence, in context

  • P(damaging frost this season) = 0.15
  • No fans: a frost costs $40 000.
  • Fans: $5000, and a frost costs $8000.
  • Should the grower hire the fans?
Worked example

Q7: expected cost each way

  • Fans cost $5000 for sure.
  • The frost loss is a chance.
  1. no fans: 0.15 × 40 000 = $6000
  2. fans: 5000 + 0.15 × 8000
  3. = 5000 + 1200 = $6200
  4. 6200 − 6000 = $200fans cost more
  5. 40 000p = 5000 + 8000pbreak-even frost risk
  6. 32 000p = 5000− 8000p
  7. p = 5000 ÷ 32 000 = 0.15625
Worked example

Q7: decide, with reasons

  • E(cost) is a long-run average.
  • One season: look at the worst case.
  1. 0.15 is just under 0.156
  2. no fans: $200 cheaper on average
  3. no fans: lose $0 or $40 000
  4. fans: lose $5000 or $13 000
  5. fans cost $200 more, cap the loss
Summary

The method that solved all seven

  • Multiply, add, subtract the cost, then explain.
StepHow
Numbern × p, rounded in context
Checkprobabilities add to 1
ValueΣ x × P(X = x)
Netsubtract the cost to play
FairE(winnings) = cost
Meaninga long-run average

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