Discrete random variables: 7 exam questions solved step by step

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What this lesson covers

  • The mean of a discrete random variable is each value times its probability, added up.
  • The probabilities in a table must sum to 1, which finds any missing value.
Key idea

7 questions, one method

  • 1. Probabilities sum to 1.
  • 2. E(X) = Σ x × P(X = x).
  • 3. Var(X) = E(X²) − [E(X)]².
  • 4. Say what it means in context.
Your turn

Question 1 · Achieved

  • X = tries a club winger scores
  • in one game of rugby.
  • Find k, then E(X).
Tries, x0123
P(X = x)0.450.3k0.05
Worked example

Q1: k, then the mean

  • Sum to 1 finds k.
  • Then value × probability.
  1. 0.45 + 0.3 + k + 0.05 = 1sum to 1
  2. 0.8 + k = 1
  3. k = 0.2
  4. E(X) = Σ x × P(X = x)
  5. = 0 × 0.45 + 1 × 0.3
  6. + 2 × 0.2 + 3 × 0.05
  7. = 0 + 0.3 + 0.4 + 0.15
  8. = 0.85 tries
  9. average over many gamesnot one game
Your turn

Question 2 · Achieved

  • The same winger: E(X) = 0.85.
  • Find the standard deviation of X.
  • Careful: E(X²) is not E(X) squared.
Tries, x0123
P(X = x)0.450.30.20.05
Worked example

Q2: the standard deviation

  • Square the value first.
  • Then minus the mean squared.
  1. E(X²) = Σ x² × P(X = x)
  2. = 0² × 0.45 + 1² × 0.3
  3. + 2² × 0.2 + 3² × 0.05
  4. = 0 + 0.3 + 0.8 + 0.45 = 1.55
  5. E(X²) ≠ [E(X)]²1.55, not 0.7225
  6. Var(X) = E(X²) − [E(X)]²
  7. = 1.55 − 0.85² = 1.55 − 0.7225
  8. = 0.8275
  9. σ = √0.8275 = 0.9097 tries
Your turn

Question 3 · Merit

  • X = attempts to pass a driving test.
  • P(X = x) = k(5 − x), x = 1, 2, 3, 4.
  • Find k, E(X) and the SD of X.
Worked example

Q3: find k

  • Put each x into the rule.
  • Then sum to 1.
  1. P(X = x) = k(5 − x)
  2. 4k + 3k + 2k + k = 1sum to 1
  3. 10k = 1, k = 0.1
  4. P(X = 1) = 0.1 × 4 = 0.4
  5. P(X = 2) = 0.1 × 3 = 0.3
  6. P(X = 3) = 0.1 × 2 = 0.2
  7. P(X = 4) = 0.1 × 1 = 0.1
  8. 0.4 + 0.3 + 0.2 + 0.1 = 1
Worked example

Q3: mean and SD

  • Write every product.
  • Clean numbers: check them.
  1. P: 0.4, 0.3, 0.2, 0.1
  2. E(X) = 1 × 0.4 + 2 × 0.3
  3. + 3 × 0.2 + 4 × 0.1
  4. = 0.4 + 0.6 + 0.6 + 0.4 = 2
  5. E(X²) = 1² × 0.4 + 2² × 0.3
  6. + 3² × 0.2 + 4² × 0.1
  7. = 0.4 + 1.2 + 1.8 + 1.6 = 5
  8. Var(X) = 5 − 2² = 1
  9. σ = √1 = 1 attempt
Your turn

Question 4 · Merit, in context

  • Kayaks hired an hour at Raglan, X:
  • E(X) = 2.4, SD(X) = 1.1.
  • Takings T = 25X − 15 dollars.
  • Find the mean and SD of T.
Worked example

Q4: T = 25X − 15

  • The booklet rules.
  • − 15 moves it, not spreads.
  1. T = 25X − 15
  2. E(T) = 25 × E(X) − 15E(aX + b)
  3. = 25 × 2.4 − 15 = 60 − 15
  4. = $45 an hour
  5. Var(T) = 25² × Var(X)− 15 drops out
  6. = 625 × 1.1² = 625 × 1.21
  7. = 756.25
  8. SD(T) = √756.25 = $27.50
  9. 25 × 1.1 − 15 = 12.50no − 15 on spread
Your turn

Question 5 · Merit, in context

  • A school fair spinner costs $2 a go.
  • Prizes and their probabilities below.
  • How much will the school expect
  • to make from 300 plays?
Prize$10$5$1$0
P0.050.10.25?
Worked example

Q5: the school fair

  • Expected prize per play.
  • Then cost minus prize.
  1. P($0) = 1 − 0.05 − 0.1 − 0.25
  2. = 0.6
  3. E(prize) = 10 × 0.05 + 5 × 0.1
  4. + 1 × 0.25 + 0 × 0.6
  5. = 0.5 + 0.5 + 0.25 = $1.25
  6. per play: 2 − 1.25 = $0.75cost minus prize
  7. 300 × 0.75 = $225
  8. about $225 from 300 playsan expected value
Your turn

Question 6 · Excellence

  • X = people in a car on a motorway
  • at peak time. E(X) = 2.5.
  • Find a and b, then the SD of X.
People, x1234
P(X = x)0.1ab0.2
Worked example

Q6: two unknowns, two facts

  • Sum to 1 is one equation.
  • The mean is the other.
  1. 0.1 + a + b + 0.2 = 1sum to 1
  2. a + b = 0.7
  3. 0.1 + 2a + 3b + 0.8 = 2.5E(X) = 2.5
  4. 2a + 3b = 1.6
  5. a = 0.7 − bfrom the first
  6. 2(0.7 − b) + 3b = 1.6
  7. 1.4 + b = 1.6, b = 0.2
  8. a = 0.7 − 0.2 = 0.5
Worked example

Q6: check, then the SD

  • Check the mean first.
  • Then E(X²) − 2.5².
  1. P: 0.1, 0.5, 0.2, 0.2
  2. E(X) = 0.1 + 1 + 0.6 + 0.8 = 2.5
  3. E(X²) = 1² × 0.1 + 2² × 0.5
  4. + 3² × 0.2 + 4² × 0.2
  5. = 0.1 + 2 + 1.8 + 3.2 = 7.1
  6. Var(X) = 7.1 − 2.5²
  7. = 7.1 − 6.25 = 0.85
  8. σ = √0.85 = 0.92204 d.p.
  9. about 0.92 people per car
Your turn

Question 7 · Excellence

  • The school fair spinner again: $2 a go.
  • Keep the $5 and $1 prizes the same.
  • What top prize gives the school
  • an expected $0.60 a play?
Prize$T$5$1$0
P0.050.10.250.6
Worked example

Q7: set the top prize

  • Form the equation from words.
  • Then check it.
  1. top prize = $T
  2. E(prize) = 0.05T + 0.5 + 0.25
  3. = 0.05T + 0.75
  4. 2 − (0.05T + 0.75) = 0.60profit per play
  5. 1.25 − 0.05T = 0.60
  6. 0.05T = 0.65
  7. T = 0.65 ÷ 0.05 = $13
  8. 2 − (0.65 + 0.5 + 0.25) = 0.60
  9. a long-run averagenot every play
Summary

The method that solved all seven

  • Sum to 1, the mean, E(X²) minus the mean squared, then say it in context.
StepHow
Missing pprobabilities sum to 1
MeanΣ x × P(X = x)
SpreadE(X²) − [E(X)]², then √
aX + bmean a × E(X) + b, SD |a| × σ
Two unknownsum to 1, and the mean
Fair gameexpected payout against cost

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