NCEA Level 3Discrete random variablesAS 91586

Discrete random variables: mean and standard deviation

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

More from the same table

  • Level 2: the mean as a long-run average.
  • Level 3: the spread as well.
  • And what scaling does to both.
Worked example

Bikes per booking

  • X = number of bikes in one booking.
  • Three probabilities are known.
  • The one for 3 bikes is k.
Bikes, xP(X = x)
10.35
20.3
3k
40.15
Worked example

The probabilities add to 1

  • Probabilities add to 1.
  • Each is between 0 and 1.
  • If not, find the error.
  1. 0.35 + 0.3 + k + 0.15 = 1sum to 1
  2. 0.8 + k = 1
  3. k = 1 − 0.8 = 0.2
  4. 0 ≤ 0.2 ≤ 1
Worked example

The distribution, drawn

  • One bike is the most likely.
  • The bars step down to 0.15.
  • The mean is the balance point.
The distribution, drawn
Worked example

The mean: Σ x × P(X = x)

  • Value × probability.
  • Write every product.
  1. E(X) = Σ x × P(X = x)
  2. = 1 × 0.35 + 2 × 0.3
  3. + 3 × 0.2 + 4 × 0.15
  4. = 0.35 + 0.6 + 0.6 + 0.6
  5. = 2.15 bikes
Worked example

One more sum: E(X²)

  • Square each value first.
  • Then times its probability.
  1. E(X²) = Σ x² × P(X = x)
  2. = 1² × 0.35 + 2² × 0.3
  3. + 3² × 0.2 + 4² × 0.15
  4. = 0.35 + 1.2 + 1.8 + 2.4
  5. = 5.75
Worked example

Variance and standard deviation

  • Var(X) = E(X²) − μ².
  • Square root for σ.
  1. Var(X) = E(X²) − [E(X)]²
  2. = 5.75 − 2.15²
  3. = 5.75 − 4.6225
  4. = 1.1275
  5. σ = √1.1275 = 1.0618 bikes
Worked example

Rounding early: the 2025 warning

  • Round μ to 2.2: Var = 0.91.
  • Nearly 20% too small.
  • Keep 4 decimal places.
  1. μ ≈ 2.2: 2.2² = 4.84
  2. Var = 5.75 − 4.84 = 0.91rounded too early
  3. Var = 5.75 − 4.6225 = 1.1275
  4. 0.91/1.1275 = 0.8071about 19% too small
Key idea

The expectation rules in the booklet

  • E(aX + b) = aE(X) + b.
  • Var(aX + b) = a²Var(X).
  • Adding b does not change the spread.
RuleGives
E(aX + b)aE(X) + b
Var(aX + b)a²Var(X)
SD(aX + b)|a| × SD(X)
Worked example

The charge: T = 30X + 10

  • E(T) = 30E(X) + 10.
  • No + 10 in the spread.
  1. T = 30X + 10
  2. E(T) = 30 × E(X) + 10
  3. = 30 × 2.15 + 10
  4. = 64.50 + 10 = $74.50
  5. Var(T) = 30² × Var(X)
  6. = 900 × 1.1275 = 1014.75
  7. SD(T) = √1014.75 = 31.8551
  8. ≈ $31.86
  9. 30 × 1.1275 + 10a not squared, + 10
Key idea

What the mean means

  • Over many bookings: 2.15 bikes on average.
  • No single booking hires 2.15 bikes.
  • 200 × $74.50 = $14,900.
Key idea

Where the marks are

  • Achieved: missing probability and the mean.
  • Merit: SD, the rules, meaning in context.
  • Excellence: accurate through a long problem.
Summary

What to take away

  • Sum to 1, then the mean, then E(X²) − μ², with four decimal places throughout.
Missing psum to 1, then check
E(X)Σ x × P(X = x)
Var(X)E(X²) − μ²
SD√Var(X)
aX + bmean: aμ + b
variance: a²Var(X)
Workingfour decimal places
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • X = hockey goals: x = 0, 1, 2, 3.
  • Question 3: $50 a goal plus $20.
GivenFind
1P: 0.1, 0.4, k, 0.2k
2the same XE(X), SD
3Y = 50X + 20 dollarsE(Y), SD(Y)
4E(X) = 1.6 goalsmeaning
Show answer

1. 0.1 + 0.4 + k + 0.2 = 1, so k = 0.3. 2. E(X) = 1.6; E(X²) = 3.4; Var(X) = 3.4 − 2.56 = 0.84; SD = 0.9165. 3. E(Y) = 50 × 1.6 + 20 = $100; Var(Y) = 2500 × 0.84 = 2100; SD(Y) = $45.83. 4. Over many games, the team averages 1.6 goals a game; no single game has 1.6.

Worked example

The answers: questions 1 and 2

  1. 1. 0.1 + 0.4 + k + 0.2 = 1
  2. 0.7 + k = 1, so k = 0.3
  3. 2. E(X) = 0 × 0.1 + 1 × 0.4
  4. + 2 × 0.3 + 3 × 0.2
  5. = 0 + 0.4 + 0.6 + 0.6 = 1.6
  6. E(X²) = 0 × 0.1 + 1 × 0.4
  7. + 4 × 0.3 + 9 × 0.2
  8. = 0 + 0.4 + 1.2 + 1.8 = 3.4
Worked example

The answers: questions 2 to 4

  1. E(X) = 1.6, E(X²) = 3.4
  2. Var(X) = 3.4 − 1.6² = 3.4 − 2.56
  3. = 0.84
  4. SD = √0.84 = 0.9165
  5. 3. E(Y) = 50 × 1.6 + 20 = $100
  6. Var(Y) = 50² × 0.84 = 2500 × 0.84
  7. = 2100
  8. SD(Y) = √2100 = $45.83
  9. 4. 1.6 goals a game, on average

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